These MP3 players contain circuits that are dc, at least in part. (The audio signal is ac.) The circuit diagram below shows a possible amplifier circuit for each stereo channel. We have already met two of the circuit elements shown: resistors and capacitors, and we discuss them in circuits in this Chapter. (The large triangle is an amplifier chip containing transistors, discussed in Chapter 40.) We also discuss voltmeters and ammeters, and how they are built and used to make measurements.

DC Circuits

T £

CHAPTER-OPENING QUESTION—Guess now! The automobile headlight bulbs shown in the circuits here are identical. The connection which produces more light is

(a) circuit 1.(b) circuit 2.(c) both the same.(d) not enough information.

r

Circuit 1

3 ,___ £+■

Circuit 2

Electric circuits are basic parts of all electronic devices from radio and TV sets to computers and automobiles. Scientific measurements, from physics to biology and medicine, make use of electric circuits. In Chapter 25, we discussed the basic principles of electric current. Now we will apply these

principles to analyze dc circuits involving combinations of batteries, resistors, and capacitors. We also study the operation of some useful instruments.*

CONTENTS26-1 EMF and Terminal Voltage26-2 Resistors in Series and in

Parallel26-3 Kirchhoff’s Rules26-4 Series and Parallel EMFs;

Battery Charging26-5 Circuits Containing Resistor

and Capacitor (RC Circuits)26-6 Electric Hazards

* 26-7 Ammeters and Voltmeters

f AC circuits that contain only a voltage source and resistors can be analyzed like the dc circuits in thisPliontAr o/' tViat ^nntoin panapitnrc anrl ntliAr r»irr»nit arp mnrp

TABLE 26-1 Symbols for Circuit Elements

Symbol Device

"I1"

-W \r

BatteryCapacitor

ResistorWire with negligible

resistanceSwitch

Ground

A CAUTI ONWhy battery voltage isn’t perfectly

constant

FIGURE 26-1 Diagram for an electric cell or battery.

FIGURE 26-2 Example 26-1.

R = 65.0 Q— vwv-----

When we draw a diagram for a circuit, we represent batteries, capacitors, and resistors by the symbols shown in Table 26-1. Wires whose resistance is negligible compared with other resistance in the circuit are drawn simply as straight lines. Some circuit diagrams show a ground symbol ( i or which may mean a real connection to the ground, perhaps via a metal pipe, or it may simply mean a common connection, such as the frame of a car.

For the most part in this Chapter, except in Section 26-5 on RC circuits, we will be interested in circuits operating in their steady state. That is, we won’t be looking at a circuit at the moment a change is made in it, such as when a battery or resistor is connected or disconnected, but rather later when the currents have reached their steady values.

26—1 EMF and Terminal VoltageTo have current in an electric circuit, we need a device such as a battery or an electric generator that transforms one type of energy (chemical, mechanical, or light, for example) into electric energy. Such a device is called a source of electromotive force or of emf. (The term “electromotive force” is a misnomer since it does not refer to a “force” that is measured in newtons. Hence, to avoid confusion, we prefer to use the abbreviation, emf.) The potential difference between the terminals of such a source, when no current flows to an external circuit, is called the emf of the source. The symbol % is usually used for emf (don’t confuse it with E for electric field), and its unit is volts.

A battery is not a source of constant current—the current out of a battery varies according to the resistance in the circuit. A battery is, however, a nearly constant voltage source, but not perfectly constant as we now discuss. You may have noticed in your own experience that when a current is drawn from a battery, the potential difference (voltage) across its terminals drops below its rated emf. For example, if you start a car with the headlights on, you may notice the headlights dim. This happens because the starter draws a large current, and the battery voltage drops as a result. The voltage drop occurs because the chemical reactions in a battery (Section 25-1) cannot supply charge fast enough to maintain the full emf. For one thing, charge must move (within the electrolyte) between the electrodes of the battery, and there is always some hindrance to completely free flow. Thus, a battery itself has some resistance, which is called its internal resistance; it is usually designated r.

A real battery is modeled as if it were a perfect emf % in series with a resistor r, as shown in Fig. 26-1. Since this resistance r is inside the battery, we can never separate it from the battery. The two points a and b in the diagram represent the two terminals of the battery. What we measure is the terminal voltage Kb = Va ~ Vb' When no current is drawn from the battery, the terminal voltage equals the emf, which is determined by the chemical reactions in the battery: Vah = %. However, when a current I flows naturally from the battery there is an internal drop in voltage equal to Ir. Thus the terminal voltage (the actual voltage) isf

Vah = % ~ Ir. (26-1)For example, if a 12-V battery has an internal resistance of 0.1 fl, then when 10 A flows from the battery, the terminal voltage is 12 V - (10 A) (0.1 fl) = 11 V. The internal resistance of a battery is usually small. For example, an ordinary flashlight battery when fresh may have an internal resistance of perhaps 0.05 fl. (However, as it ages and the electrolyte dries out, the internal resistance increases to many ohms.) Car batteries have lower internal resistance.

Battery with internal resistance. A 65.0-0 resistor is connected to the terminals of a battery whose emf is 12.0 V and whose internal resistance is 0.5 0 , Fig. 26-2. Calculate (a) the current in the circuit, (b) the terminal voltage of the battery, K b, and (c) the power dissipated in the resistor R and in the battery’s internal resistance r.APPROACH We first consider the battery as a whole, which is shown in Fig. 26-2 as an emf % and internal resistance r between points a and b. Then we applyV = IR to the circuit itself.

SOLUTION (a) From Eq. 26-1, we have Vab = % - Ir.

We apply Ohm’s law (Eqs. 25-2) to this battery and the resistance R of the circuit: Vab = IR. Hence IR = % — Ir or % = I(R + r), and so

= % = 12‘0 Y = 12,0 V R + r 65.0 ft + 0.5ft 65.5ft

(b) The terminal voltage isVah = % - Ir = 12.0 V - (0.183 A) (0.5 ft) = 11.9 V.

(c) The power dissipated (Eq. 25-7) in R isPR = I 2R = (0.183 A )2(65.0 ft) = 2.18 W,

and in r isPr = I 2r = (0.183 A )2(0.5 ft) = 0.02 W.

I EXERCISE A Repeat Example 26-1 assuming now that the resistance R = 10.0 ft, I whereas % and r remain as before.In much of what follows, unless stated otherwise, we assume that the battery’s internal resistance is negligible, and that the battery voltage given is its terminal voltage, which we will usually write simply as V rather than Kb- Be careful not to confuse V (italic) for voltage and V (not italic) for the volt unit.

2 6 —2 Resistors in Series and in ParallelWhen two or more resistors are connected end to end along a single path as shown in Fig. 26-3a, they are said to be connected in series. The resistors could be simple resistors as were pictured in Fig. 25-12, or they could be lightbulbs (Fig. 26-3b), or heating elements, or other resistive devices. Any charge that passes through R t in Fig. 26-3a will also pass through R2 and then R3. Hence the same current I passes through each resistor. (If it did not, this would imply that either charge was not conserved, or that charge was accumulating at some point in the circuit, which does not happen in the steady state.)

We let V represent the potential difference (voltage) across all three resistors in Fig. 26-3a. We assume all other resistance in the circuit can be ignored, so V equals the terminal voltage supplied by the battery. We let Vl f V2f and V3 be the potential differences across each of the resistors, R1,R 2, and R3, respectively. From Ohm’s law, V = IR, we can write Vx = IR X, V2 = IR2, and V3 = IR3. Because the resistors are connected end to end, energy conservation tells us that the total voltage V is equal to the sum of the voltages* across each resistor:

v = Vi + v2 + V3 = IR X + IR2 + IR3. [series] (26-2)Now let us determine the equivalent single resistance Req that would draw the

same current I as our combination of three resistors in series; see Fig. 26-3c. Such a single resistance Req would be related to V by

V = IReq.We equate this expression with Eq. 26-2, V = l(R 1 + R2 + R3), and find

Req = R t + R2 + R3. [series] (26-3)

This is, in fact, what we expect. When we put several resistances in series, the total or equivalent resistance is the sum of the separate resistances. (Sometimes we may also call it the “net resistance.”) This sum applies to any number of resistances in series. Note that when you add more resistance to the circuit, the current through the circuit will decrease. For example, if a 12-V battery is connected to a 4-ft resistor, the current will be 3 A. But if the 12-V battery is connected to three 4-ft resistors in series, the total resistance is 12 ft and the current through the entire circuit will be only 1 A.

tTo see in more detail why this is true, note that an electric charge q passing through Rj loses an

FIGURE 26-3 (a) Resistances connected in series, (b) Resistances could be lightbulbs, or any other type of resistance, (c) Equivalent single resistance Req that draws the same current: Req = Ri + R2 + R3.

i _____f t f t

Batlery(b)

w w -

(C)

FIGURE 2 6 -4 (a) Resistances connected in parallel, (b) The resistances could be lightbulbs.(c) The equivalent circuit with R eq obtained from Eq. 26-4:1R,eq

JL jl JLR i R 2 R 3

FIGURE 2 6 -5 Water pipes in parallel— analogy to electric currents in parallel.

Another simple way to connect resistors is in parallel so that the current from the source splits into separate branches or paths, as shown in Fig. 26-4a and b. The wiring in houses and buildings is arranged so all electric devices are in parallel, as we already saw in Chapter 25, Fig. 25-20. With parallel wiring, if you disconnect one device (say, in Fig. 26-4a), the current to the other devices is not interrupted. Compare to a series circuit, where if one device (say, Rt in Fig. 26-3a) is disconnected, the current is stopped to all the others.

In a parallel circuit, Fig. 26-4a, the total current I that leaves the battery splits into three separate paths. We let Ix, I2, and I3 be the currents through each of the resistors, RX,R 2, and R3, respectively. Because electric charge is conserved, the current I flowing into junction A (where the different wires or conductors meet, Fig. 26-4a) must equal the current flowing out of the junction. Thus

I = Ix + I2 + I3. [parallel]When resistors are connected in parallel, each has the same voltage across it. (Indeed, any two points in a circuit connected by a wire of negligible resistance are at the same potential.) Hence the full voltage of the battery is applied to each resistor in Fig. 26-4a. Applying Ohm’s law to each resistor, we have

and

Let us now determine what single resistor Req (Fig. 26-4c) will draw the same current I as these three resistances in parallel. This equivalent resistance Req must satisfy Ohm’s law too:

i - k We now combine the equations above:

I = h + I2 + I3,V_RP

V V V — — + — + — ■ R i R n R q■-eq

When we divide out the V from each term, we have 1 Req

1 1 1--- + --- + ---i?i R2 R3

[parallel] (26-4)

For example, suppose you connect two 4-H loudspeakers to a single set of output terminals of your stereo amplifier or receiver. (Ignore the other channel for a moment—our two speakers are both connected to the left channel, say.) The equivalent resistance of the two 4-H “resistors” in parallel is

J _ - J _ 1 - 2 _ 14 a + 4 a“■eq 4 0 20,

and so Req = 2 0 . Thus the net (or equivalent) resistance is less than each single resistance. This may at first seem surprising. But remember that when you connect resistors in parallel, you are giving the current additional paths to follow. Hence the net resistance will be less.

Equations 26-3 and 26-4 make good sense. Recalling Eq. 25-3 for resistivity, R = pl/A, we see that placing resistors in series increases the length and therefore the resistance; putting resistors in parallel increases the area through which current flows, thus reducing the overall resistance.

An analogy may help here. Consider two identical pipes taking in water near the top of a dam and releasing it below as shown in Fig. 26-5. The gravitational potential difference, proportional to the height h, is the same for both pipes, just as the voltage is the same for parallel resistors. If both pipes are open, rather than only one, twice as much water will flow through. That is, with two equal pipes open, the net resistance to the flow of water will be reduced, by half, just as for electrical resistors in parallel. Note that if both pipes are closed, the dam offers infinite resistance to the flow of water. This corresponds in the electrical case to an open circuit—when the path is

CONCEPTUAL EXAMPLE 26-2 I Series or parallel? (a)ThelightbulbsinFig.26-6 are identical. Which configuration produces more light? (ft) Which way do you think the headlights of a car are wired? Ignore change of filament resistance R with current. RESPONSE (a) The equivalent resistance of the parallel circuit is found from Eq. 26-4, 1 /R eq = 1/R + 1/R = 2/R. Thus Req = R/2. The parallel combination then has lower resistance (= R/2) than the series combination (Req = R + R = 2R). There will be more total current in the parallel configuration (2), since I = V /R eq and V is the same for both circuits. The total power transformed, which is related to the light produced, is P = IV, so the greater current in (2) means more light produced. (b) Headlights are wired in parallel (2), because if one bulb goes out, the other bulb can stay lit. If they were in series (1), when one bulb burned out (the filament broke), the circuit would be open and no current would flow, so neither bulb would light.NOTE When you answered the Chapter-Opening Question on page 677, was your answer circuit 2? Can you express any misconceptions you might have had?

