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MyChapter 18

LectureOutline

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Chapter 18: Electric Current and Circuits

•Electric current

•EMF

•Current & Drift Velocity

•Resistance & Resistivity

•Kirchhoff’s Rules

•Series & Parallel Circuit Elements

•Applications of Kichhoff’s Rules

•Power & Energy

•Ammeters & Voltmeters

•RC Circuits

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§18.1 Electric Current

e-

e-

e-

e-

e- e-

e- e-A metal wire.

Assume electrons flow to the right.

Current is a measure of the amount of charge that passes though an area perpendicular to the flow of charge.

t

qI

ΔΔ

=: Current

Current units: 1C/sec = 1 amp

4

A current will flow until there is no potential difference.

The direction of current flow in a wire is opposite the flow of the electrons. (In the previous drawing the current is to the left.)

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Example: If a current of 80.0 mA exists in a metal wire, how many electrons flow past a given cross-section of the wire in 10.0 minutes?

( )( ) C 0.48sec 600A 100.80 3 =×=Δ=Δ

Δ

Δ=

−tIq

t

qI

electrons 1000.3

C/electron101.60

C 0.48

electronper charge

qelectrons of #

20

19

×=×

=

=

−

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§18.2 EMF and Circuits

An ideal battery maintains a constant potential difference. This potential difference is called the battery’s EMF().

The work done by an ideal battery in pumping a charge q is W = q.

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The circuit symbol for a battery (EMF source) is

+

At high potential

At low potential

Batteries do work by converting chemical energy into electrical energy. A battery dies when it can no longer sustain its chemical reactions and so can do no more work to move charges.

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§18.3 Microscopic View of Current in a Metal

Electrons in a metal might have a speed of ~106 m/s, but since the direction of travel is random, an electron has vdrift

= 0.

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Only when the ends of a wire are at different potentials (E 0) will there be a net flow of electrons along the wire (vdrift 0). Typically, vdrift < 1 mm/sec.

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Calculate the number of charges (Ne) that pass through the shaded region in a time Δt:

)(

)(

tvnA

AlnN

d

e

Δ==

The current in the wire is: de neAvt

eN

t

qI =

Δ=

ΔΔ

=

l

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Example (text problem 18.19): A copper wire of cross-sectional area 1.00 mm2 has a constant current of 2.0 A flowing along its length. What is the drift speed of the conduction electrons? Copper has 1.101029 electrons/m3.

( )( )( )mm/sec 0.11m/sec 101.1

m 1000.1C 1060.1m 101.10

A 0.2

4

2619329

=×=

×××==

=

−

−−−neA

Iv

neAvI

d

d

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§18.4 Resistance and Resistivity

A material is considered ohmic if ΔVI, where

IRV =Δ

The proportionality constant R is called resistance and is measured in ohms (; and 1 = 1 V/A).

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The resistance of a conductor is:A

LR ρ=

where ρ is the resistivity of the material, L is the length of the conductor, and A is its cross sectional area.

With Rρ a material is considered a conductor if ρ is “small” and an insulator if ρ is “large”.

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The resistivity of a material depends on its temperature:

( )( )00 1 TT −+= αρρ

where ρ0 is the resistivity at the temperature T0, and is the temperature coefficient of resistivity.

A material is called a superconductor if ρ = 0.

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( )( ) (2) 1 00 TT −+= αρρ

Example (text problem 18.28): The resistance of a conductor is 19.8 at 15.0 C and 25.0 at 85.0 C. What is the temperature coefficient of resistivity?

Values of R are given at different temperatures, not values of ρ. But the two quantities are related.

(1) A

LR ρ=

Multiply both sides of equation (2) by L/A and use equation (1) to get:

( )( ) (3) 1 00 TTRR −+= α

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Example continued:

Solve equation (3) for and evaluate using the given quantities:

130 C 1075.3C 0.15C 0.85

1 8.19 0.251

−− °×=°−°

−

=Δ

−=

TRR

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§18.5 Kirchhoff’s Rules

Junction rule: The current that flows into a junction is the same as the current that flows out. (Charge is conserved)

A junction is a place where two or more wires (or other components) meet.

