physics beyond 2000 chapter 15 electric circuits

Physics Beyond 2000

Chapter 15

Electric Circuits

Electric current• An electric current is a flow of electric

charges through a conductor.

• Current = rate of flow of charges

unit: ampere A

Constant current

Changing current

Definition of ampere and coulomb

• Definition of 1 Ampere of current:

I = 1A I = 1A

1. Two infinitely longparallel wires 1 m apart

2. The same amount ofcurrents are in the wire

3. The forces onthe wires are2 × 10-7N m-1

4. The current ineach wire is 1 ampere

Definition of ampere and coulomb

• Definition of 1 coulomb (C):

Q = I.t

• 1 C = The amount of charges flowing through in 1 second with a current of 1 A.

I = 1A

t = 1 s

Charge-carriers

• The charge-carriers move and form the current.

• The type of charges carried by a charge-carrier may either be negative or positive depending on the material.

Charge-carriers

Substance charge-carrier Type of charge

Metal Free electrons Negative charge

Electrolyte Positive ions Positive charges

negative ions Negative charges

Gas discharge tube

Thermionic electrons

Negative charges

Charge-carriers

• Charge-carriers in electrolyte: positive and negative ions.

positiveelectrode

negativeelectrode

positive ion

negative ion

Electric field in a circuit

1. Cell with emf ξ2. Switch S

3. Resistor R

The switch is open.The p.d. of AB is zero. Not any current to flow.

Close the switch S. An electric field is established inside the circuit.

The electric field exerts force on the charge-carriers(electrons). Charge-carriers are accelerated.

The charge-carriers are also retarded by the collisionwith atoms.

Electric field in a circuitAs a result, the charge-carriers are moving atconstant speed.

Drift velocity• Charge-carriers in a metal move with constant sp

eed under an electric field.

• The constant speed is called the drift velocity vD.

E = electric field strength

Drift velocity• Consider a conductor with n charge-carriers

per unit volume.

E = electric field strength

Drift velocity• In time Δt , all the charge-carriers in the

shaded region have passed the imaginary cross-section.

E = electric field strengthvD. Δt

cross-sectionalarea A

Drift velocity• What is the number of charge-carriers ΔN

in the shaded area?

cross-sectionalarea AΔN = nA.(vD.Δt)

Drift velocity• Suppose each charge-carrier has charge q. What is

the amount of charges ΔQ passing the cross-sectional area in time Δt?

ΔQ = q.ΔN = q.nA.(vD. Δt)

Drift velocity• What is the current I in the conductor?

electric field strengthvD. Δt

DnAqvdt

Drift velocity• So the drift velocity of the charge-carriers

in a conductor is

electric field strengthvD. ΔtnAq

Drift velocity• So the drift velocity of the conductor is

• In small cross-sectional area, charge-carriers would move faster if other quantities are the same.• In semi-conductor, n is small. The drift velocity is fast.

Drift velocity• So the drift velocity of the conductor is

• In metal, the charge-carriers are electrons which are in random motion at very fast speed.• This random speed does not contribute tothe current.

Example 1

• Drift velocity of the charge-carriers in a copper wire. (10-5 ms-1)

Signal and drift velocity• The drift velocity of electrons in a circuit is

very small.

• The electric signal travels at a very fast speed in the circuit. It is close to the speed of light.

Electromotive force1. Suppose that the charge-carriers are positively charged.

2. The charge-carriers gain energy from the batterywhen they flow from the low potential to high potentialinside the battery.

3. The charge-carrierslose energy when theymove from high potentialto low potential in thecircuit.

4. The charge-carrierslose all energy whenthey return to thebattery.

Electromotive force1. Suppose that the charge-carriers are positively charged.

2. The battery does work on the charge-carriers in the battery.The energy transferred to one coulomb of charge is calledthe electromotive force of the battery.

Electromotive forceThe electromotive force ξof a sourceis the energy transferred into electricalenergy per unit charge within the source.

where Q is the amount of chargethrough the battery andU is the total energy provided bythe battery to the charge Q.

unit: J C-1 or V, volt

Electromotive force

The electromotive force ξof a sourceexists though there is not any current.

Example 2

•The e.m.f. (electromotive force) of an AA cell.•Ah (ampere-hour) is a unit for charge. Q = I.t

Combination of cells

• Cells in series

• Cells in parallel

Cells in series

• ξ= ξ1 + ξ2 +ξ3

• Produce a high e.m.f.