CONCEPTUAL EXAMPLE 26-3 I An illuminating surprise. A 100-W. 120-V lightbulb and a 60-W, 120-V lightbulb are connected in two different ways as shown in Fig. 26-7. In each case, which bulb glows more brightly? Ignore change of filament resistance with current (and temperature).RESPONSE (a) These are normal lightbulbs with their power rating given for 120 V. They both receive 120 V, so the 100-W bulb is naturally brighter.(b) The resistance of the 100-W bulb is less than that of the 60-W bulb (calculated from P = V 2/R at constant 120 V). Here they are connected in series and receive the same current. Hence, from P = I 2R ( / = constant) the higher-resistance “60-W” bulb will transform more power and thus be brighter.NOTE When connected in series as in (b), the two bulbs do not dissipate 60 W and 100 W because neither bulb receives 120 V.

Note that whenever a group of resistors is replaced by the equivalent resistance, current and voltage and power in the rest of the circuit are unaffected.

Circuit with series and parallel resistors. How much current is drawn from the battery shown in Fig. 26-8a?APPROACH The current I that flows out of the battery all passes through the 400-0 resistor, but then it splits into Ix and I2 passing through the 500-0 and 700-0 resistors. The latter two resistors are in parallel with each other. We look for something that we already know how to treat. So let’s start by finding the equivalent resistance, Rp, of the parallel resistors, 500 O and 700 O. Then we can consider this Rp to be in series with the 400-0 resistor.SOLUTION The equivalent resistance, Rp, of the 500-0 and 700-0 resistors in parallel is given by

JL _ 1Rf ~ 5000

17000

= 0.0020 O 1 + 0.0014 O 1 = 0.00340

This is l/i?P, so we take the reciprocal to find RP. It is a common mistake to forget to take this reciprocal. Notice that the units of reciprocal ohms, O-1, are a reminder. Thus

RP = „ * „ , = 2900.0.0034 r r '

This 290 O is the equivalent resistance of the two parallel resistors, and is in series with the 400-0 resistor as shown in the equivalent circuit of Fig. 26-8b. To find the total equivalent resistance Req, we add the 400-0 and 290-0 resistances together, since they are in series, and find

Req = 4000 + 2900 = 6900.The total current flowing from the battery is then

V 12.0 V/ =Leq 6900

= 0.0174 A 17 mA.

StSt

(1) Series (2) ParallelFIGURE 26-6 Example 26-2.

FIGURE 26-7 Example 26-3.

60 W

(a)100 w

VWW120 V

(b)

60 W 100 W-* m >— vwh

120 V

FIGURE 26-8 (a) Circuit for Examples 26-4 and 26-5.(b) Equivalent circuit, showing the equivalent resistance of 290 O for the two parallel resistors in (a).

a 400 Q bH /W V ^ 1

500 0 7l— V W V ^

700 0 72

12.0 V (a)

a 400 O I3 290 O q— vwv— •— v w w -r

12.0 V (b)

A CAUTI ONKinrc Ti-ii Remember to take the reciprocal

jj 400 £2 I3f-VWV-*-11

500 Q 7l— V W V ^

700 Q— v w v

12.0 V(a)

a 400 Q— v w v

Rp =290 Q c

AAAA/—

12.0 Y(b)

FIGURE 26-8 (repeated)(a) Circuit for Examples 26-4 and 26-5. (b) Equivalent circuit, showing the equivalent resistance of 290 O for the two parallel resistors in (a).

FIGURE 26-9 Example 26-6, three identical lightbulbs. Each yellow circle with -WAr- inside represents a lightbulb and its resistance.

EXAMPLE 26-5 Current in one branch. What is the current through the 500-0 resistor in Fig. 26-8a?APPROACH We need to find the voltage across the 500-0 resistor, which is the voltage between points b and c in Fig. 26-8a, and we call it V c- Once Vbc is known, we can apply Ohm’s law, V = IR, to get the current. First we find the voltage across the 400-0 resistor, Vab, since we know that 17.4 mA passes through it (Example 26-4).SOLUTION Vah can be found using V = IR:

Vab = (0.0174 A) (4000) = 7.0 V.Since the total voltage across the network of resistors is Vac = 12.0 V, then Vbc must be 12.0 V - 7.0 V = 5.0 V. Then Ohm’s law applied to the 500-0 resistor tells us that the current Ix through that resistor is

5.0Vh ~ = 1-0 X 10 A = 10 mA.5000

This is the answer we wanted. We can also calculate the current I2 through the 700-0 resistor since the voltage across it is also 5.0 V:

_ 5.0V h ~ 7000

= 7 mA.

NOTE When Ix combines with I2 to form the total current I (at point c in Fig. 26-8a), their sum is 10 mA + 7 mA = 17 mA. This equals the total current I as calculated in Example 26-4, as it should.

CONCEPTUAL EXAMPLE 26-6 I Bulb brightness in a circuit The circuit shown in Fig. 26-9 has three identical lightbulbs, each of resistance R. (a) When switch S is closed, how will the brightness of bulbs A and B compare with that of bulb C? (b) What happens when switch S is opened? Use a minimum of mathematics in your answers.RESPONSE (a) With switch S closed, the current that passes through bulb C must split into two equal parts when it reaches the junction leading to bulbs A and B. It splits into equal parts because the resistance of bulb A equals that of B. Thus, bulbs A and B each receive half of C’s current; A and B will be equally bright, but they will be less bright than bulb C (P = I 2R). (b) When the switch S is open, no current can flow through bulb A, so it will be dark. We now have a simple one-loop series circuit, and we expect bulbs B and C to be equally bright. However, the equivalent resistance of this circuit (= R + R) is greater than that of the circuit with the switch closed. When we open the switch, we increase the resistance and reduce the current leaving the battery. Thus, bulb C will be dimmer when we open the switch. Bulb B gets more current when the switch is open (you may have to use some mathematics here), and so it will be brighter than with the switch closed; and B will be as bright as C.

ESTIMATE I A two-speed fan. One way a multiple-speedEXAMPLE 26-7ventilation fan for a car can be designed is to put resistors in series with the fan motor. The resistors reduce the current through the motor and make it run more slowly. Suppose the current in the motor is 5.0 A when it is connected directly across a 12-V battery, (a) What series resistor should be used to reduce the current to 2.0 A for low-speed operation? (b) What power rating should the resistor have? APPROACH An electric motor in series with a resistor can be treated as two resistors in series. The power comes from P = IV.SOLUTION (a) When the motor is connected to 12 V and drawing 5.0 A, its resistance is R = V /I = (12V)/(5.0A) = 2.4 0 . We will assume that this is the motor’s resistance for all speeds. (This is an approximation because the current through the motor depends on its speed.) Then, when a current of 2.0 A is flowing, the voltage across the motor is (2.0 A) (2.40) = 4.8 V. The remaining12.0 V - 4.8 V = 7.2 V must appear across the series resistor. When 2.0 A flows through the resistor, its resistance must be R = (7.2 W f 2.0 A) = 3.6 0 . (b) The

EXAMPLE 26-8 Analyzing a circuit. A 9.0-V battery whose internal resis­tance r is 0.50 O is connected in the circuit shown in Fig. 26-10a. (a) How much current is drawn from the battery? (b) What is the terminal voltage of the battery? (c) What is the current in the 6.0-0 resistor?APPROACH To find the current out of the battery, we first need to determine the equivalent resistance Req of the entire circuit, including r, which we do by identi­fying and isolating simple series or parallel combinations of resistors. Once we find I from Ohm’s law, I = %/Req, we get the terminal voltage using ^ab = « - Ir. For (c) we apply Ohm’s law to the 6.0-0 resistor.SOLUTION (a) We want to determine the equivalent resistance of the circuit. But where do we start? We note that the 4.0-0 and 8.0-0 resistors are in parallel, and so have an equivalent resistance Reql given by

1 1 1 3

10.0 0

leql .0 0 + 4 .0 0 .0 0

so i?eql = 2.7 O. This 2.7 O is in series with the 6.0-0 resistor, as shown in the equivalent circuit of Fig. 26-10b. The net resistance of the lower arm of the circuit is then

leq2 6.0 O + 2.7 O — 8.7 O,as shown in Fig. 26-10c. The equivalent resistance R eq3 of the 8.7-0 and 10.0-0 resistances in parallel is given by

1 1 1

Leq3 10.00 8.7 0= 0 .2 1 0 -l

so Req3 = (1/0.210 -1) = 4.8 O. This 4.8 O is in series with the 5.0-0 resistor and the 0.50-0 internal resistance of the battery (Fig. 26-10d), so the total equivalent resistance Req of the circuit is R eq = 4.8 O + 5.0 O + 0.50 O = 10.3 O. Hence the current drawn is

9.0 VI =

Req 10.3 O= 0.87 A.

(b) The terminal voltage of the battery isVab = % - l r = 9.0 V - (0.87 A) (0.50 0 ) = 8.6 V.

(c) Now we can work back and get the current in the 6.0-0 resistor. It must be the same as the current through the 8.7 O shown in Fig. 26-10c (why?). The voltage across that 8.7 O will be the emf of the battery minus the voltage drops across r and the 5.0-0 resistor: V8J = 9.0 V - (0.87 A )(0.500 + 5.0 0 ) . Applying Ohm’s law, we get the current (call it / ')

9.0 V - (0.87 A) (0.500 + 5 .00)r =

5.7 0= 0.48 A.

This is the current through the 6.0-0 resistor.

2 6 —3 Kirchhoffs RulesIn the last few Examples we have been able to find the currents in circuits by combining resistances in series and parallel, and using Ohm’s law. This technique can be used for many circuits. However, some circuits are too complicated for that analysis. For example, we cannot find the currents in each part of the circuit shown in Fig. 26-11 simply by combining resistances as we did before.

To deal with such complicated circuits, we use Kirchhoff’s rules, devised by G. R. Kirchhoff (1824-1887) in the mid-nineteenth century. There are two rules, and they are simply convenient applications of the laws of conservation of charge and energy.

10.0 0

10.0 0 ■AAAAr

- i Re q2=8.7 0 H------ VWV------

5.0 0

(c)

r = 0.50 0-AAAAr-

= 9.0 V

FIGURE 26-10 Circuit for Example 26-8, where r is the internal resistance of the battery.

FIGURE 26-11 Currents can be calculated using Kirchhoff s rules.

Ir30 0 h

- m N — -r =

40 0 h IO 45VH W W ^ ^ A A A A H H

©i = r = 80 V IO

20 0

Kirchhoff’s first rule or junction rule is based on the conservation of electric charge that we already used to derive the rule for parallel resistors. It states that

Junction rule (conservation o f charge)

Loop rule (conservation o f energy)

a 400 Q — VW\r

290 Q P-vwv—

e 12.0 V d(a)

(b)

FIGURE 2 6-12 Changes in potential around the circuit in (a) are plotted in (b).

P R O B L E M S O L V I N GBe consistent with signs when

applying the loop rule

at any junction point, the sum of all currents entering the junction must equal the sum of all currents leaving the junction.

That is, whatever charge goes in must come out. We already saw an instance of this in the n o te at the end of Example 26-5.

Kirchhoff’s second rule or loop rule is based on the conservation of energy. It states that

the sum of the changes in potential around any closed loop of a circuit must be zero.

To see why this rule should hold, consider a rough analogy with the potential energy of a roller coaster on its track. When the roller coaster starts from the station, it has a particular potential energy. As it climbs the first hill, its potential energy increases and reaches a maximum at the top. As it descends the other side, its potential energy decreases and reaches a local minimum at the bottom of the hill. As the roller coaster continues on its path, its potential energy goes through more changes. But when it arrives back at the starting point, it has exactly as much potential energy as it had when it started at this point. Another way of saying this is that there was as much uphill as there was downhill.

Similar reasoning can be applied to an electric circuit. We will analyze the circuit of Fig. 26-11 shortly but first we consider the simpler circuit in Fig. 26-12. We have chosen it to be the same as the equivalent circuit of Fig. 26-8b already discussed. The current in this circuit is I = (12.0 V)/(690fl) = 0.0174 A, as we calculated in Example 26-4. (We keep an extra digit in I to reduce rounding errors.) The positive side of the battery, point e in Fig. 26-12a, is at a high potential compared to point d at the negative side of the battery. That is, point e is like the top of a hill for a roller coaster. We follow the current around the circuit starting at any point. We choose to start at point d and follow a positive test charge completely around this circuit. As we go, we note all changes in potential. When the test charge returns to point d, the potential will be the same as when we started (total change in potential around the circuit is zero). We plot the changes in potential around the circuit in Fig. 26-12b; point d is arbitrarily taken as zero.

