Chpt. 32 – 5,18,25,27,36,42,51,57,61,76
32.5.
Model: Assume ideal connecting wires and an ideal battery for which
∆
V bat
=
E .
Visualize: Please refer to Figure EX32.5. We will choose a clockwise direction for I . Note that the choice of the current’s direction is arbitrary because, with two batteries, we may not be sure of the actual current direction.
The 3 V battery will be labeled 1 and the 6 V battery will be labeled 2.
Solve: (a) Kirchhoff’s loop law, going clockwise from the negative terminal of the 3-V battery is
∆
V closed loop
=
i
( ) i
= ∆
V bat 1
+ ∆
V
R
+ ∆
V bat 2
=
0
⇒ +
3 V – (18
Ω
) I
+
6 V
=
0
⇒
I
=
9 V
18
Ω
=
0.5 A
Thus, the current through the 18
Ω
resistor is 0.5 A. Because I is positive, the current is left to right (i.e., clockwise).
(b)
Assess: The graph shows a 3 V gain in battery 1, a
−
9 V loss in the resistor, and a gain of 6 V in battery 2. The final potential is the same as the initial potential, as required.
32.18.
Model: Assume ideal connecting wires but not an ideal battery.
Visualize: The circuit for an ideal battery is the same as the circuit in Figure EX32.18, except that the 1
Ω resistor is not present.
Solve: In the case of an ideal battery, we have a battery with E
Ω
and 20
Ω
resistance. Because the equivalent resistance is R eq
=
15 V connected to two series resistors of 10
=
10
Ω +
20
Ω =
30
Ω
and the potential difference across R eq
is 15 V, the current in the circuit is
I
=
∆
V
=
E
R eq
R eq
=
15 V
30
Ω
=
0.50 A
The potential difference across the 20
Ω
resistor is
∆
V
20
=
IR
=
(0.50 A)(20
Ω
)
=
10.0 V
In the case of a real battery, we have a battery with E
=
15 V connected to three series resistors: 10
Ω
, 20
Ω
, and an internal resistance of 1.0
Ω
. Now the equivalent resistance is
R
′ = Ω + Ω + Ω = Ω
The potential difference across R eq
is the same as before ( E
=
15 V). Thus,
I
′ =
∆
V
′
R
′ eq
=
E
R
′ eq
=
15 V
31
Ω
=
0.4839 A
Therefore, the potential difference across the 20
Ω
resistor is
1

∆
V
′
20
=
I R
= (
0.4839 A
)(
20
)
9.68 V
That is, the potential difference across the 20
Ω
resistor is reduced from 10.0 V to 9.68 V due to the internal resistance
1
Ω
of the battery. The percentage change in the potential difference is
10.0 V
−
9.68 V
10.0 V
100 3.2% of
32.25.
Model: The connecting wires are ideal with zero resistance.
Solve:
In the first step, the resistors 100
Ω
, 100
Ω
, and 100
Ω
in the top branch are in series. Their combined resistance is 300
Ω
. In the middle branch, the two resistors, each 100
Ω
, are in series. So, their equivalent resistance is 200
Ω
. In the second step, the three resistors are in parallel. Their equivalent resistance is
1
R eq
=
1
Ω
+
1
Ω
+
1
Ω
⇒
R eq
=
54.5
Ω
The equivalent resistance of the circuit is 54.5
Ω
.
32.27.
Model: Grounding does not affect a circuit’s behavior.
Visualize: Please refer to Figure EX32.27.
Solve: Because the earth has V earth
=
0 V, point d has a potential of zero. In going from point d to point a, the potential increases by 9 V. Thus, point a is at a potential of 9 V. Let us calculate the current I in the circuit before calculating the potentials at points b and c. Applying Kirchhoff’s loop rule, starting clockwise from point d,
i
( ) i
= ∆
V
9 V bat
+ ∆
V
2
Ω
+ ∆
V
6 V bat
+ ∆
V
1
Ω
=
0
⇒ +
9 V – I (2
Ω
) – 6 V – I (1
Ω
)
=
0
⇒
I
=
3 V
3
Ω
=
1 A
There is a drop in potential from point a to point b by an amount IR
=
(1 A)(2
Ω
)
=
2 V. Thus, the potential at point b is
9 V
−
2 V
=
7 V. The potential decreases from 7 V at point b to 7 V
−
6 V
=
1 V at point c. There is a further decrease in potential across the 1
Ω
resistor of IR
=
(1 A)(1
Ω
)
=
1 V. That is, the potential of 1 V at c becomes 0
V at point d, as it must. In summary, the potentials at a, b, c, and d are 9 V, 7 V, 1 V, and 0 V.
