Chapter 23 Textbook Problems
Answers to odd problems are found in the back of the textbook
RESISTORS
Section 3
Series and Parallel Circuits
10, 11, 12, 13, 15
Section 5
More Complex Circuits
17, 18, 19, 20, 21, 23, 24
General Problems
56, 61
P23.10. Prepare: Please refer to Figure P23.10. The three resistances in both (a) and (b) are series resistors. We will thus use
Equation 23.8 to find the equivalent resistance.
Solve: (a) The equivalent resistance is Req  2   3   6   11 .
(b) The equivalent resistance is Req  3   3   3   9 .
Assess: We must learn how to combine series and parallel resistors.
P23.12.
Prepare: The copper wire and the iron wire are connected in series. The composite resistance is simply the equivalent
8
8
resistance of RCu and RFe. The resistivities of copper and iron from Table 22.1 are 1.7 10  m and 9.7 10  m, respectively.
Solve:
R  RCu  RFe 
 CuLCu
ACu

FeLFe
AFe
Using the data for lengths and areas of cross section,
R
(1.7 108  m)(0.20 m) (9.7 108  m)(0.60 m)
3
3

 4.3310   74.1 10   78 m
(0.5 10 3 m)2
(0.5 103 m) 2
Assess: As obtained above, we didn’t expect a large resistance for the two highly conducting wires of copper
and iron.
P23.18.
Prepare: Assume ideal connecting wires and an ideal battery. The three resistors are in series and we can use Equation
23.8 to find the equivalent resistance.
Solve: As shown in Figure P23.18, a potential difference of 5.0 V causes a current of 100 mA through the three series resistors.
The situation is the same if we replace the three resistors with an equivalent resistor Req. That is, a potential difference of 5.0 V
across Req causes a current of 100 mA through it. From Ohm’s law,
Req 
VR
5.0 V
 R 15  10  
 R  25   50   R  25 
I
100 mA
Assess: Learn how to combine resistors and apply Ohm’s law.
Prepare: The connecting wires are ideal with zero resistance. For the first step, the 10  and 30  resistors are in
series; for the second step, the 60  and 40  resistors are in parallel; and for the third step, the
24  and 10  resistors are in series.
Solve:
P23.20.
For the first step, the 10  and 30  resistors are in series and the equivalent resistance is 40 . For the second step, the 60  and
40  resistors are in parallel and the equivalent resistance is
1
 1
1 


  24 
 40  60  
For the third step, the 24  and 10  resistors are in series and the equivalent resistance is 34 .
Assess: Have a good understanding of how series and parallel resistances are combined to get an equivalent
resistance.
P23.24.
Prepare: The battery and the connecting wires are ideal. The figure shows how to simplify the circuit in Figure P23.24
using the laws of series and parallel resistances. Having reduced the circuit to a single equivalent resistance, we will reverse the
procedure and “build up” the circuit using the loop law and the junction law to find the current and potential difference of each
resistor.
Solve:
From the last circuit in the diagram,
E
12 V
I

 2.0 A
6.0

6.0


Thus, the current through the battery is 2.0 A. As we rebuild the circuit, we note that series resistors must have the same current I
and that parallel resistors must have the same potential difference V.
In Step 1, the 6.0  resistor is returned to a 3.0  and 3.0  resistor in series. Both resistors must have the same
2.0 A current as the 6.0  resistance. We then use Ohm’s law to find
V3  (2.0 A)(3.0 )  6.0 V
As a check, 6.0 V  6.0 V  12 V, which was V of the 6.0  resistor. In Step 2, one of the two 3.0  resistances is returned to the
4.0 , 48 , and 16  resistors in parallel. The three resistors must have the same V  6.0 V. From Ohm’s law,
I4 
6V
 1.5 A
4
Resistor
I48 
6V
 0.125 A
48 
Potential difference (V)
I16 
6V
 0.375 A
16 
Current (A)
3
6
2
4
6
1.5
48 
6
0.13
16 
6
0.38
Assess: The larger the resistance, the smaller the current it carries. This is what we would have expected.
P23.56.
Prepare: Please refer to Figure P23.56. Assume the batteries and the connecting wires are ideal.
Solve: (a) The two batteries in this circuit are oriented to “oppose” each other. The direction of the current is counterclockwise
because the 12 V battery “wins.”
(b) There are no junctions, so the same current I flows through all circuit elements. Applying Kirchhoff’s loop law in the
counterclockwise direction and starting at the lower right corner, Vi  12 V  I(12 )  I(6.0 )  6.0 V – IR  0.
Note that the IR terms are all negative because we’re applying the loop law in the direction of current flow, and the potential
decreases as current flows through a resistor. We can easily solve to find the unknown resistance R:
6.0 V  I(18 ) – IR  0  R 
6.0 V  (18 )I 6.0 V  (18 )(0.25 A)

 6.0 
I
0.25 A
(c) The power is P  I2R  (0.25 A)2(6 )  0.38 W.
(d)
The potential difference across a resistor is V  IR, giving V6  1.5 V, and V12  3.0 V. Starting from the lower
left corner, the graph goes around the circuit clockwise, opposite from the direction in which we applied the
loop law. In this direction, we speak of potential as lost in the batteries and gained in the resistors.