Loop rule: The sum of the voltage dropped around a closed loop is zero. (Energy is conserved.)

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For a resistor: If you cross a resistor in the direction of the current flow, the voltage drops by an amount IR (write as IR). There is a voltage rise if you cross the other way (write as +IR).

B

A

I

If the current flows from A to B, then the potential decreases from A to B. The potential difference between A and B is < 0 (ΔV = IR) .

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For batteries (or other sources of EMF): If you move from the positive to the negative terminal the potential drops by (write as ). The potential rises if you cross in the other direction (write as +).

+

At high potential

At low potential

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A current will only flow around a closed loop.

0

0AB

=−−=−

IRIrIRV

B

VAB is the terminal voltage.

A

Applying the loop rule:

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In a circuit, if the current always flows in the same direction it is called a direct current (DC) circuit.

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§18.6 Series and Parallel Circuits

The current through the two resistors is the same. It is not “used up” as it flows around the circuit!

These resistors are in series.

Apply Kirchhoff’s loop rule:

eq2121

21

)(

0

IRRRIIRIR

IRIR

=+=+==−−

Resistors:

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The pair of resistors R1 and R2 can be replaced with a single equivalent resistor provided that Req = R1 + R2

.

In general, for resistors in series

.1

21eq

∑=

=

+++=n

ii

n

R

RRRR K

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Current only flows around closed loops. When the current reaches point A it splits into two currents. R1 and R2 do not have the same current through them, they are in parallel.

Apply Kirchhoff’s loop rule:

0

0

22

11

=−=−

RIRI

The potential drop across each resistor is the same.

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Applying the junction rule at A: I = I1+ I2.

eq21

21

111

RRR

I

RRI

=+=

+=

From the loop rules: 2211 RIRI ==

Substituting for I1 and I2 in the junction rule:

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In general, for resistors in parallel

.1

1111

1

21eq

∑=

=

+++=

n

i i

n

R

RRRRK

The pair of resistors R1 and R2 can be replaced with a single equivalent resistor provided that

.111

21eq RRR+=

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Example (text problem 18.40): In the given circuit, what is the total resistance between points A and B?

R2 and R3 are in parallel. Replace with an equivalent resistor R23.

=

+=

8

111

23

3223

R

RRR

R3 = 24

R2 =

12

R1 = 15

A

B

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R23 = 8

R1 = 15

A

B

The resistors R23 and R1 are in series:

eq

231123

23 R

RRR

==+=

The circuit can now be redrawn:

B

R123 =

23

AIs the equivalent circuit and the total resistance is 23 .

Example continued:

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Capacitors:

C2C1

For capacitors in series the charge on the plates is the same.

Apply Kirchhoff’s loop rule:

eq21

21

111

0

CCCQ

C

Q

C

Q

=+=

=−−

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In general, for capacitors in series

.1

1111

1

21eq

∑=

=

+++=

n

i i

n

C

CCCCK

The pair of capacitors C1 and C2 can be replaced with a single equivalent capacitor provided that

.111

21eq CCC+=

31

C2

C1

For capacitors in parallel the charge on the plates may be different. Here

Apply Kirchhoff’s loop rule:

0

0

2

2

1

1

=−

=−

CQCQ

21eq

21eq

CCC

QQQ

+=

+=

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In general, for capacitors in parallel

.1

21eq ∑=

=+++=n

iin CCCCC K

The pair of capacitors C1 and C2 can be replaced with a single equivalent capacitor provided that Ceq= C1 + C2.

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Example (text problem 18.49): Find the value of a single capacitor that replaces the three in the circuit below ifC1 = C2 = C3 = 12 F.

C2

C1

C3

B

A

C2 and C3 are in parallel

F 243223

=+= CCC

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The circuit can be redrawn:C1

C23

B

A The remaining two capacitors are in series.