ξ1 ξ2 ξ3

Cells in parallel

• Total e.m.f. = ξ

• It can supply a larger current.

• It lasts longer.

• It decreases the internal resistance of the source.

Potential difference (p.d.)• The p.d. between two points in the circuit is

the amount of energy change as one coulomb of charge passes from one point to another.

• Unit: V (volt)

1. An amount of chargeQ passes from X to Y

2. The charge loses electrical energy U

Voltage

• Voltage is a general term.

• It may refer to the e.m.f. of a source or the p.d. between two points.

• Unit: V (volt)

Measurement of voltage

• Use voltmeter

• Use CRO

to voltmeter or CRO to voltmeter or CRO

Example 3

• First find the amount of charge of one mole of electrons.

• U = QV

Voltage and electric field strength

1. A p.d. V between points X and Y.

2. The separationXY is d.

3. There is anelectric field Ebetween X and Y.

4. The average electricfield strength is E = V/d.

Example 4

• Use E = V/d and F = e.E

Note that the electrons move with uniform drift speed.

Resistance

1. The p.d. betweenthe ends of the conductoris V

I2. A current I passesthrough the conductor 3. The resistance of

the conductor is R = V/I

4. Unit of R is ohm, Ω

I-V graph

• Pure metal

• R = constant or V I (Ohm’s law)

• The slope givesI

http://www.fed.cuhk.edu.hk/sci_lab/Simulations/phe/ohmslaw.htm

I-V graph• Bulb filament.

• The filament resistance R increases with temperature.

• Ohm’s law is not obeyed. I

I-V graph• Semi-conducting diode• Current increases sharply in forward bias.• Current is zero in reverse bias.

current startsto increase

Dynamic resistance Rd

• Find the slope from the I-V graph

• UsedV

Dependence of resistance

• Depending on the physical dimension such as length ( ) and cross-sectional area (A).

• Depending on the material which is described by its resistivity (ρ).

Resistivity

Material Resistivity ρ (Ωm)

silver 1.62 × 10-8

copper 1.69 × 10-8

tungsten 5.25 × 10-8

pure silicon 2.5 × 103

Effect of temperature(metals)

• Resistance increases uniformly with temperature for metals.

θ/oC0

resistivity ρ

ρ= ρo (1 + α.θ)

αis called the temperaturecoefficient.

Effect of temperature(metals)

• An increase in temperature increases the thermal agitation of atoms.

• The interaction between atoms and electrons increases.

• The resistance increases. atom

electron

Superconductivity

• Some metals or alloys lose the resistance at very low temperature.

• No dissipation of energy when current flows in a superconductor.

Internal resistance of battery

• When current flows inside a battery, energy is also lost.

• The battery has internal resistance.

• The e.m.f. of a battery > The terminal voltage of a battery.

r = e.m.f.

V= terminal voltage

1. e.m.f. describesthe provision of power

2. terminal voltagedescribes the lossof power

To measure the internal resistance r of a battery

1. A battery with emf andinternal resistance r

2. An ammeterto measure thecurrent I.

3. A variable resistor Rto adjust the current

Express R in terms of , I and r. rI

Plot the graph of R-I

How to find r from the graph?

The internal resistance r of a battery

• We assume that the internal resistance is a constant.

Combination of resistors

• Resistors in series

• Resistors in parallel

• Resistors in series

• A common current passing through all resistors

• The equivalent resistance R of the system is

R = R1 + R2 + R3

R1 R2 R3

http://www.lightlink.com/sergey/java/java/resist2/index.html

• Resistors in parallel

• Same voltage across each resistor

• The equivalent resistor R gives

http://www.lightlink.com/sergey/java/java/resist4/index.html

Power and heating effect• When a current I passes through a resistor

of resistance R and the p.d. is V, some electric energy is converted into internal energy, the power output is

VRIIVP

Power of a cell

• If the e.m.f. of a cell is ξand it provides a current I for an external circuit, then the power of the cell is

rRrRIIPo

2 ).(.

Example 5• Note that the internal resistor r is not

counted in the power output.

power outputV

Power output and external resistance

• The power output Pout is

power output

• The power output is maximum when R = r.

power output

• What is the efficiency when R = r?

Efficiency

power output

Hint: efficiency = Pout/Po

Example 6

• Use a cell of r = 2 Ω

Power transmission

• High voltage is used for transmission of electric power.

Power transmission• What is the loss of power in the cable? (Use IL, the

current in the cable, and R, the resistance of the cable.)

