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**preview**shows pages 1-3. to view the full**24 pages of the document.**CHAPTER 28 ELECTRIC CIRCUITS

ActivPhysics can help with these problems: Section 12, “DC Circuits”

Section 28-1: Circuits and Symbols

Problem

1. Sketch a circuit diagram for a circuit that includes a resistor

R

1

connected to the positive terminal of a battery, a pair of

parallel resistors

R

R

2

3

and

connected to the lower-voltage end of

R

1

, then returned to the battery’s negative terminal,

and a capacitor across

R

2

.

Solution

A literal reading of the circuit specifications results in connections like those in sketch (a). Because the connecting wires are

assumed to have no resistance (a real wire is represented by a separate resistor), a topologically equivalent circuit diagram is

shown in sketch (b).

Problem 1 Solution (a).

Problem 1 Solution (b).

Problem

2. A circuit consists of two batteries, a resistor, and a capacitor, all in series. Sketch this circuit. Does the description

allow any flexibility in how you draw the circuit?

Solution

In a series circuit, the same current must flow through all elements. One possibility is shown. The order of elements and the

polarity of the battery connections are not specified.

Problem 2 Solution.

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660 CHAPTER 28

Problem

3. Resistors

R

1

and

R

2

are connected in series, and this series combination is in parallel with

R

3

. This parallel

combination is connected across a battery whose internal resistance is

R

int

.

Draw a diagram representing this circuit.

Solution

The circuit has three parallel branches: one with

R

1

and

R

2

in series; one with just

R

3

; and one with the battery (an ideal

emf in series with the internal resistance).

Problem 3 Solution.

Section 28-2: Electromotive Force

Problem

4. What is the emf of a battery that delivers 27 J of energy as it moves 3.0 C between its terminals?

Solution

From the definition of emf (as work per unit charge), E

=

=

=

W

q

= =

27

3

9

J

C

V

.

Problem

5. A 1.5-V battery stores 4.5 kJ of energy. How long can it light a flashlight bulb that draws 0.60 A?

Solution

The average power, supplied by the battery to the bulb, multiplied by the time equals the energy capacity of the battery. For

an ideal battery,

P

E

=

I

,

therefore

E

It

=

4

5

.

,

kJ

or t= = × =4 5 15 0 60 5 10 139

3

.(.)( .). kJ V A s h.=

Problem

6. If you accidentally leave your car headlights (current drain 5 A) on for an hour, how much of the 12-V battery’s

chemical energy is used up?

Solution

The power delivered by an emf is

E

I

,

so (if the voltage and current remain constant) the energy converted is

E

It

=

=

(

)(

)(

)

.

12

5

3600

216

V

A

s

kJ

Problem

7. A battery stores

50

W

h

⋅

of chemical energy. If it uses up this energy moving

3 0

10

4

.

×

C

through a circuit, what is its

voltage?

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CHAPTER 28 661

Solution

The emf is the energy (work done going through the source from the negative to the positive terminal) per unit charge:

E=⋅× =()( ) ( ) .50 3600 3 10 6

4

W h s/h C V= (This is the average emf; the actual emf may vary with time.)

Section 28-3: Simple Circuits: Series and Parallel Resistors

Problem

8. A

47-

k

Ω

resistor and a

39-

k

Ω

resistor are in parallel, and the pair is in series with a

22-

k

Ω

resistor. What is the

resistance of the combination?

Solution

From Equations 28-1 and 3c,

R

=

+

+

=

22

47

39

47

39

43

3

k

k

k

Ω

Ω

Ω

(

)

(

)

.

.

=

Problem

9. What resistance should be placed in parallel with a

56-

k

Ω

resistor to make an equivalent resistance of

45 k

Ω

?

Solution

The solution for R2 in Equation 28-3a is

R

R

R

R

R

2 1 1 56 45 56 45 229

=

−

=

−

=

parallel parallel k k .= =( ) ( )( ) ( )

Ω

Ω

Problem

10. In Fig. 28-49 all resistors have the same value, R. What will be the resistance measured (a) between A and B or

(b) between A and C?

FIGURE 28-49 Problems 10 and 11.

Solution

(a) The resistance between A and B is equivalent to two resistors of value R in series with the parallel combination of

resistors of values R and 2R. Thus,

R

R

R

R

R

R

R

R

AB

=

+

+

+

=

(

)

(

)

.

2

2

8

3

= = (b)

R

AC

is equivalent to just one resistor of

value R in series with the parallel combination of R and 2R (since the resistor at point B carries no current, i.e., its branch is

an open circuit). Thus

R

R

R

R

R

R

AC

=

+

=

(

)

.

2

3

5

3

= =

Problem

11. In Fig. 28-49, take all resistors to be

10.

Ω

. If a 6.0-V battery is connected between points A and B, what will be the

current in the vertical resistor?

Solution

The circuit in Fig. 28-49, with a battery connected across points A and B, is similar to the circuit analysed in Example 28-4.

In this case, R|| ()( ) ( ) ( ) ,= + =1 2 1 2 2

3

Ω Ω= and Rtot = + + =1 1 2

38

3

Ω Ω Ω Ω. The total current (that through the battery)

is IR

tot tot V A= = =E= =68

39

4

( ) ( ) .Ω The voltage across the parallel combination is IR

tot A V

|| ()( ),= =

9

42

33

2

Ω which is

the voltage across the vertical

1 Ω

resistor. The current through this resistor is then ( ) ( ) . .

3

2

1 15V A

=Ω=

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