32.36.
Visualize: Please refer to Figure P32.36.
Solve: Bulbs D and E are in series, so the same current will go through both and make them equally bright (D
=
E). Bulbs B and C are in parallel, so they have the same potential difference across them. Because they are identical bulbs with equal resistances, they will have equal currents and be equally bright (B
=
C). Now the equivalent resistance of B
+
C in parallel is less than the resistance of E, so the total resistance along the path through A is less than the total resistance along path through D. The two paths have the same total potential difference—the emf of the battery—so more current will flow through the A path than through the D path.
Consequently, A will have more current than D and E and will be brighter than D and E (A > D
=
E). Bulbs B and C each have half the current of A, because the current splits at the junction, so A is also brighter than B and
C (A > B
=
C).
The remaining issue is how B and C compare to D and E. Suppose B and C were replaced by wires with zero resistance, leaving just bulb A in the middle path. Then the resistance of the path through A would be half of the
2

resistance of the path through D. This would mean that the current through A would be twice the current through D, so I
A
=
2 I
D
. When B and C are present, their resistance adds to the resistance of A to lower the current through the middle path. So in reality, I
A
< 2 I
D
. We already know that I
B
=
I
C
=
1
2
I
A
, so we can conclude that I
B
=
I
C
< I
D
. Since the current through B and C is less than the current through D and E, D and E are brighter than B and C. The final result of our analysis is A > D
=
E > B
=
C.
32.42.
Model: Use the laws of series and parallel resistances.
Visualize:
Solve: Despite the diagonal orientation of the 12
Ω
resistor, the 6
Ω
, 12
Ω
, and 4
Ω
resistors are in parallel because they have a common connection at both the top end and at the bottom end. Their equivalent resistance is
R eq
=
⎜
⎛
⎝
1
Ω
+
1
Ω
+
1
Ω
⎟
⎞
⎠
−
1
=
2
Ω
The trickiest issue is the 10
Ω
resistor. It is in parallel with a wire , which is the same thing as a resistor with R
=
0
Ω
. The equivalent resistance of 10
Ω
in parallel with 0
Ω
is
R eq
=
⎜
⎛
⎝
1
Ω
+
1
Ω
⎟
⎞
⎠
−
1
( ) −
1 =
1
∞
=
0
Ω
In other words, the wire is a short circuit around the 10
Ω
, so all the current goes through the wire rather than the resistor. The 10
Ω
resistor contributes nothing to the circuit. So the total circuit is equivalent to a 2
Ω
resistor in series with the 2
Ω
equivalent resistance in series with the final 3
Ω
resistor. The equivalent resistance of these three series resistors is
R ab
=
2
Ω +
2
Ω +
3
Ω =
7
Ω
32.51.
Model: Assume an ideal battery and ideal connecting wires.
Visualize:
3

Solve: (a) Grounding one point doesn’t affect the basic analysis of the circuit. In Figure P32.51, there is a single loop with a single current I flowing in the clockwise direction. Applying Kirchhoff’s loop law clockwise from the lower right corner gives
∑∆
V i
= −
8 I
+
12 V – 4 I – 12 I
=
0 V
⇒
I
=
12 V
24
Ω
=
0.50 A
Knowing the current, we can use
∆
V
=
IR to find the potential difference across each resistor:
∆
V
8
=
4 V
∆
V
4
=
2 V
∆
V
12
=
6 V
The purpose of grounding one point in the circuit is to establish that point as the specific potential V
Grounding point d makes that potential there V d
=
0 V.
=
0 V. Then we can use the known potential differences to find the potential at other points in the circuit. Point a is 4 V less than point d (because potential decreases in the direction of current flow), so point b, so V c
=
V
V a
=
V d
– 4 V
=
–4 V. Point b is 12 V more than point a because of the battery. So V b
= b
– 2 V
=
6 V. Point d is 6 V less than point c, so V
V a d
=
+
12 V
V c
=
8 V. Point c is 2 V less than
– 6 V
=
0 V. This is a consistency check—making one complete loop brings us back to the potential at which we started, namely 0 V.