As our positive test charge goes from point d, which is the negative or low potential side of the battery, to point e, which is the positive terminal (high potential side) of the battery, the potential increases by 12.0 V. (This is like the roller coaster being pulled up the first hill.) That is,

Ved = + 12.0 V.When our test charge moves from point e to point a, there is no change in potential since there is no source of emf and we assume negligible resistance in the connecting wires. Next, as the charge passes through the 400-0 resistor to get to point b, there is a decrease in potential of V = IR = (0.0174 A )(400 fl) = 7.0 V. The positive test charge is flowing “downhill” since it is heading toward the negative terminal of the battery, as indicated in the graph of Fig. 26-12b. Because this is a decrease in potential, we use a negative sign:

Vba = Vb - V* = -7.0 V.As the charge proceeds from b to c there is another potential decrease (a “voltage drop”) of (0.0174 A) X (2900) = 5.0 V, and this too is a decrease in potential:

Vcb = -5.0 V.There is no change in potential as our test charge moves from c to d as we assume negligible resistance in the wires.

*O

Bt Kirchhoff s Rules

1. Label the current in each separate branch of the given circuit with a different subscript, such as / i , I2,h (see Fig- 26-11 or 26-13). Each current refers to a segment between two junctions. Choose the direction of each current, using an arrow. The direction can be chosen arbitrarily: if the current is actually in the opposite direction, it will come out with a minus sign in the solution.

2. Identify the unknowns. You will need as many inde­pendent equations as there are unknowns. You may write down more equations than this, but you will find that some of the equations will be redundant (that is, not be independent in the sense of providing new information). You may use V = IR for each resistor, which sometimes will reduce the number of unknowns.

3. Apply Kirchhoff’s junction rule at one or more junctions.

4. Apply Kirchhoff’s loop rule for one or more loops: follow each loop in one direction only. Pay careful attention to subscripts, and to signs:(a) For a resistor, apply Ohm’s law; the potential differ­

ence is negative (a decrease) if your chosen loop direction is the same as the chosen current direc­tion through that resistor; the potential difference is positive (an increase) if your chosen loop direction is opposite to the chosen current direction.

(b) For a battery, the potential difference is positive if your chosen loop direction is from the negative terminal toward the positive terminal; the potential difference is negative if the loop direction is from the positive terminal toward the negative terminal.

5. Solve the equations algebraically for the unknowns. Be careful when manipulating equations not to err with signs. At the end, check your answers by plugging them into the original equations, or even by using any addi­tional loop or junction rule equations not used previously

EXAMPLE 26-9 Using Kirchhoff's rules. Calculate the currents Ix, I2, and I3in the three branches of the circuit in Fig. 26-13 (which is the same as Fig. 26-11).APPROACH AND SOLUTION1. Label the currents and their directions. Figure 26-13 uses the labels Ix, I2, and

I3 for the current in the three separate branches. Since (positive) current tends to move away from the positive terminal of a battery, we choose I2 and I3 to have the directions shown in Fig. 26-13. The direction of Ix is not obvious in advance, so we arbitrarily chose the direction indicated. If the current actually flows in the opposite direction, our answer will have a negative sign.

2. Identify the unknowns. We have three unknowns and therefore we need three equations, which we get by applying Kirchhoff’s junction and loop rules.

3. Junction rule: We apply Kirchhoff’s junction rule to the currents at point a, where /3 enters and I2 and It leave:

h = h + h- (a) This same equation holds at point d, so we get no new information by writing an equation for point d.

4. Loop rule: We apply Kirchhoff’s loop rule to two different closed loops. First we apply it to the upper loop ahdcba. We start (and end) at point a. From a to h we have a potential decrease Vha = - ( / 1)(30 fl). From h to d there is no change, but from d to c the potential increases by 45 V: that is, Vcd = +45 V. From c to a the potential decreases through the two resistances by an amount Vac = - ( / 3)(40n + l f l ) = - (4 1 f l) /3. Thus we have yha + V* + Vac = 0, or

-30 /i + 45 - 41/3 = 0, (b)where we have omitted the units (volts and amps) so we can more easily do the algebra. For our second loop, we take the outer loop ahdefga. (We could have chosen the lower loop abcdefga instead.) Again we start at point a and have Vha = ~(/i)(3Ofl), and = 0. But when we take our positive test charge from d to e, it actually is going uphill, against the current—or at least against the assumed direction of the current, which is what counts in this calculation. Thus Ved = I2(20 fi) has a positive sign. Similarly, Vfe = I2( 1 fl). From f to g there is a decrease in potential of 80 V since we go from the high potential terminal of the battery to the low. Thus VPf = -80 V. Finally,

i P R O B L E M S O L V I N GChoose current directions arbitrarily

20 Q

FIGURE 26-13 Currents can be calculated using Kirchhoff’s rules. See Example 26-9.

(c)

12 V

12 V

FIGURE 2 6 -1 4 R e tte r ie s in se r ie s

5. Solve the equations. We have three equations—labeled (a), (b), and (c)—and three unknowns. From Eq. (c) we have

80 + 30/i21

= 3. 1.4/!

From Eq. (b) we have45 - 30/i

h ~ A \~We substitute Eqs. (d) and (e) into Eq. (a):

= 1 . 1 - 0.73/!.

id)

(*)

h = h ~ 12 = 1-1 - 0.73/! - 3.8 - 1 .4 /j.

We solve for / i , collecting terms:3.1/i = -2.7

/1 = -0.87 A.The negative sign indicates that the direction of Ix is actually opposite to that initially assumed and shown in Fig. 26-13. The answer automatically comes out in amperes because all values were in volts and ohms. From Eq. (d) we have

/2 = 3.8 + 1.4/i = 3.8 + 1.4(—0.87) = 2.6 A, and from Eq. (e)

/3 = 1.1 - 0.73/i = 1.1 - 0.73(—0.87) = 1.7A.This completes the solution.

NOTE The unknowns in different situations are not necessarily currents. It might be that the currents are given and we have to solve for unknown resistance or voltage. The variables are then different, but the technique is the same.

EXERCISE C Write the equation for the lower loop abcdefga of Example 26-9 and show, assuming the currents calculated in this Example, that the potentials add to zero for this lower loop.

2 6 —4 Series and Parallel EMFs; Battery Charging

When two or more sources of emf, such as batteries, are arranged in series as in Fig. 26-14a, the total voltage is the algebraic sum of their respective voltages. On the other hand, when a 20-V and a 12-V battery are connected oppositely, as shown in Fig. 26-14b, the net voltage VCSL is 8 V (ignoring voltage drop across internal resistances). That is, a positive test charge moved from a to b gains in potential by 20 V, but when it passes from b to c it drops by 12 V. So the net change is 20 V — 12 V = 8 V. You might think that connecting batteries in reverse like this would be wasteful. For most purposes that would be true. But such a reverse arrangement is precisely how a battery charger works. In Fig. 26-14b, the 20-V source is charging up the 12-V battery. Because of its greater voltage, the 20-V source is forcing charge back into the 12-V battery: electrons are being forced into its negative terminal and removed from its positive terminal.

An automobile alternator keeps the car battery charged in the same way. A voltmeter placed across the terminals of a (12-V) car battery with the engine running fairly fast can tell you whether or not the alternator is charging the battery. If it is, the voltmeter reads 13 or 14 V. If the battery is not being charged, the voltage will be 12 V, or less if the battery is discharging. Car batteries can be recharged, but other batteries may not be rechargeable, since the chemical reactions in many cannot be reversed. In such cases, the arrangement of Fig. 26-14b would simply waste energy.

Sources of emf can also be arranged in parallel, Fig. 26-14c. With equal emfs, a parallel arrangement can provide more energy when large currents are needed. Each of the cells in parallel has to produce only a fraction of the total current, so the enerev loss due to internal resistance is less than for a sinele cell: and the

EXAMPLE 26-10 Jump starting a car. A good car battery is being used to jump start a car with a weak battery. The good battery has an emf of 12.5 V and internal resistance 0.020 fl. Suppose the weak battery has an emf of 10.1 V and internal resistance 0.10 fl. Each copper jumper cable is 3.0 m long and 0.50 cm in diameter, and can be attached as shown in Fig. 26-15. Assume the starter motor can be represented as a resistor Rs = 0.15 fl. Determine the current through the starter motor (a) if only the weak battery is connected to it, and (b) if the good battery is also connected, as shown in Fig. 26-15.APPROACH We apply Kirchhoff’s rules, but in (b) we will first need to determine the resistance of the jumper cables using their dimensions and the resistivity (p = 1.68 X 10 8fl*m for copper) as discussed in Section 25-4.SOLUTION (a) The circuit with only the weak battery and no jumper cables is simple: an emf of 10.1 V connected to two resistances in series, 0.10 fl + 0.15 fl = 0.25 0 . Hence the current is I = V /R = (10.1 V)/(0.25 fl) = 40 A.(b) We need to find the resistance of the jumper cables that connect the good battery. From Eq. 25-3, each has resistance Rj = p i/A = (1.68 X 10_8fl-m)(3.0m)/(7r)(0.25 X 10-2 m)2 = 0.0026 fl. Kirchhoffs loop rule for the full outside loop gives

12.5 V - /i(2 i? j + r j - I3Rs = 012.5 V - / i (0.025 fl) - J3(0.15 fl) = 0 (a)

since (2Rj + r) = (0.0052 fl + 0.020 fl) = 0.025 fl.The loop rule for the lower loop, including the weak battery and the starter, gives

10.1 V - 73(0.15 fl) - J2 (0.100) = 0. (b)The junction rule at point B gives

h + h = h - (c)

We have three equations in three unknowns. From Eq. (c), = I3 - I2 and we substitute this into Eq. (a):

12.5 V - (J3 - /2)(0.025 fl) - /3(0.15 fl) = 0,12.5 V - /3(0.175 fl) + /2(0.025 fl) = 0.

Combining this last equation with (b) gives I3 = 71 A, quite a bit better than in (fl).The other currents are I2 = - 5 A and Ix = 76 A. Note that I2 = —5 A is in the opposite direction from that assumed in Fig. 26-15. The terminal voltage of the weak 10.1-V battery is now VBA = 10.1 V - ( -5 A)(0.10fl) = 10.6 V. NOTE The circuit shown in Fig. 26-15, without the starter motor, is how a battery can be charged. The stronger battery pushes charge back into the weaker battery.

EXERCISE D If the jumper cables of Example 26-10 were mistakenly connected in reverse, the positive terminal of each battery would be connected to the negative terminal of the other battery (Fig. 26-16). What would be the current I even before the starter motor is engaged (the switch S in Fig. 26-16 is open)? Why could this cause the batteries to explode?

26—5 Circuits Containing Resistor and Capacitor Circuits)

Our study of circuits in this Chapter has, until now, dealt with steady currents that do not change in time. Now we examine circuits that contain both resistance and capacitance. Such a circuit is called an R C circuit. RC circuits are common in everyday life: they are used to control the speed of a car’s windshield wiper, and the timing of the change of traffic lights. They are used in camera flashes, in heart pacemakers, and in many other electronic devices. In RC circuits, we are not so interested in the final “steady state” voltage and charge on the capacitor, but

FIGURE 26-15a jump start.

Example 26-10,

FIGURE 26-16 Exercise D.

r Good battery

Cj—|-VWV------1|—I r\ = g = [0_020Q__ 12.5 Vj

Jumper cables

Starterswitch

(closed)

Weak battery

Af— fVWV-------11—

jo.w n i f f y

s R S = 0A5Q

AAAArStartermotor

R■AAAAr

(a)

(b) Time

(c) Time

FIGURE 2 6 -1 7 After the switch S closes in the R C circuit shown in (a), the voltage across the capacitor increases with time as shown in (b), and the current through the resistor decreases with time as shown in (c).

Let us now examine the RC circuit shown in Fig. 26-17a. When the switch S is closed, current immediately begins to flow through the circuit. Electrons will flow out from the negative terminal of the battery, through the resistor R, and accumulate on the upper plate of the capacitor. And electrons will flow into the positive terminal of the battery, leaving a positive charge on the other plate of the capacitor. As charge accumulates on the capacitor, the potential difference across it increases (Vc = QIC ), and the current is reduced until eventually the voltage across the capacitor equals the emf of the battery, %. There is then no potential difference across the resistor, and no further current flows. Potential difference Vc across the capacitor thus increases in time as shown in Fig. 26-17b. The mathematical form of this curve—that is, Vc as a function of time—can be derived using conservation of energy (or Kirchhoff’s loop rule). The emf % of the battery will equal the sum of the voltage drops across the resistor (IR) and the capacitor (Q/C):

% = IR (26-5)

The resistance R includes all resistance in the circuit, including the internal resistance of the battery; I is the current in the circuit at any instant, and Q is the charge on the capacitor at that same instant. Although % R, and C are constants, both Q and I are functions of time. The rate at which charge flows through the resistor (I = dQ /dt) is equal to the rate at which charge accumulates on the capacitor. Thus we can write

dQ 1= R

dt + c Q-This equation can be solved by rearranging it:

dQ dt“ ~RC'C% — Q

We now integrate from capacitor:

t = 0, when <2 = 0, to time t when a charge Q is on the

[Q dQ 1RC t

RC

•ln(C8 - Q) - ( - I n C%) =

Jo C% - Q

- I n (C% - Q)

[d tJo

In(C* - Q) ~ In{C%) = ~ j g

In1 - ^ 1 = -

tRC

We take the exponential of both sidesQ _ c%

= ~ -t/R C

or(26-6a)Q = C%( 1 - e~t,RC).