Domestic Electricity

1. Live wire: with potential changing between positive andnegative

2. Neutral wire: maintainingat zero potential by earthingin the power station.

3. Electrical devicesare connected in parallel. Same p.d. foreach device.

Ring circuit1. Same p.d.for each device.

live neutral

Ring circuitlive neutral

1. Same p.d.for each device.

2. If a part is broken, can itbe operated?

YES, it can.

Combination of two cells in parallelwith same e.m.f.

1. A cell with emf andinternal resistance r1

2. A cell with emf andinternal resistance r2

3. Connect themin parallel

4. Note that there is not any currentflow between the cells because theiremf are the same.

Combination of two cells in parallelwith different e.m.f.

1. A cell with emf 1 andinternal resistance r1

2. A cell with emf 2 andinternal resistance r2

3. 1 > 2 5. There is current flowingfrom the cell of high emfto that of low emf

Example 6 (p.327)

1. A cell with emf 14V andinternal resistance 5.

2. A cell with emf 12V andinternal resistance 15 .

3. 1(14V) > 2(12V) 5. There is current flowingfrom the cell of high emfto that of low emf. Find the current.

Combination of two cells in parallelwith same emf

1. Two cells of emf areconnected in parallel.

3. Set up equations forthese quantities.

2. Both cells provide current I for the externalresistor R.

I1 + I2 = I ................. (1)

= I.R + I1.r1 ............ (2)

I1 + I2 = I ................. (1)

= I.R + I1.r1 ............ (2)

= I.R + I2.r2 ............ (3)

Example 6

Given: = 1.5V, r1 = 2 , r2 = 4 and R = 6 .

Find the currentsand the power output.

I-V characteristics

• Linear: Obey Ohm’s law. Ohmic device.

• Non-linear: Not obey Ohm’s law. Non-ohmic device.

Non-ohmic

Note that R = I

I-V characteristics

= 3 Vr = 1 Find R if I = 1 A.

I-V characteristics

• Junction diodeI/mA

0.8VD is 0.8 V for a currentof 20 mA.If the current is too large, the diode would be damaged.

I-V characteristics

• Junction diodeI/mA

= 3V, r = 0

Find the suitable value of R.What would happen if R is too big or too small?

I-V characteristics

• Thermionic diodeI

The maximum current is 2.0 mA.This current is called the saturation current.

I-V characteristics

• Thermionic diode

I = 2 mA

=12V, r = 0

R I/mA

Find the minimum value for R.

Kirchhoff’s rules

• 1. Kirchhoff's Junction Rule,

• 2. Kirchhoff's Loop Rule and

• 3. Ohm's Law (i.e. V/I = R, where I is the current, V is the voltage, and R is the resistance).

Kirchhoff’s rules• 1. Kirchhoff's Junction Rule

The algebraic sum of the currents at any branch point or junction in a circuit is zero.

I1 + I2 + I3 - I4 - I5 = 0

Note that it is necessaryto add a minus sign beforeI4 and I5 because they areleaving the junction.

Kirchhoff’s rules

• 2. Kirchhoff's Loop Rule

The algebraic sum of the potential changes around any complete loop in the network is zero.

Note the the potentialdrops by IR when current passing a resistor. It is necessary to add a minus before I.R following the current.

Kirchhoff’s rules

• 2. Kirchhoff's Loop Rule

The algebraic sum of the potential changes around any complete loop in the network is zero.

Consider the loop ABEF:

1-I1R1-I2R2 - 2

-I2R4 = 0

Kirchhoff’s rules• circuit 1

• circuit 2

• circuit 3

• circuit 4

• circuit 5

http://www.lightlink.com/sergey/java/java/kirch5/index.html

Bridge circuits

• 4 resistors are connected as a bridge circuit.

• Two currents in the bridge circuit.

I2I = I1 + I2

Bridge circuits• If the potentials at B and C are equal,• Connect BD with a galvanometer. Is there any

current passing the galvanometer?

I = I1 + I2

Example 7• The galvanometer shows no reflection

• No current flows between B and D

• The potentials at B and D are equal

Measuring devices

• Potentiometer

Measuring devices

• Potentiometer– To measure small p.d.– To measure the internal resistance of a cell– To calibrate a voltmeter– To measure resistance

•Advantage: It can measure the voltage of a device without drawing any current from the device.

http://www.plus2physics.com/current_electricity/study_material.asp?chapter=5

Potentiometer• An accumulator provides a steady current for a

long period of time.• A resistance wire with uniform cross-section.