(b) The information about the potentials is shown in the graph above.
(c) Moving the ground to point a doesn’t change the basic analysis of part (a) or the potential differences found there. All that changes is that now V a
+
12 V
V c
– 6 V
=
4 V. Point a is 4 V less
=
0 V. Point b is 12 V more than point a because of the battery. So, V b
=
=
12 V. Point c is 2 V less than point b, so V
than point d, so V c
=
V a
=
V b
– 2 V
=
10 V. Point d is 6 V less than point c, so V d
– 4 V d
V
= a
=
0 V. This brings us back to where we started.
The information about the potentials is shown in the graph above.
32.57.
Model: The voltage source/battery and the connecting wires are ideal.
Visualize: Please refer to Figure P32.57.
Solve: Let us first apply Kirchhoff’s loop law starting clockwise from the lower left corner:
+
V in
– IR – I (100
Ω
)
=
0 V
⇒
I
=
R
V in
+
100
Ω
The output voltage is
V out
= ( ) ( )
⎝ R
V in
+
100
Ω ⎠
⇒
V out
=
V in
R
100
Ω
+
100
Ω
For V out
=
V in
10 , the above equation can be simplified to obtain R :
V in
V in
10
=
100
Ω
R
+
100
Ω
⇒
R
+
100
Ω =
1000
Ω ⇒
R
=
900
Ω
32.61.
Model: The battery and the connecting wires are ideal.
Visualize:
The figure shows how to simplify the circuit in Figure P32.61 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,
I
=
E
Ω
=
12 V
Ω
=
6 6
2 A
Thus, the current through the battery is 2 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 .
4

In Step 1, the 6
Ω
resistor is returned to a 3
Ω
and 3
Ω
resistor in series. Both resistors must have the same 2 A current as the 6
Ω
resistance. We then use Ohm’s law to find
∆
V
3
=
(2 A)(3
Ω
)
=
6 V
As a check, 6 V
+
6 V
=
12 V, which was
∆
V of the 6
Ω
resistor. In Step 2, one of the two 3
Ω
resistances is returned to the 4
Ω
, 48
Ω
, and 16
Ω
resistors in parallel. The three resistors must have the same
∆
V
=
6 V. From
Ohm’s law,
I
4
=
6 V
4
Ω
=
1.5 A I
48
=
6 V
=
1
A I
16
=
6 V
=
3
A
Resistor
3
Ω
4
Ω
48
Ω
16
Ω
Potential difference (V)
6
6
6
6
Current (A)
2
1.5
1.2
3.8
32.76.
Model: The connecting wires are ideal. The capacitors discharge through the resistors.
Visualize:
The figure shows how to simplify the circuit in Figure P32.76 using the laws of series and parallel resistors and the laws of series and parallel capacitors.
Solve: The
Ω
and 20
Ω
resistors are in parallel and are equivalent to a 12
Ω
resistor. This 12
Ω
resistor is in series with the 8
Ω
resistor so the equivalent resistance of the circuit R eq
=
20
Ω
. The two 60 F capacitors are in series producing an equivalent capacitance of 30
µ
µ
F. This 30
µ
F capacitor is in parallel with the 20
µ
F capacitor so the equivalent capacitance C eq of the circuit is 50
µ
F. The time constant of this circuit is
τ =
R eq
C eq
=
(20
Ω
)(50
µ
F)
=
1.0 ms
The current due to the three capacitors through the 20
Ω
equivalent resistor is the same as through the 8
Ω resistor. So, the voltage across the 8
Ω
resistor follows the decay equation V
=
V
0 e
− t/
τ
. For V
=
V
0
/2, we get
V
0
/2
=
V
0 e
− t/ 1.0 ms
⇒ ln t
1.0 ms
⇒ t
=
0.69 ms
5