The potential difference across the capacitor is Vc = Q IC , soVc = « (1 - e~t/RC). (26-6b)

From Eqs. 26-6 we see that the charge Q on the capacitor, and the voltage Vc across it, increase from zero at t = 0 to maximum values Qmax = C% and Vc = % after a very long time. The quantity RC that appears in the exponent is called the time constant r of the circuit:

t = RC. (26-7)It represents the time^ required for the capacitor to reach (l - e~x) = 0.63 or 63% of its full charge and voltage. Thus the product RC is a measure of how quickly the

capacitor gets charged. In a circuit, for example, where R = 200 kO and C = 3.0 jiF, the time constant is (2.0 X 105 fl)(3.0 X 10-6F) = 0.60 s. If the resistance is much lower, the time constant is much smaller. This makes sense, since a lower resistance will retard the flow of charge less. All circuits contain some resistance (if only in the connecting wires), so a capacitor never can be charged instantaneously when connected to a battery.

From Eqs. 26-6, it appears that Q and Vc never quite reach their maximum values within a finite time. However, they reach 86% of maximum in 2RC, 95% in 3RC, 98% in 4RC, and so on. Q and Vc approach their maximum values asymptoti­cally. For example, if R = 20 kO and C = 0.30 /xF, the time constant is (2.0 X 104 n)(3.0 X 10-7 F) = 6.0 X 10-3 s. So the capacitor is more than 98% charged in less than ^ of a second.

The current I through the circuit of Fig. 26-17a at any time t can be obtained by differentiating Eq. 26-6a:

= dG = l e-,/RC (26_8)dt R

Thus, at t = 0, the current is I = %/R, as expected for a circuit containing only a resistor (there is not yet a potential difference across the capacitor). The current then drops exponentially in time with a time constant equal to RC, as the voltage across the capacitor increases. This is shown in Fig. 26-17c. The time constant RC represents the time required for the current to drop to 1/e « 0.37 of its initial value.

EXAMPLE 26-11 RC circuit, with emf. The capacitance in the circuit of Fig. 26-17a is C = 0.30 /jlF, the total resistance is 20 kft, and the battery emf is 12 V. Determine (a) the time constant, (b) the maximum charge the capacitor could acquire, (c) the time it takes for the charge to reach 99% of this value, (d) the current I when the charge Q is half its maximum value,(e) the maximum current, and (f) the charge Q when the current I is 0.20 its maximum value.APPROACH We use Fig. 26-17 and Eqs. 26-5, 6,7, and 8.SOLUTION (a) The time constant is RC = (2.0 x 104fl)(3.0 x 1 (T7F) = 6.0 x 10“3s.(b) The maximum charge would be Q = C% = (3.0 X 10-7 F)(12V) = 3.6 ji.C.(c) In Eq. 26-6a, we set Q = 0.99C%\

0.99C% = C%( 1 - e~'/RC),or

Then

so

, - t /R C = l _ a99 = 0 .0 L

= —ln(O.Ol) = 4.6

t = 4.6RC = 28 X 10“ 3 s or 28 ms (less than ^s).(d) From part (b) the maximum charge is 3.6 /jlC. When the charge is half this value, 1.8 /jlC, the current I in the circuit can be found using the original differential equation, or Eq. 26-5:

1 ( Q \ 1 / 1.8 X 10-6C \I = N - -77 = ----------- T ~ 12 V ----------------- t - = 300 fiA .R \ CJ 2.0 X 104n V 0.30 X 10“6F / ^

(e) The current is a maximum when there is no charge on the capacitor (Q = 0):12 V

R 2 .0 X 1 0 4H if) Again using Eq. 26-5, now with I = 0.20/max = 120 /jlA , we have

Time(b)

FIGURE 26-18 For the RC circuit shown in (a), the voltage Vc across the capacitor decreases with time, as shown in (b), after the switch S is closed at t = 0. The charge on the capacitor follows the same curve since Vc oc Q.

The circuit just discussed involved the charging of a capacitor by a battery through a resistance. Now let us look at another situation: when a capacitor is already charged (say to a voltage V0), and it is then allowed to discharge through a resistance R as shown in Fig. 26-18a. (In this case there is no battery.) When the switch S is closed, charge begins to flow through resistor R from one side of the capacitor toward the other side, until the capacitor is fully discharged. The voltage across the resistor at any instant equals that across the capacitor:

The rate at which charge leaves the capacitor equals the negative of the current in the resistor, I = - dQ /dt, because the capacitor is discharging (Q is decreasing). So we write the above equation as

- W - R = Q .dt C

We rearrange this to

dQ = dt Q RC

and integrate it from t = 0 when the charge on the capacitor is Q0, to some time t later when the charge is Q:

i Q- - _ _ Ln Go RC

orQ = Q0e-,,RC.

The voltage across the capacitor (Vc

Vc = V0e ^ RC,

QIC) as a function of time is

(26-9a)

(26-9b)

where the initial voltage V0 = QJC. Thus the charge on the capacitor, and the voltage across it, decrease exponentially in time with a time constant RC. This is shown in Fig. 26-18b. The current is

/ = = dt

Q ° _ e - t /R C _ ,

RC V-t/R C (26-10)

and it too is seen to decrease exponentially in time with the same time constant RC. The charge on the capacitor, the voltage across it, and the current in the resistor all decrease to 37% of their original value in one time constant t = r = RC.

EXERCISE E In 10 time constants, the charge on the capacitor in Fig. 26-18 will be about(a) Qo/20,000, (b) Q0/ 5000, (c) Q0/1000, (d) Q J 10, (e) Q J 3?

FIGURE 26-19 Example 26-12.EXAMPLE 26-12 Discharging RC circuit. In the RC circuit shown in Fig. 26-19, the battery has fully charged the capacitor, so Q0 = C%. Then at t = 0 the switch is thrown from position a to b. The battery emf is 20.0 V, and the capacitance C = 1.02 / jlF . The current I is observed to decrease to 0.50 of its initial value in 40 /jls. (a) What is the value of Q, the charge on the capacitor, at t = 0? (b) What is the value of R ? (c) What is Q at t = 60 ts?

APPROACH At t = 0, the capacitor has charge Q0 = C%, and then the battery is removed from the circuit and the capacitor begins discharging through the resistor, as in Fig. 26-18. At any time t later (Eq. 26-9a) we have

Q = Qoe~t/RC = C%e~t/RC.

SOLUTION (a) At t = 0,

Q = Q0 = C% = (1.02 X 1(T6F)(20.0V) = 2.04 X 10“5 C = 20.4 ^C.

(b) To find R, we are given that at t = 40 /j l s , I = 0.5010. Hence

0.50 IQ = I0e~t/RC.

Taking natural logs on both sides (In 0.50 = -0.693):

t0.693 =

soRC

t (40 X 10“ 6 s)(0.693)C (0.693)(1.02 X KT6F)

= 57 a

(c) At t = 60 /j l s ,

60xl0~6 sQ = Q0e~,/RC = (20.4 X 10-6C)e (57“Xi-02xio-‘f) = 7 3

CONCEPTUAL EXAMPLE 26-13 I Bulb in ffC circuit In the circuit of Fig. 26-20, the capacitor is originally uncharged. Describe the behavior of the lightbulb from the instant switch S is closed until a long time later.

RESPONSE When the switch is first closed, the current in the circuit is high and the lightbulb burns brightly. As the capacitor charges, the voltage across the capacitor increases causing the current to be reduced, and the lightbulb dims. As the potential difference across the capacitor approaches the same voltage as the battery, the current decreases toward zero and the lightbulb goes out.

1FIGURE 2 6 -2 0 Example 26-13.

* Applications of RC CircuitsThe charging and discharging in an RC circuit can be used to produce voltage pulses at a regular frequency. The charge on the capacitor increases to a particular voltage, and then discharges. One way of initiating the discharge of the capacitor is by the use of a gas-filled tube which has an electrical breakdown when the voltage across it reaches a certain value V0. After the discharge is finished, the tube no longer conducts current and the recharging process repeats itself, starting at a lower voltage V'0. Figure 26-21 shows a possible circuit, and the “sawtooth” voltage it produces.

A simple blinking light can be an application of a sawtooth oscillator circuit. Here the emf is supplied by a battery; the neon bulb flashes on at a rate of perhaps1 cycle per second. The main component of a “flasher unit” is a moderately large capacitor.

The intermittent windshield wipers of a car can also use an RC circuit. The RC time constant, which can be changed using a multi-positioned switch for different values of R with fixed C, determines the rate at which the wipers come on.

EXAMPLE 26-14 ESTIMATE! Resistor in a turn signal. Estimate the order of magnitude of the resistor in a turn-signal circuit.

APPROACH A typical turn signal flashes perhaps twice per second, so the time constant is on the order of 0.5 s. A moderate capacitor might have C = 1 [jlF .

SOLUTION Setting r = RC = 0.5 s, we find

0.5 s

( j p P H Y S I C S A P P L I E DSawtooth, blinkers, windshield wipers

FIGURE 26-21 (a) An R C circuit, coupled with a gas-filled tube as a switch, can produce a repeating “sawtooth” voltage, as shown in (b).

R = — 500 kO.

@ P H Y S I C S A P P L I E DHeart pacemaker

FIGURE 26-22 Electronic battery- powered pacemaker can be seen on the rib cage in this X-ray.

^ P H Y S I C S A P P L I E DDangers of electricity

FIGURE 26-23 A person receives an electric shock when the circuit is completed.

An interesting medical use of an RC circuit is the electronic heart pacemaker, which can make a stopped heart start beating again by applying an electric stimulus through electrodes attached to the chest. The stimulus can be repeated at the normal heartbeat rate if necessary. The heart itself contains pacemaker cells, which send out tiny electric pulses at a rate of 60 to 80 per minute. These signals induce the start of each heartbeat. In some forms of heart disease, the natural pacemaker fails to function properly, and the heart loses its beat. Such patients use electronic pacemakers which produce a regular voltage pulse that starts and controls the frequency of the heartbeat. The electrodes are implanted in or near the heart (Fig. 26-22), and the circuit contains a capacitor and a resistor. The charge on the capacitor increases to a certain point and then discharges a pulse to the heart. Then it starts charging again. The pulsing rate depends on the values of R and C.

2 6 —6 Electric HazardsExcess electric current can heat wires in buildings and cause fires, as discussed in Section 25-6. Electric current can also damage the human body or even be fatal. Electric current through the human body can cause damage in two ways: (1) Electric current heats tissue and can cause burns; (2 ) electric current stimulates nerves and muscles, and we feel a “shock.” The severity of a shock depends on the magnitude of the current, how long it acts, and through what part of the body it passes. A current passing through vital organs such as the heart or brain is especially serious for it can interfere with their operation.

Most people can “feel” a current of about 1 mA. Currents of a few mA cause pain but rarely cause much damage in a healthy person. Currents above 10 mA cause severe contraction of the muscles, and a person may not be able to let go of the source of the current (say, a faulty appliance or wire). Death from paralysis of the respiratory system can occur. Artificial respiration, however, can sometimes revive a victim. If a current above about 80 to 100 mA passes across the torso, so that a portion passes through the heart for more than a second or two, the heart muscles will begin to contract irregularly and blood will not be properly pumped. This condition is called ventricular fibrillation. If it lasts for long, death results. Strangely enough, if the current is much larger, on the order of 1 A, death by heart failure may be less likely,* but such currents can cause serious burns, especially if concentrated through a small area of the body.

The seriousness of a shock depends on the applied voltage and on the effective resistance of the body. Living tissue has low resistance since the fluid of cells contains ions that can conduct quite well. However, the outer layer of skin, when dry, offers high resistance and is thus protective. The effective resistance between two points on opposite sides of the body when the skin is dry is in the range of 104 to 106 ft. But when the skin is wet, the resistance may be 103 O or less. A person who is barefoot or wearing thin-soled shoes will be in good contact with the ground, and touching a 120-V line with a wet hand can result in a current

120 VI = = 120mA.iooo a

As we saw, this could be lethal.A person who has received a shock has become part of a complete circuit.

Figure 26-23 shows two ways the circuit might be completed when a person

1 Larger currents apparently bring the entire heart to a standstill. Upon release of the current, the heart returns to its normal rhythm. This may not happen when fibrillation occurs because, once started, it can be hard to stop. Fibrillation may also occur as a result of a heart attack or during heart surgery. AHaviVa tnmim oo a A n t s w ^HAcmKAH in QAMmn OA_A\ nan onnlu a Kripf ViirtVi MirrAnt t n tliA liAart

ru

(a) (b)

FIGURE 26-24 (a) An electric oven operating normally with a 2-prong plug.(b) Short to the case with ungrounded case: shock, (c) Short to the case with the case grounded by a 3-prong plug.

accidentally touches a “hot” electric wire—“hot” meaning a high potential such as 120 V (normal U.S. household voltage) relative to ground. The other wire of building wiring is connected to ground—either by a wire connected to a buried conductor, or via a metal water pipe into the ground. In Fig. 26-23a, the current passes from the high-voltage wire, through the person, to the ground through his bare feet, and back along the ground (a fair conductor) to the ground terminal of the source. If the person stands on a good insulator—thick rubber-soled shoes or a dry wood floor—there will be much more resistance in the circuit and conse­quently much less current through the person. If the person stands with bare feet on the ground, or is in a bathtub, there is lethal danger because the resistance is much less and the current greater. In a bathtub (or swimming pool), not only are you wet, which reduces your resistance, but the water is in contact with the drain pipe (typically metal) that leads to the ground. It is strongly recommended that you not touch anything electrical when wet or in bare feet. Building codes that require the use of non-metal pipes would be protective.