1. accumulator Vo (2V)

2. A uniform resistancewire of length

To measure an unknown voltage (p.d.)

V1. Devicewith p.d.

2. A galvanometerto detect current throughthe device

3. A protectiveresistor

4. A slide tocontact wire AB.

5. Both +ends areconnected

To measure an unknown voltage (p.d.)

1. Move the slide on ABto find a balance point.

2. The galvanometerdoes not deflect when Pis the balance point.

3.Show that

Using the protective resistor

1.Before the balance pointis found, the current passingthe galvanometer is large. Use the protective resistor to decrease the current.

2. When the balance point isnear, the current passing thegalvanometer is small. Shortthe protective resistor to increase the current.

Calibrating a potentiometer• Use a standard cell (e.g. Weston cell with emf = 1.

0186 V at 20oC) to calibrate a potentiometer.

standardcell

Calibrating a potentiometer• Find the balance point P and measure the length AP.• Find the voltage per unit length of the resistance wire.

standardcell

1. What is the voltageacross AP?2. So what is ?

3. = AP

0186.1

Calibrating a potentiometer

• Use this calibrated potentiometer to measure the p.d. V of a device.

+ -P’

device

1. V = .AP’

No balance point is found

• The terminals of the device are reversed.

• The p.d. of the device is too large. V > Vo.

+ -P’

device

Measure the emf of a cell

cell withemf V

1. Find the balance point P2. emf V = AP where is the voltage per unit length.

3. Why is the result the emf?Why not the p.d.?

It is because there isnot any current throughthe cell. The pd is equalto the emf.

Example 8

• The SI unit of is V m-1.

Measuring very small p.d.• AP would be too short if the measured V is very small because

V = .AP.• The error of measurement is too big.

device with verysmall V

Measuring very small p.d.• Add a variable resistor R in series with the resistance wire. Adjust R so

that VAB is just greater than V.• Must re-calibrate the potentiometer when R is changed.

device with verysmall V

Examples 9-10• The emf of a thermocouple is very small.• Adjust R so that the balance point P is near B.

cold junction

hotjunction

Measure the internal resistance r

of a cell

3. R is a known resistor

4. S is a switch

2. r is the internal resistance

1. The cell

of a cell

1. With switch S open, locate the balance point P1.2. Measure lengthAP1.3. The length AP1

represents the emf’ of the cell.

4. There is notany current flowingthrough the cell

of a cell

1. With switch S closed, locate the balance point P2.2. Measure lengthAP2.3. There is currentI2 passing the cell and the resistor R.4. Length AP2

represents the voltage V acrossR.

of a cell

2. We have ’ = I2.R + I2.r

1. Note that the two currentsI and I2 do not mix up.

Measure the internal resistance r of a cell

‘ = AP1.............(1)

V = AP2.............(2)

’ = I2.R + I2.r.........(3)

V = I2.R ................ (4)

Find r in terms of AP1, AP2 and R.

APAPr .

Example 11

• First find the voltage per unit length .

• The terminal voltage V is the voltage across the cell with current flowing.

• The e.m.f. is the voltage across the cell without current flowing.

• The above two are not equal if the cell has internal resistance r.

Calibrating a voltmeter

• To check that the reading of a voltmeter is correct.

R R’

1. V = . AP

1. V = . APA B

R R’

2. Connect a voltmeterto measure the voltageacross R. Compare itsreading with the abovevalue

3. Adjust R for other readings of the voltage.Each time locate the new balance point

Measure resistance R

• Measure V across R using the potentiometer. V= ξ.AP

• Read the current I through the resistor.

• R =

Electrical meters

• Ideal measuring instruments: should not change the system being measured.

• Ammeter: its resistance should be as small as possible.

• Voltmeter: its resistance should be as large as possible.

Moving-coil galvanometer

• It can measure a small current.

• http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/galvan.html#c1

Centre-zero galvanometer

• Can be changed into an ammeter or voltmeter.

• Typical data of a moving-coil galvanometer:

f.s.d. current = 1.0 mA

f.s.d. voltage = 0.15 V

resistance = 150 Ω

Ammeter

• It is connected in series in the circuit.

Note that the total resistance of thecircuit is R + r where r is the resistanceof the ammeter.

Adapt a galvanometer to measure a large current (an ammeter)

• Shunt: add a resistor in parallel with the galvanometer.• The adapted galvanometer has a smaller resistance.

moving-coil galvanometer

shuntthe ammeter

Adapt a galvanometer to measure a large current (an ammeter)

• Example: A moving-coil galvanometer with f.s.d. current 1.0 mA is adapted to measure 1.0A at f.s.d.. What is the resistance of the shunt?