F 8

F 24

1

F 12

1

111

123

231123

=

+=

+=

C

CCC

C123

B

A

Is the final, equivalent circuit.

Example continued:

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§18.8 Power and Energy in Circuits

VIVt

q

t

UP Δ=Δ

Δ=

ΔΔ

=

IP =

The energy dissipation rate is:

For an EMF source:

For a resistor:R

VRIVIP

22 Δ

==Δ=

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Example: Use the results of the example starting on slide 35 to determine the power dissipated by the three resistors in that circuit.

Resistance () Current (A) Power (W)

122 0.199 4.83

5.6 0.123 0.0847

75 0.0760 0.433

RIP 2=

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§18.9 Measuring Currents and Voltages

Current is measured with an ammeter. An ammeter is placed in series with a circuit component.

R2

R1

A1

A3

A2

A1 measures the current through R1.

A2 measures the current through R2.

An ammeter has a low internal resistance.

A3 measures the current drawn from the EMF.

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A voltmeter is used to measure the potential drop across a circuit element. It is placed in parallel with the component. A voltmeter has a large internal resistance.

R2

R1

V The voltmeter measures the voltage drop across R1.

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§18.10 RC Circuits

0=−−CQ

IR

Close the switch at t = 0 to start the flow of current. The capacitor is being charged.

Apply Kirchhoff’s loop rule:

C

R Switch

+

+

t

QI

ΔΔ

= :Note

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The current I(t) that satisfies Kirchhoff’s loop rule is:

( )

. and 0

0

RCR

I

eItIt

==

=−

τε

τ

where

is the RC time constant and is a measure of the charge (and discharge) rate of a capacitor.

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The voltage drop across the capacitor is: ( ) ⎟

⎠⎞⎜

⎝⎛ −=

− t

C etV 1

The voltage drop across the resistor is: ( ) ( )RtItVR =

The charge on the capacitor is:

( ) ( )tCVtQ CC =

Note: Kirchhoff’s loop rule must be satisfied for all times.

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Plots of the voltage drop across the (charging) capacitor and current in the circuit.

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C

R S1

S2 +

I

While the capacitor is charging S2 is open. After the capacitor is fully charged S1 is opened at the same time S2 is closed: this removes the battery from the circuit. Current will now flow in the right hand loop only, discharging the capacitor.

0=+−CQ

IR

Apply Kirchhoff’s loop rule:

( ) .0τt

eItI−

=The current in the circuit is

But the voltage drop across the capacitor is now ( ) . τεt

C etV−

=

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The voltage drop across the discharging capacitor:

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Example (text problem 18.85): A capacitor is charged to an initial voltage of V0 = 9.0 volts. The capacitor is then discharged through a resistor. The current is measured and is shown in the figure.

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(a) Find C, R, and the total energy dissipated in the resistor.

Use the graph to determine . I0 = 100 mA; the current is I0/e = 36.8 mA at t = 13 msec.

RCR

I

==

==

msec 13

mA 1000

Since = V0 = 9.0 volts, R = 90 and C = 144 F.

All of the energy stored in the capacitor is eventually dissipated by the resistor.

J 108.52

1 320

−×== CVU

Example continued:

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(b) At what time is the energy in the capacitor half of the initial value?

20

2

2

1)0( and )(

2

1)( CVtUtCVtU ===

0

20

2

20

2

1)(

4

1)(

2

14

1)0(

2

1)(

VtV

CVtCV

CVtUtU

=

=

===Want:

Example continued:

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( ) msec 5.42lnmsec 132ln

2

1)( 00

===

==−

τ

τ

t

VeVtVt

Solve for t:

Example continued:

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Summary

•Current & Drift Velocity

•Resistance & Resistivity

•Ohm’s Law

•Kirchhoff’s Rules

•Series/Parallel Resistors/Capacitors

•Power

•Voltmeters & Ammeters

• RC Circuits