In Fig. 26-23b, a person touches a faulty “hot” wire with one hand, and the other hand touches a sink faucet (connected to ground via the pipe). The current is particularly dangerous because it passes across the chest, through the heart and lungs. A useful rule: if one hand is touching something electrical, keep your other hand in your pocket (don’t use it!), and wear thick rubber-soled shoes. It is also a good idea to remove metal jewelry, especially rings (your finger is usually moist under a ring).

You can come into contact with a hot wire by touching a bare wire whose insulation has worn off, or from a bare wire inside an appliance when you’re tinkering with it. (Always unplug an electrical device before investigating1" its insides!) Another possibility is that a wire inside a device may break or lose its insulation and come in contact with the case. If the case is metal, it will conduct electricity. A person could then suffer a severe shock merely by touching the case, as shown in Fig. 26-24b. To prevent an accident, metal cases are supposed to be connected directly to ground by a separate ground wire. Then if a “hot” wire touches the grounded case, a short circuit to ground immediately occurs internally, as shown in Fig. 26-24c, and most of the current passes through the low-resistance ground wire rather than through the person. Furthermore, the high current should open the fuse or circuit breaker. Grounding a metal case is done by a separate ground wire connected to the third (round) prong of a 3-prong plug. Never cut off the third prong of a plug—it could save your life.

ttM rnL _ J ^ _ J_120 VCurrent

(c)

/ j \ CAUTI ON________Keep one hand in your pocket when other touches electricity

0 P H Y S I C S A P P L I E DGrounding and shocks

FIGURE 2 6 -2 5 (a) A 3-prong plug, and (b) an adapter (gray) for old-fashioned 2-prong outlets— be sure to screw down the ground tab. (c) A polarized 2-prong plug.

FIGURE 2 6 -2 6 Four wires entering a typical house. The color codes for wires are not always as shown here—be careful!

____ Black (or red) Hot± 120 V;-------- i ----T ^ ± 120VFromelectric \ \ white company; U

ite |

+ 120 V-

120 v n Black {

Neutral,

— + 120 V

GreenGround at - L Ground at

A three-prong plug, and an adapter, are shown in Figs. 26-25a and b.Why is a third wire needed? The 120 V is carried by the other two wires—one

hot (120 V ac), the other neutral, which is itself grounded. The third “dedicated” ground wire with the round prong may seem redundant. But it is protection for two reasons: (1 ) it protects against internal wiring that may have been done incorrectly; (2) the neutral wire carries normal current (“return” current from the 120 V) and it does have resistance; so there can be a voltage drop along it—normally small, but if connections are poor or corroded, or the plug is loose, the resistance could be large enough that you might feel that voltage if you touched the neutral wire some distance from its grounding point.

Some electrical devices come with only two wires, and the plug’s two prongs are of different widths; the plug can be inserted only one way into the outlet so that the intended neutral (wider prong) in the device is connected to neutral in the wiring (Fig. 26-25c). For example, the screw threads of a lightbulb are meant to be connected to neutral (and the base contact to hot), to avoid shocks when changing a bulb in a possibly protruding socket. Devices with 2-prong plugs do not have their cases grounded; they are supposed to have double electric insulation. Take extra care anyway.

The insulation on a wire may be color coded. Hand-held meters may have red (hot) and black (ground) lead wires. But in a house, black is usually hot (or it may be red), whereas white is neutral and green is the dedicated ground, Fig. 26-26. But beware: these color codes cannot always be trusted. [In the U.S., three wires normally enter a house: two hot wires at 120 V each (which add together to 240 V for appliances or devices that run on 240 V) plus the grounded neutral (carrying return current for the two hots). See Fig. 26-26. The “dedicated” ground wire (non-current carrying) is a fourth wire that does not come from the electric company but enters the house from a nearby heavy stake in the ground or a buried metal pipe. The two hot wires can feed separate 120-V circuits in the house, so each 120-V circuit inside the house has only three wires, including ground.]

Normal circuit breakers (Sections 25-6 and 28-8) protect equipment and buildings from overload and fires. They protect humans only in some circumstances, such as the very high currents that result from a short, if they respond quickly enough. Ground fault circuit interrupters (GFCI), described in Section 29-8, are designed to protect people from the much lower currents (10 mA to 100 mA) that are lethal but would not throw a 15-A circuit breaker or blow a 20-A fuse.

It is current that harms, but it is voltage that drives the current. 30 volts is sometimes said to be the threshhold for danger. But even a 12-V car battery (which can supply large currents) can cause nasty burns and shock.

Another danger is leakage current, by which we mean a current along an unintended path. Leakage currents are often “capacitively coupled.” For example, a wire in a lamp forms a capacitor with the metal case; charges moving in one conductor attract or repel charge in the other, so there is a current. Typical electrical codes limit leakage currents to 1 mA for any device. A 1-mA leakage current is usually harmless. It can be very dangerous, however, to a hospital patient with implanted electrodes connected to ground through the apparatus. This is due to the absence of the protective skin layer and because the current can pass directly through the heart as compared to the usual situation where the current enters at the hands and spreads out through the body. Although 100 mA may be needed to cause heart fibrillation when entering through the hands (very little of it actually passes through the heart), as little as 0.02 mA has been known to cause fibrillation when passing directly to the heart. Thus, a “wired” patient is in considerable danger from leakage current even from as simple an act as touching a lamp.

Finally, don’t touch a downed power line (lethal!) or even get near it. A hot power line is at thousands of volts. A huge current can flow along the ground or pavement, from where the high-voltage wire touches the ground along its path to the grounding

4-1______

*26—7 Ammeters and VoltmetersAn ammeter is used to measure current, and a voltmeter measures potential differ- P H Y S I C S A P P L I E Dence or voltage. Measurements of current and voltage are made with meters that j ) c metersare of two types: (1 ) analog meters, which display numerical values by the position 1of a pointer that can move across a scale (Fig. 26-27a); and (2) digital meters, t wwhich display the numerical value in numbers (Fig. 26-27b). We now discuss themeters themselves and how they work, then how they are connected to circuits tomake measurements. Finally we will discuss how using meters affects the circuit ■being measured, possibly causing erroneous results—and what to do about it. g J

* Analog Ammeters and VoltmetersThe crucial part of an analog ammeter or voltmeter, in which the reading is by a pointer on a scale (Fig. 26-27a), is a galvanometer. The galvanometer works on theprinciple of the force between a magnetic field and a current-carrying coil of wire, ^and will be discussed in Chapter 27. For now, we merely need to know that the deflection of the needle of a galvanometer is proportional to the current flowing through it. The full-scale current sensitivity of a galvanometer, /m, is the electric current needed to make the needle deflect full scale.

A galvanometer can be used directly to measure small dc currents. For example, a galvanometer whose sensitivity Im is 50 /jlA can measure currents from about 1 jjlA (currents smaller than this would be hard to read on the scale) up to 50 /jlA. To measure larger currents, a resistor is placed in parallel with the galvanometer. Thus, an analog ammeter, represented by the symbol consists of a galvanometer (•-©-•) in parallel with a resistor called the shunt resistor, as shown in Fig. 26-28.(“Shunt” is a synonym for “in parallel”) The shunt resistance is Rsh, and the resistance of the galvanometer coil, through which current passes, is r. The value of i?sh is chosen according to the full-scale deflection desired; R&h is normally very small—giving an ammeter a very small net resistance—so most of the current passes through Rsh and very little (^ 50 /iA ) passes through the galvanometer to deflect the needle.

Ammeter/_ ^ - © _ = 7.

FIGURE 26-28 An ammeter is a galvanometer in parallel with a (shunt)

V\M— *----- • resistor with low resistance, Rsh.I r

EXAMPLE 26-15 Ammeter design. Design an ammeter to read 1.0 A at full scale using a galvanometer with a full-scale sensitivity of 50 /jlA and a resistance r = 30 fl. Check if the scale is linear.APPROACH Only 50/jlA (= Ig = 0.000050 A) of the 1.0-A current must pass through the galvanometer to give full-scale deflection. The rest of the current (.IR = 0.999950 A) passes through the small shunt resistor, R sh, Fig. 26-28. The potential difference across the galvanometer equals that across the shunt resistor (they are in parallel). We apply Ohm’s law to find Rsh.SOLUTION Because I = IG + IR, when I = 1.0 A flows into the meter, we want IR through the shunt resistor to be IR = 0.999950 A. The potential difference across the shunt is the same as across the galvanometer, so Ohm’s law tells us

Ir R%h = Ig r\then

= (5.0 x 10~5 A )(30tl) sh IR (0.999950 A)

or 0.0015 ft. The shunt resistor must thus have a very low resistance and most of the current passes through it.

1 \ (b)FIGURE 26-27 (a) An analog multimeter being used as a voltmeter, (b) An electronic digital meter.

An analog voltmeter («-©-) also consists of a galvanometer and a resistor. But the resistor R &er is connected in series, Fig. 26-29, and it is usually large, giving a voltmeter a high internal resistance.

FIGURE 2 6 -2 9 A voltmeter is a galvanometer in series with a resistor with high resistance, R ser.

^ ser y•A^A/V

•—measured

FIGURE 2 6 -3 0 An ohmmeter.

FIGURE 2 6-31 Measuring current and voltage.

—VW\/—•—VW\/—

(a)

a b cf-JWV\r-M/W\r-t^1 ^2

(b)

— ® — I

-JWW--R/W\r4

EXAMPLE 26-16 Voltmeter design. Using a galvanometer with internal resistance r = 3011 and full-scale current sensitivity of 50 /jlA, design a volt­meter that reads from 0 to 15 V. Is the scale linear?APPROACH When a potential difference of 15 V exists across the terminals of our voltmeter, we want 50 /jlA to be passing through it so as to give a full-scale deflection. SOLUTION From Ohm’s law, V = IR, we have (see Fig. 26-29)

15 V = (50 M X ' + *ser),SO

15 V5.0 X 10“ 5 A

= 300 kO - 30II = 300 k a

Notice that r = 3011 is so small compared to the value of 2?ser that it doesn’t influence the calculation significantly. The scale will again be linear: if the voltage to be measured is 6.0 V, the current passing through the voltmeter will be (6.0 V)/(3.0 X 105 fi) = 2.0 X 10“5 A, or 20 /jlA. This will produce two-fifths of full-scale deflection, as required (6.0V/15.0V = 2/5).

The meters just described are for direct current. A dc meter can be modified to measure ac (alternating current, Section 25-7) with the addition of diodes (Chapter 40), which allow current to flow in one direction only. An ac meter can be calibrated to read rms or peak values.

Voltmeters and ammeters can have several series or shunt resistors to offer a choice of range. Multimeters can measure voltage, current, and resistance. Sometimes a multimeter is called a VOM (Volt-Ohm-Meter or Volt-Ohm-Milliammeter).

An ohmmeter measures resistance, and must contain a battery of known voltage connected in series to a resistor (i?ser) and to an ammeter (Fig. 26-30). The resistor whose resistance is to be measured completes the circuit. The needle deflection is inversely proportional to the resistance. The scale calibration depends on the value of the series resistor. Because an ohmmeter sends a current through the device whose resistance is to be measured, it should not be used on very deli­cate devices that could be damaged by the current.

The sensitivity of a meter is generally specified on the face. It may be given as so many ohms per volt, which indicates how many ohms of resistance there are in the meter per volt of full-scale reading. For example, if the sensitivity is 30,000 H/V, this means that on the 10-V scale the meter has a resistance of 300,000 O, whereas on a 100-V scale the meter resistance is 3 MO. The full-scale current sensitivity, Im, discussed earlier, is just the reciprocal of the sensitivity in 1 1 / V.

* H owto Connect MetersSuppose you wish to determine the current I in the circuit shown in Fig. 26-31a, and the voltage V across the resistor R t . How exactly are ammeters and volt­meters connected to the circuit being measured?

Because an ammeter is used to measure the current flowing in the circuit, it must be inserted directly into the circuit, in series with the other elements, as shown in Fig. 26-31b. The smaller its internal resistance, the less it affects the circuit.