I = 1.0A I1=1.0mA

shuntI2 R

Note that the p.d. across R = the p.d. across the galvanometer.

R = 0.15V

voltmeter• It is connected in parallel in the circuit.

Note that the total resistance of the circuit is R.r/(R + r) where r is the resistance of the voltmeter.

Adapt a galvanometer to measure a large voltage (a voltmeter)

• Multiplier: add a resistor in series with the galvanometer.

• The adapted galvanometer has a larger resistance.

multiplier

voltmeter

Adapt a galvanometer to measure a large voltage (a voltmeter)

• Example: A moving-coil galvanometer with f.s.d. voltage 0.1 V is adapted to measure 15 V at f.s.d.. What is the resistance of the multiplier?

15V0.1V

I=1.0mA I

Note that the same current is passingthrough the multiplier and the galvanometer.

R = 15 k

Load effect of a voltmeter: a voltmeter draws current

1. Without voltmeter

I = 0.6 mA

2. With a voltmeter of resistance 15 k

I = 0.686 mA

0.171 mA

0.515 mA

Load effect of a voltmeter: a voltmeter draws current

• The best solution is to add a voltage follower.• A voltage follower draws a current as small as 0.1

nA from the circuit.

< 0.1nA

The voltmeterdraws currentfrom the voltagefollower. Not fromthe circuit.

Ohm-meter (digital)

• To read the resistance directly from the ohm-meter.

Ohm-meter (analog)

2. Moving-coilgalvanometer

3. Variable resistancefor setting zero

1. cell withemf

4. Connect resistor to these terminalsfor measurement

Ohm-meter (analog)

1. When measuring a resistor R,

2. The deflection of the galvanometeris proportional to I.

3. Adjust the scale of thegalvanometer to read R fromthe deflection.

4. The scale is non-linear.

Measuring resistance by voltmeter-ammeter method

1. For small R 2. For big R

Set up either of the following circuits. Measure V and I.Calculate R from

Measuring resistance by voltmeter-ammeter method

1. For small R 2. For big R

Discuss why there are two different circuits. Assume values for R in your discussion.

Multimeter

• A-V-O meter: ammeter, voltmeter and ohm-meter.

http://www.fed.cuhk.edu.hk/sci_lab/ntnujava/electronics/multimeter.html

(Analog) (Digital)

Multimeter

shunts

multiplier

sunknown resistor

10A1A100mA

To setzero

Shunts and currents

Given: f.s.d. current of the galvanometer = 1.0 mA resistance of the galvanometer = 150 resistance of each shunt = 0.15 Find: the f.s.d current in each of the following adapted galvanometer.

R R RResistance of the shuntis smaller.

Resistance of the shuntis larger.

1A 0.5A

Multimeter

shunts

Z10A1A

To setzero

•To measure largecurrent

I I1. external currentflows into the multimeter

2. externalcurrent throughthe shunt (oneresistor, biggercurrent).

3. external current throughthe galvanometer

4. external currentflows out of themultimeter

Multimeter

shunts

Z10A1A

To setzero

•To measure small current

I I1. external currentflows into the multimeter

2. externalcurrent throughthe shunt (threeresistors, less current).

Multiplier and voltagesGiven: f.s.d. current of the galvanometer = 1.0mA f.s.d. voltage of the galvanometer = 0.15V resistance of the galvanometer = 150 resistance of each multiplier = 15 kFind: the f.s.d voltage in each of the following adapted galvanometer.

Resistance of the multiplieris smaller.

Resistance of the multiplieris larger.

15V 30V

Multimeter

•To measure a high voltage

multiplier

To setzero

V1. external currentflows into the multimeter

2. externalcurrent throughthe multiplier(three resistors, higher voltage).

Multimeter

•To measure a low voltage

multiplier

To setzero

V1. external currentflows into the multimeter

2. externalcurrent throughthe multiplier(one resistor, lower voltage).

Multimeter

shunts

multiplier

s2. connect an unknown resistor R to the inputs

To setzero

1. short theinputs and setzero with Z.

3. The internalcell providesthe current.

4. The galvanometer deflects to showthe resistance R.

5. The currentdepends on R.

• To measure R

physics beyond 2000 chapter 15 electric circuits

electric current

s slide

vdvd slide

ampere slide

drift velocity chargecarriers

n chargecarriers

electric field strength

chargecarriers electrons

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