A voltmeter, on the other hand, is connected “externally,” in parallel with the circuit element across which the voltage is to be measured. It is used to measure the potential difference between two points. Its two wire leads (connecting wires) are connected to the two points, as shown in Fig. 26-31c, where the voltage across Ri is beine measured. The larger its internal resistance. (R*„ + r) in Fie. 26-29. the

* Effects o f Meter ResistanceIt is important to know the sensitivity of a meter, for in many cases the resistance of the meter can seriously affect your results. Take the following Example.

Voltage reading versus true voltage. Suppose you are testing an electronic circuit which has two resistors, R1 and R2, each 15 kfi, connected in series as shown in Fig. 26-32a. The battery maintains 8.0 V across them and has negligible internal resistance. A voltmeter whose sensitivity is 10,000 ft/V is put on the 5.0-V scale. What voltage does the meter read when connected across Rx, Fig. 26-32b, and what error is caused by the finite resistance of the meter? APPROACH The meter acts as a resistor in parallel with R1. We use parallel and series resistor analyses and Ohm’s law to find currents and voltages.SOLUTION On the 5.0-V scale, the voltmeter has an internal resistance of (5.0 V) (10,000 fi/V ) = 50,000 H. When connected across R1, as in Fig. 26-32b, we have this 50 kfi in parallel with R1 given by l ^

+

15 kfi. The net resistance Req of these two is

1350 m,veq 15 kO 150 kfi

so Req = 11.5 kfi. This Req = 11.5 kfi is in series with R2 = 15 kfi, so the total resistance of the circuit is now 26.5 kfi (instead of the original 30 kfi). Hence the current from the battery is

I = = 3.0 x 10-4 A = 0.30mA.26.5 kfiThen the voltage drop across R1, which is the same as that across the voltmeter, is (3.0 X 10“4A )(ll.5 X 103 Xl) = 3.5 V. [The voltage drop across R2 is (3.0 X 10“ 4 A)(l5 X 103 fi) = 4.5 V, for a total of 8.0 V.] If we assume the meter is precise, it will read 3.5 V. In the original circuit, without the meter, R1 = R2 so the voltage across R 1 is half that of the battery, or 4.0 V. Thus the voltmeter, because of its internal resistance, gives a low reading. In this case it is off by 0.5 V, or more than 10%.

Example 26-17 illustrates how seriously a meter can affect a circuit and give a misleading reading. If the resistance of a voltmeter is much higher than the resistance of the circuit, however, it will have little effect and its readings can be trusted, at least to the manufactured precision of the meter, which for ordinary analog meters is typically 3% to 4% of full-scale deflection. An ammeter also can interfere with a circuit, but the effect is minimal if its resistance is much less than that of the circuit as a whole. For both voltmeters and ammeters, the more sensitive the galvanometer, the less effect it will have. A 50,000-fl/V meter is far better than a 1000-fl/V meter.

* Digital MetersDigital meters (see Fig. 26-27b) are used in the same way as analog meters: they are inserted directly into the circuit, in series, to measure current (Fig. 26-31b), and connected “outside,” in parallel with the circuit, to measure voltage (Fig. 26-31c).

The internal construction of digital meters, however, is different from that of analog meters in that digital meters do not use a galvanometer. The electronic circuitry and digital readout are more sensitive than a galvanometer, and have less effect on the circuit to be measured. When we measure dc voltages, a digital meter’s resistance is very high, commonly on the order of 10 to 100 Mfl (107-108 fi), and doesn’t change significantly when different voltage scales are selected. A 100-Mfl digital meter draws off very little current when connected across even a 1-Mfi resistance.

The precision of digital meters is exceptional, often one part in 104 (=0.01%) or better. This precision is not the same as accuracy, however. A precise meter of internal resistance 10 8 fi will not give accurate results if used to measure a voltage across a 10 8-fi resistor—in which case it is necessary to do a calculation like that in Example 26-17.

Whenever we make a measurement on a circuit, to some degree we affect that circuit (Example 26-17). This is true for other types of measurement as well: when we make a measurement on a svstem. we affect that svstem in some wav. On a

0 P H Y S I C S A P P L I E DCorrecting for meter resistance

w— vwv—•— VWV—

(a)

FIGURE 26-32 Example 26-17.

SummaryA device that transforms another type of energy into electrical energy is called a source of emf. A battery behaves like a source of emf in series with an internal resistance. The emf is the potential difference determined by the chemical reactions in the battery and equals the terminal voltage when no current is drawn. When a current is drawn, the voltage at the battery’s terminals is less than its emf by an amount equal to the potential decrease Ir across the internal resistance.

When resistances are connected in series (end to end in a single linear path), the equivalent resistance is the sum of the individual resistances:

Re q = Ri + R2 + •••. (26-3)

In a series combination, Req is greater than any component resistance.

When resistors are connected in parallel, the reciprocal of the equivalent resistance equals the sum of the reciprocals of the individual resistances:

In a parallel connection, the net resistance is less than any of the individual resistances.

Kirchhoff’s rules are helpful in determining the currents and voltages in circuits. Kirchhoff’s junction rule is based on conservation of electric charge and states that the sum of all currents entering any junction equals the sum of all currents leaving that junction. The second, or loop rule, is based on conservation of energy and states that the algebraic sum of the

changes in potential around any closed path of the circuit must be zero.

When an R C circuit containing a resistor R in series with a capacitance C is connected to a dc source of emf, the voltage across the capacitor rises gradually in time characterized by an exponential of the form (l - e~t/RC), where the time constant,

t = RC, (26-7)

is the time it takes for the voltage to reach 63 percent of its maximum value. The current through the resistor decreases as e~t/RC.

A capacitor discharging through a resistor is characterized by the same time constant: in a time r = RC, the voltage across the capacitor drops to 37 percent of its initial value. The charge on the capacitor, and voltage across it, decreases as e~t/RC, as does the current.

Electric shocks are caused by current passing through the body. To avoid shocks, the body must not become part of a complete circuit by allowing different parts of the body to touch objects at different potentials. Commonly, shocks are caused by one part of the body touching ground and another part touching a high electric potential.

[*An ammeter measures current. An analog ammeter consists of a galvanometer and a parallel shunt resistor that carries most of the current. An analog voltmeter consists of a galvanometer and a series resistor. An ammeter is inserted into the circuit whose current is to be measured. A voltmeter is external, being connected in parallel to the element whose voltage is to be measured. Digital voltmeters have greater internal resistance and affect the circuit to be measured less than do analog meters.]

Questions1. Explain why birds can sit on power lines safely, whereas

leaning a metal ladder up against a power line to fetch a stuck kite is extremely dangerous.

2. Discuss the advantages and disadvantages of Christmas tree lights connected in parallel versus those connected in series.

3. If all you have is a 120-V line, would it be possible to light several 6-V lamps without burning them out? How?

4. Two lightbulbs of resistance Ri and R2 (R2 > R\) and a battery are all connected in series. Which bulb is brighter? What if they are connected in parallel? Explain.

5. Household outlets are often double outlets. Are these connected in series or parallel? How do you know?

6. With two identical lightbulbs and two identical batteries, how would you arrange the bulbs and batteries in a circuit to get the maximum possible total power to the lightbulbs? (Assume the batteries have negligible internal resistance.)

7. If two identical resistors are connected in series to a battery, does the battery have to supply more power or less power than when only one of the resistors is connected? Explain.

8. You have a single 60-W bulb on in your room. How does the overall resistance n f vonr rnnm ’s electric circuit change

9. When applying Kirchhoff’s loop rule (such as in Fig. 26-33), does the sign (or direction) of a battery’s emf depend on the direction of current through the battery? What about the terminal voltage?

r = 1.0 Q

FIGURE 26-33Question 9. □r

<9 = 12 V

10. Compare and discuss the formulas for resistors and for capacitors when connected in series and in parallel.

11. For what use are batteries connected in series? For what use are they connected in parallel? Does it matter if the batteries are nearly identical or not in either case?

12. Can the terminal voltage of a battery ever exceed its emf? Explain.

13. Explain in detail how you could measure the internal resis­tance of a battery.

14. In an RC circuit, current flows from the battery until the capacitor is completely charged. Is the total energy suppliedhv the hatterv ennal to the total enerav stored hv the canac-

15. Given the circuit shown in Fig. 26-34, use the words “increases,” “decreases,” or “stays the same” to complete the following statements:(a) If R7 increases, the potential difference between A and

E ____ . Assume no resistance in ® and %.(ib) If R7 increases, the potential difference between A and

E ____ . Assume @ and % have resistance.(c) If R7 increases, the voltage drop across RA_________ .(id) If R2 decreases, the current through _________ .(e) If R2 decreases, the current through R$_________ .( /) If R2 decreases, the current

through R3____ .(g) If Rs increases, the voltage

drop across R2 ____ .(h) If R5 increases, the voltage

drop across R4____ .(i) If R2,R 5, and R7 increase,

% (r = 0) ____ .

FIGURE 26-34Question 15. R2,R s, and Rj are variable resistors (you can change their resistance), given the symbol .

16. Figure 26-35 is a diagram of a capacitor (or condenser) microphone. The changing air pressure in a sound wave causes one plate of the capacitor C to move back and forth. Explain how a current of the same frequency as the sound wave is produced.

FIGURE 26-35 Diagram of a capacitor microphone. Question 16.

Sound*C•

pressure 1 ^1 *n

/ Tt L i

M livable - J L

17. Design a circuit in which two different switches of the type shown in Fig. 26-36 can be used to operate the same light­bulb from opposite sides of a room.

FIGURE 26-36Question 17.

*18. What is the main difference between an analog voltmeter and an analog ammeter?

*19. What would happen if you mistakenly used an ammeter where you needed to use a voltmeter?

*20. Explain why an ideal ammeter would have zero resistance and an ideal voltmeter infinite resistance.

*21. A voltmeter connected across a resistor always reads less than the actual voltage across the resistor when the meter is not present. Explain.

*22. A small battery-operated flashlight requires a single 1.5-V battery. The bulb is barely glowing, but when you take the battery out and check it with a voltmeter, it registers 1.5 V. How would you explain this?

23. Different lamps might have batteries connected in either of the two arrangements shown in Fig. 26-37. What would be the advantages of each scheme?

outpui-vl

(diaphragm IFIGURE 26-37Question 23.

| Problems26-1 Emf and Terminal Voltage1. (I) Calculate the terminal voltage for a battery with an

internal resistance of 0.900 fi and an emf of 6.00 V when the battery is connected in series with (a) an 81.0-0 resistor, and (ib) an 810-0 resistor.

2. (I) Four 1.50-V cells are connected in series to a 12-0 light­bulb. If the resulting current is 0.45 A, what is the internal resistance of each cell, assuming they are identical and neglecting the resistance of the wires?

3. (II) A 1.5-V dry cell can be tested by connecting it to a low- resistance ammeter. It should be able to supply at least 25 A. What is the internal resistance of the cell in this case, assuming it is much greater than that of the ammeter?

4. (II) What is the internal resistance of a 12.0-V car battery w h ose term inal voltage drons to 8 .4 V when the starter

2 6-2 Resistors in Series and ParallelIn these Problems neglect the internal resistance of a battery unless the Problem refers to it.5. (I) A 650-0 and a 2200-0 resistor are connected in series with

a 12-V battery. What is the voltage across the 2200-0 resistor?6. (I) Three 45-0 lightbulbs and three 65-0 lightbulbs are

connected in series, (a) What is the total resistance of the circuit? (b) What is the total resistance if all six are wired in parallel?

7. (I) Suppose that you have a 680-0, a 720-0, and a 1.20-k0 resistor. What is (a) the maximum, and (b) the minimum resistance you can obtain by combining these?

8. (I) How many 10-0 resistors must be connected in series to give an equivalent resistance to five 10 0 -0 resistors connected in oarallel?

9. (II) Suppose that you have a 9.0-V battery and you wish to apply a voltage of only 4.0 V. Given an unlimited supply of1 .0-0 resistors, how could you connect them so as to make a “voltage divider” that produces a 4.0-V output for a 9.0-V input?

10. (II) Three 1.70-k0 resistors can be connected together in four different ways, making combinations of series and/or parallel circuits. What are these four ways, and what is the net resistance in each case?

11. (II) A battery with an emf of 12.0 V shows a terminal voltage of 11.8 V when operating in a circuit with two light­bulbs, each rated at 4.0 W (at 12.0 V), which are connected in parallel. What is the battery’s internal resistance?

12. (II) Eight identical bulbs are connected in series across a 110-V line, (a) What is the voltage across each bulb? (b) If the current is 0.42 A, what is the resistance of each bulb, and what is the power dissipated in each?

13. (II) Eight bulbs are connected in parallel to a 110-V source by two long leads of total resistance 1.4 0 . If 240 mA flows through each bulb, what is the resistance of each, and what fraction of the total power is wasted in the leads?

14. (II) The performance of the starter circuit in an automobile can be significantly degraded by a small amount of corrosion on a battery terminal. Figure 26-38a depicts a properly functioning circuit with a battery (12.5-V emf, 0.02-0 internal resistance)attached via corrosion-free cables to a starter motor of resistance R$ = 0.15 O. Suppose that later, corro­sion between a battery terminal and a starter cable introduces an extra series resistance of just R c = 0.100 into the circuit as suggested in Fig. 26-38b. Let P0 be the power delivered to the starter in the circuit free of corrosion, and let P be the power delivered to the circuit with corrosion. Determine the ratio P /P q-

FIGURE 26-38Problem 14.

(a)

tf, = 0 ,10 1 1 r— V M —

A W r-

15. (II) A close inspection of an electric circuit reveals that a 480-0 resistor was inadvertently soldered in the place where a 370-0 resistor is needed. How can this be fixed without removing anything from the existing circuit?

16. (II) Determine (a) the equivalent resistance of the circuit shown in Fig. 26-39, and (b) the voltage across each resistor.

820 Q 680 Q ► 960 £2

17.

18.

19.

20.

s \

(II) A 75-W, 110-V bulb is connected in parallel with a 25-W, 110-V bulb. What is the net resistance?(II) (a) Determine the equivalent resistance of the “ladder” of equal 125-0 resistors shown in Fig. 26-40. In other words, what resistance would an ohmmeter read if connected between points A and B? (b) What is the current through each of the three resistors on the left if a50.0-V battery is connected between points A and B?

FIGURE 26-40Problem 18.

(II) What is the net resistance of the circuit connected to the battery in Fig. 26-41?

R

FIGURE 26-41Problems 19 and 20.

(II) Calculate the current through each resistor in Fig. 26-41 if each resistance R = 1.20 kO and V = 12.0 V. What is the potential difference between points A and B?

21. (II) The two terminals of a voltage source with emf % and internal resistance r are connected to the two sides of a load resistance R. For what value of R will the maximum power be delivered from the source to the load?

22. (II) Two resistors when connected in series to a 110-V line use one-fourth the power that is used when they are connected in parallel. If one resistor is 3.8 kO, what is the resistance of the other?

23. (Ill) Three equal resistors (R) are connected to a battery as shown in Fig. 26-42. Qualitatively, what happens to (a) the voltage drop across each of these resistors, (b) the current flow through each, and (c) the terminal voltage of the battery, when the switch S is opened, after having been closed for a long time? (d) If the emf of the battery is9.0 V, what is its terminal voltage when the switch is closed if the internal resistance r is 0.50 0 and R = 5.500?(e) What is the terminal ^voltage when the switch is open?

FIGURE 26-42Problem 23. fv ■

FIGURE 26-39

24. (Ill) A 2.8-kO and a 3.7-kO resistor are connected in parallel; this combination is connected in series with a 1.8-kO resistor. If each resistor is rated at |W (maximumw ithout nvp.rhpntino^ what is the maYirrmm vnltaap that ran

25. (Ill) Consider the network of resistors shown in Fig. 26-43. Answer qualitatively: (a) What happens to the voltage across each resistor when the switch S is closed? (b) What happens to the current through each when the switch is closed?(c) What happens to the power output of the battery when the switch is closed? (d) Let R 1 = R2 = R3 = R4 = 125 D, and V = 22.0 V. Determine the current through each resistor before and after closing the switch. Are your qualita­tive predictions confirmed?

*1

Ra

26. (Ill) You are designing a wire resistance heater to heat an enclosed volum e of gas. For the apparatus to function prop­erly, this heater must transfer heat to the gas at a very constant rate. While in operation, the resistance of the heater will always be close to the value R = R 0 , but may fluctuate slightly causing its resistance to vary a small amount AR ( « R 0). To maintain the heater at constant power, you design the circuit shown in Fig. 26-44 , which includes two resistors, each of resistance r. D eterm ine the value for r so that the heater power will remain constant even if its resistance R fluctuates by a small amount. [Hint: If A/? « Rq , then \ P «

R

29. (II) For the circuit shown in Fig. 26-47 , find the potential difference between points a and b. Each resistor has R = 130 XI and each battery is 1.5 V.

FIGURE 26-47Problem 29.

30. (II) (a) A network of five equal resistors R is connected to a battery % as shown in Fig. 26-48 . D eterm ine the current I that flows out of the battery, (b) U se the value determined for I to find the single resistor i?eq that is equivalent to the five-resistor network.

R

31. (II) (a) What is the potential difference between points a and d in Fig. 26 -49 (similar to Fig. 26-13, Example 26-9), and (b) what is the

34 Q hf P ------------ v w v — —

terminal voltage of each battery?

FIGURE 26-49Problem 31.

*1 , r= § 2 =4 7 Q 3 I Q 4 5 V

a f—v w v —f — v w v H H db c

I t<§1 = r =75 V IQ

-H h H W W -

26-3 Kirchhoffs Rules27. (I) Calculate the current in the circuit o f Fig. 26-45 , and

show that the sum of all thevoltage changes around the ____ 11 r Jw V~~circuit is zero.

9.0 V9 .5 Q ‘

FIGURE 26-45Problem 27.

-V W \r12.0 £2

28. (II) Determ ine the terminal voltage of each battery in Fig. 26-46.

r2 = 2.0 Q

FIGURE

<9 1= 18R = 6.6 Q

rx = 1.0

32. (II) Calculate the currents in each resistor of Fig. 26-50.

58 V

“ O

25 Q- N m -

:64Q

82 Q

FIGURE 26-50 Problem 32.

33. (II) D eterm ine the magnitudes and directions of the currents through R x and R 2 in

Fig-26"51- 1 HI-------W\A/—

FIGURE 26-51

Z?2 = 1 8 Q- J W \A r-

V3 = 6.0V

34. (II) D eterm ine the magnitudes and directions of the currents in each resistor shown in Fig. 26-52 . The batteries have emfs of %\ = 9.0 V and %i = 12.0 V and the resistors have values of R j = 25 fl,R 2 = 4 8 fl, and R 3 = 35 fl.(a) Ignore internal resistance of the batteries. (b) Assum e each battery has internal resistance r = 1.0 fl.

FIGURE 26-52Problem 34.

35. (II) A voltage V is applied to n identical resistors connected in parallel. If the resistors are instead all connected in series with the applied voltage, show that the power transformed is decreased by a factor n2

36. (I ll) (a) Determ ine the currents I \ , I 2, and / 3 in Fig. 26-53 . Assum e the internal resistance of each battery is r = 1.0 fl.(b) What is the terminal voltage of the 6.0-V battery?

h\K

22 Q

15 Q

12 Q

18 Q

FIGURE 26-53Problems 36 and 37.

37. (I ll) What would the current Ii be in Fig. 26-53 if the 12-fl resistor is shorted out (resistance = 0)? Let r = 1.0 fl.

38. (I ll) Determ ine the current through each of the resistors in Fig. 26-54.

FIGURE 26-54Problems 38 and 39.

39. (I ll) If the 25-ft resistor in Fig. 26 -54 is shorted out (resistance = 0), what then would be the current through the 15-fl resistor?

40. (I ll) Twelve resistors, each of resistance R, are connected as the edges of a cube as shown in Fig. 26-55. Determ ine the equivalent resistance (a) between points a and b, the ends of a side; (b) between points a and c, the ends of a face diagonal; (c) between points a and d, the ends of the volume diagonal.[Hint: Apply an em f and determine currents; use symmetry at junctions.]

41. (I ll) D eterm ine the net resistance in Fig. 26 -56 (a) between points a and c, and (b) between points a and b. Assume R ' ^ R. [Hint: Apply an em f and determine currents; use symmetry at junctions.]

FIGURE 26-56Problem 41.

26-4 Emfs Combined, Battery Charging42. (II) Suppose two batteries, with unequal emfs of 2.00 V and

3.00 V, are connected as shown in Fig. 26-57. If each internal resistance is r = 0.450 f l, and i? = 4 00 QR = 4.00 fl , what is the ________ V W \rvoltage across the resistor R?

<§= 2.00 V

FIGURE 26-57Problem 42.

- W i—r

H p \ M -------<§ = 3 .00V r

26-5 RC Circuits43. (I) Estimate the range of resistance needed to make a variable

timer for typical intermittent windshield wipers if the capacitor used is on the order of 1 pF.

44. (II) In Fig. 26 -58 (same as Fig. 26-17a), the total resistance is 15.0 k fl, and the battery’s em f is 24.0 V. If the time constant is measured to be 24.0 ps,calculate (a) the total capacitance of the rcircuit and (b ) the time it takes for the — — voltage across the resistor to reach16.0 V after the switch is closed.

FIGURE 26-58Problems 44 and 46.

45. (II) Two 3.8-pF capacitors, two 2.2-kft resistors, and a 12.0-V source are connected in series. Starting from the uncharged state, how long does it take for the current to drop from its initial value to 1.50 mA?

46. (II) How long does it take for the energy stored in a capac­itor in a series R C circuit (Fig. 26 -58) to reach 75% of its maximum value? Express answer in terms of the time constant r = RC.

47. (II) A parallel-plate capacitor is filled with a dielectric of dielectric constant K and high resistivity p (it conducts very slightly). This capacitor can be m odeled as a pure capaci­tance C in parallel with a resistance R. Assum e a battery places a charge +Q and —Q on the capacitor’s opposing plates and is then disconnected. Show that the capacitor discharges with a time constant r = K e0p (known as the dielectric relaxation time). Evaluate r if the dielectric is glass with p = 1.0 X 1012fl-m and K = 5.0.

48. (II) The R C circuit o f Fig. 26 -59 (same as Fig. 26-18a) has R = 8.7 k fl and C = 3.0 pF. The capacitor is at voltage V0 at t = 0, when the switch is closed.H ow long does it take the capacitor

49. (II) Consider the circuit shown in Fig. 26-60 , where all resistors have the same resistance R. A t t = 0, with the capacitor C uncharged, the switch is closed, (a) A t t = 0, the three currents can be determined by analyzing a simpler, but equivalent, circuit. Identify this simpler circuit and use it to find the values of I \ , I 2, and / 3 at t = 0.(b) A t t = oo, the currents can be determined by analyzing a simpler, equivalent circuit. Identify this simpler circuit and implement it in finding the values o f I \ , I 2, and / 3 at t = oo.(c) A t t = oo, what is the potential difference across the capacitor?

^ R< — m -

51.

<g-FIGURE 26-60Problem 49.

50. (I ll) D eterm ine the time constant for charging the capacitor in the circuit o f Fig. 26-61 . [Hint: U se K irchhoff s rules.] (b) What is the maximum charge on the capacitor?

FIGURE 26-61Problem 50.

(I ll) Two resistors and two uncharged capacitors are arranged as shown in Fig. 26-62 . Then a potential difference of 24 V is applied across the combination as shown. (a) What is the potential at point a with switch S open? (Let V = 0 at the negative terminal of the source.) (b) What is the potential at point b with the switch open? (c) W hen the switch is closed, what is the final potential o f point b? (d) How much charge flows through the switch S after it is closed?

FIGURE 26-62Problems 51 and 52.

52. (I ll) Suppose the switch S in Fig. 26 -62 is closed. What is the time constant (or time constants) for charging the capacitors after the 24 V is applied?

26-7 Ammeters and Voltmeters53. (I) A n ammeter has a sensitivity o f 35,000 f t /V . What

current in the galvanometer produces full-scale deflection?54. (I) What is the resistance of a voltmeter on the 250-V scale

if the meter sensitivity is 35,000 ft /V ?55. (II) A galvanometer has a sensitivity o f 45 k ft /V and

internal resistance 20.0 ft. H ow could you make this into (d\ an ammeter that reads 2.0 A full scale, or (H\ a voltmeter

* 56. (II) A galvanometer has an internal resistance of 32 ft anddeflects full scale for a 55-juA current. Describe how to use this galvanometer to make (a) an ammeter to read currents up to 25 A , and (b) a voltmeter to give a full scale deflection of 250 V.

* 57. (II) A particular digital meter is based on an electronic modulethat has an internal resistance of 100 M ft and a full-scale sensitivity o f 400 mV. Two resistors connected as shown in Fig. 26 -63 can be used to change the voltage range. Assume R i = 10 M ft. Find the

, Meter module r = 100 MQ

value of R 2 that will result in a voltmeter with a full-scale range of 40 V.

FIGURE 26-63Problem 57.

- V W HR i

* 58. (II) A milliammeter reads 25 m A full scale. It consists of a0.20-ft resistor in parallel with a 33-ft galvanometer. How can you change this ammeter to a voltmeter giving a full scale reading of 25 V without taking the ammeter apart? What will be the sensitivity ( f t /V ) of your voltmeter?

*59. (II) A 45-V battery of negligible internal resistance is connected to a 44-kft and a 27-kft resistor in series. What reading will a voltmeter, o f internal resistance 95 kft, give when used to measure the voltage across each resistor? What is the percent inaccuracy due to meter resistance for each case?

*60. (II) A n ammeter whose internal resistance is 53 ft reads 5.25 m A when connected in a circuit containing a battery and two resistors in series whose values are 650 ft and 480 ft. What is the actual current when the ammeter is absent?

*61. (II) A battery with % = 12.0 V and internal resistance r = 1.0 ft is connected to two 7.5-kft resistors in series. A n ammeter of internal resistance 0.50 ft measures the current, and at the same time a voltmeter with internal resistance 15 kft measures the voltage across one of the 7.5-kft resistors in the circuit. What do the ammeter and voltmeter read?

* 62. (II) A 12.0-V battery (assume the internal resistance = 0)is connected to two resistors in series. A voltmeter whose internal resistance is 18.0 kft measures 5.5 V and 4.0 V, respectively, when connected across each of the resistors. What is the resistance of each resistor?

*63. (I ll) Two 9.4-kft resistors are placed in series and connected to a battery. A voltmeter of sensitivity 1000 f t /V is on the 3.0-V scale and reads 2.3 V when placed across either resistor. What is the em f of the battery? (Ignore its internal resistance.)

*64. (I ll) W hen the resistor R in Fig. 26 -64 is 35 ft, the high-resistance voltmeter reads 9.7 V. W hen R is replaced by a 14.0-ft resistor, the voltmeter reading drops to 8.1 V. What are the em f and internal resistance of the battery?

FIGURE 26-64

- e -

-V W W -R

| General Problems65. Suppose that you wish to apply a 0.25-V potential differ­

ence between two points on the human body. The resistance is about 1800 O, and you only have a 1.5-V battery. How can you connect up one or more resistors to produce the desired voltage?

66. A three-way lightbulb can produce 50 W, 100 W, or 150 W, at 120 V. Such a bulb contains two filaments that can be connected to the 120 V individually or in parallel.(a) Describe how the connections to the two filaments are made to give each of the three wattages. (b) What must be the resistance of each filament?

67. Suppose you want to run som e apparatus that is 65 m from an electric outlet. Each of the wires connecting your appa­ratus to the 120-V source has a resistance per unit length of0.0065 n /m . If your apparatus draws 3.0 A , what will be the voltage drop across the connecting wires and what voltage will be applied to your apparatus?

68. For the circuit shown in Fig. 26-18a, show that the decrease in energy stored in the capacitor from t = 0 until one time constant has elapsed equals the energy dissipated as heat in the resistor.

69. A heart pacemaker is designed to operate at 72beats/m in using a 6.5-/xF capacitor in a simple RC circuit. What value of resistance should be used if the pacemaker is to fire (capac­itor discharge) when the voltage reaches 75% of maximum?

70. Suppose that a person’s body resistance is 950 fi. (a) What current passes through the body when the person accidentally is connected to 110 V? (b) If there is an alternative path to ground whose resistance is 35 O, what current passes through the person? (c) If the voltage source can produce at most 1.5 A, how much current passes through the person in case (b)l

71. A Wheatstone bridge is a type of “bridge circuit” used to make measurements of resistance. The unknown resistance to be measured, R x , is placed in the circuit with accurately known resistances R i ,R 2, and R 3 (Fig. 26-65). One of these,

, is a variable resistor which is adjusted so that when theswitch is closed momentarily, the ammeter current flow, (a) Determine R x in terms of R± , R 2 , and R 3. (b) If a Wheatstone bridge is “balanced” when R \ = 6300 , #2 = 9 7 2 0 , and R 3 = 78.6 fi, what is the value of the unknown resistance?

FIGURE 26-65Problems 71 and 72. W heatstone bridge.

shows zero

72. A n unknown length of platinum wire 1.22 mm in diameter is placed as the unknown resistance in a W heatstone bridge (see Problem 71, Fig. 26-65). Arms 1 and 2 have resistance of 38.0 f i and 29.2 f i , respectively. Balance is achieved when R 3 is 3.48 f i . How long is the platinum wire?

73. The internal resistance of a 1.35-V mercury cell is 0.030 fi, whereas that of a 1.5-V dry cell is 0.35 fi. Explain why threem p .rr.n rv r.p.lls ra n m n rp p ffp .rtivp .lv n n w p r a 9 S -W hp .arino

74. H ow many §-W resistors, each of the same resistance, must be used to produce an equivalent 3.2-kfl, 3.5-W resistor? What is the resistance of each, and how must they beconnected? D o not exceed P = \ W in each resistor.

75. A solar cell, 3.0 cm square, has an output of 350 m A at0.80 V when exposed to full sunlight. A solar panel that delivers close to 1.3 A of current at an em f of 120 V to an external load is needed. H ow many cells will you need to create the panel? H ow big a panel will you need, and how should you connect the cells to one another? H ow can you optimize the output o f your solar panel?

76. A power supply has a fixed output voltage of 12.0 V, but you needV x = 3.0 V output for an experiment, (a) Using the voltage divider shown in Fig. 26-66 , what should R 2 be if Ri is 14.5 O? (b) What will the terminal voltage Vt be if you connect a load to the 3.0-V output, assuming the load has a resistance of7.0 fi?

FIGURE 26-66Problem 76.

77. The current through the 4.0-kfl resistor in Fig. 26 -67 is 3.10 m A. What is theterminal voltage V a of 4.0 kQthe “unknown” battery? v ba 5 o kO(There are two answers.Why?)

FIGURE 26-67Problem 77.

12.0 V t

12.0 V

78. A battery produces 40.8 V when 7.40 A is drawn from it, and 47.3 V when 2.80 A is drawn. What are the em f and internal resistance of the battery?

79. In the circuit shown in Fig. 26-68 , the 33 -0 resistor dissipates 0.80 W. What isthe battery voltage? 68 Q

3 3 Q f------V W V ------H V W V —

FIGURE 26-68Problem 79.

—vwv—175 a

80. The current through the 2 0 -0 resistor in Fig. 26 -69 does not change whether the

20 0two switches Si and S2 are both open or both closed. U se this clue to determine the value of the unknown resis­tance R.

FIGURE 7R-RQ

* 81. (a) A voltmeter and an ammeter can be connected as shown in Fig. 26 -70a to measure a resistance R . If V is the volt­meter reading, and I is the ammeter reading, the value of R will not quite be V / I (as in Ohm ’s law) because some of the current actually goes through the voltmeter. Show that the actual value of R is given by

1 _ L _ J _R ~ V R y ’

where R y is the voltmeter resistance. N ote that R ~ V / I if R y » R . (b ) A voltmeter and an ammeter can also be connected as shown in Fig. 26-70b to measure a resistance R . Show in this case that

where V and I are the voltmeter and ammeter readings and R a is the resistance of the ammeter. N ote that R ~ V / I ifR a <s c R .

(a)

FIGURE 26-70(b)

Problem 81.

82. (a) What is the equivalent resistance of the circuit shown in Fig. 26-71? (b ) What is the current in the 18-11 resistor? (c) What is the current in the 1 2 -0 resistor? (d) What is the powerdissipation in the 4.5-11 resistor? 12 Q

FIGURE 26-71Problem 82.

-± -6 .0 V

■ vw v^4.5 Q

83. A flashlight bulb rated at 2.0 W and 3.0 V is operated by a 9.0-V battery. To light the bulb at its rated voltage and power, a resistor R is connected in series as shown in Fig. 26-72 . What value I- V W V 1 should the resistor have? R

FIGURE 26-72Problem 83.

A9.0 V

84. Some light-dimmer switches use a variable resistor as shown in Fig. 26-73 . The slide moves from position x = 0 to x = 1, and the resistance up to slide position x is propor­tional to x (the total resistance is R war = 150 XI at x = 1). What is the power expended in the lightbulb if (d) x = 1.00, (b ) x = 0.65, (c) * = 0.35?

120 V ± -

FIGURE 26-73

R yar= 150Q

85. A potentiometer is a device to precisely measure potential differences or emf, using a “null” technique. In the simple potentiometer circuit shown in Fig. 26-74 , R ' represents the total resistance of the resistor from A to B (which could be a long uniform “slide” wire), whereas R represents the resis­tance of only the part from A to the movable contact at C. W hen the unknown em f to be measured, %x , is placed into the circuit as shown, the movable contact C is moved until the galvanometer G gives a null reading (i.e., zero) when the switch S is closed. The resistance between A and C for this situation we call R x . Next, a standard emf, % , which is known precisely, is inserted into the circuit in place of %x and again the contact C is m oved until zero current flows through the galvanometer when the switch S is closed. The resistance between A and C now is called R s . (a) Show that the unknown em f is given by

•- ■ ( f hwhere R X, R S, and % are all precisely known. The working battery is assumed to be fresh and to give a constant voltage. (b ) A slide-wire potentiometer is balanced against a 1.0182-V standard cell when the slide wire is set at 33.6 cm out of a total length of 100.0 cm. For an unknown source, the setting is 45.8 cm. What is the em f of the unknown? (c) The galvanometer of a

Working battery

— H i-------------— |

---------- R ---------- *\

potentiometer has an internal resistance of 35 O and can detect a current as small as 0.012 mA. What is the minimum uncertainty possible in measuring an unknown voltage? (id) Explain the advan­tage of using this “null” method of measuring emf.

FIGURE 26-74Potentiometer circuit.Problem 85.

86. Electronic devices often use an R C circuit to protect against power outages as shown in Fig. 26-75 . (a) If the protector circuit is supposed to keep the supply voltage at least 75% of full voltage for as long as 0.20 s, how big a resistance R is needed? The capacitor is 8.5 / j lF . Assum e the attached “electronics” draws negligible current. (b) Betw een which two terminals should the device be connected, a and b, b and c, or a and c?

Protector R circuit Electronic

device

Power Power outage FIGURE 26-75

87. The circuit shown in Fig. 26 -76 is a primitive 4-bit digital- to-analog converter (DAC). In this circuit, to represent each digit (2n) of a binary number, a “1” has the nth switch closed whereas zero (“0”) has the switch open. For example, 0010 is represented by closing switch n = 1, while all other switches are open. Show that the voltage V across the 1 .0 -0 resistor for the binary numbers 0001, 0010, 0100, and 1010 (w hichrepresent i^Q kn1, 2, 4, 10) follows the pattern that you expect for a 4-bit DAC.

FIGURE 26-76Problem 87.

1.0 Q “VWV”

n= 1 ✓ 8.0 kQS . — VA\—

4.0 kQ• — vwv—

2.0 kQ•—vwv—

h— V—H

FIGURE 26-77Problem 88. 6.80 Q

3.0% ofinitialcharge?

its

12.0 V -± -

FIGURE 26-78Problem 89.

5.0 Q

FIGURE 26-79Problem 90.

100.0 V -=F33.0 kQ

C = 4.00 fiF

* 91. Measurements made on circuits that contain large resistances can be confusing. Consider a circuit powered by a battery % = 15.000 V with a 10.00-MH resistor in series with an unknown resistor R . A s shown in Fig. 26-80 , a particular voltmeter reads V\ = 366 mV when connected across the 10.00-M fl resistor, and this meter reads V2 = 7.317 V when connected across R . Determine the value of R . [Hint. Define R y as the voltmeter’s internal resistance.]

88. D eterm ine the current in each resistor of the circuit shown in Fig. 26-77 . 8_00 y

^-4.00 V

89. In the circuit shown in Fig. 26-78 , switch S is closed at time t = 0. {a) A fter the capacitor is fully charged, what is the voltage across it? H ow much charge is on it? (b) Switch S is now opened. How long does it now take for the capacitor to discharge until it has only

^ : l o q ^ 1 0 .0 Q

FIGURE 26-80 Problem 91.

*92. A typical voltmeter has an internal resistance of 10 M il and can only measure voltage differences of up to several hundred volts. Figure 26-81 shows the design of a probe to measure a very large voltage difference V using a voltmeter. If you want the voltmeter to read 50 V when V = 50 kV, what value R should be used in this probe?

90. Figure 26-79 shows the circuit for a simple sawtooth oscil­lator. A t time t = 0, its switch S is closed. The neon bulb has initially infinite resistance until the voltage across it reaches 90.0 V, and then it begins to conduct with very little resistance (essentially zero). It stops conducting (its resistance becomes essentially infinite) when the voltage drops down to 65.0 V. (a) A t what time t\ does the neon bulb reach 90.0 V and start conducting? (b) A t what time t2 does the bulb reach 90.0 V for a second time and again become conducting? (c) Sketch the sawtooth waveform between t = 0 and t = 0.70 s.

* Numerical/Computer* 93. (II) A n R C series circuit contains a resistor R = 15 kO, a

capacitor C = 0.30 /j l F , and a battery of emf % = 9.0 V. Starting at t = 0, when the battery is connected, determine the charge Q on the capacitor and the current I in the circuit from t = 0 to t = 10.0 ms (at 0.1-ms intervals). Make graphs showing how the charge Q and the current I change with time within this time interval. From the graphs find the time at which the charge attains 63% of its final value, C% and the current drops to 37% of its initial value, %/R.

Answers to Exercises

A: {a) 1.14 A; (b) 11.4 V; (c) PR = 13.1 W, Pr = 0.65 W.

B: 6 O and 25 O.

D: 180 A; this high current through the batteries could cause them to becom e very hot: the power dissipated in the weak battery would be P = I 2r = (180 A )2(0.10X1) = 3200 W!