Chapter 31
Physics 230
31.43. Visualize: Please refer to Figure P31.43.
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
resistance of the path through D. This would mean that the current through A would be twice the current through D,
so IA  2ID. 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, IA < 2ID. We already know that I B  IC  21 I A , so we can conclude that IB  IC < ID. 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.
31.49. Model: Use the laws of series and parallel resistances.
Visualize:
Solve: Despite the diagonal orientation of the 12  resistor, the 6 , 12 , and 4  resistors are in parallel
because they have a common connection at both the top end and at the bottom end. Their equivalent resistance is
1
1
1 
 1
Req  


  2
 6  12  4  
The trickiest issue is the 10  resistor. It is in parallel with a wire, which is the same thing as a resistor with R 
0  . The equivalent resistance of 10  in parallel with 0  is
1
1
1 
1
 1
Req  

      0 
 10  0  
In other words, the wire is a short circuit around the 10 , so all the current goes through the wire rather than the
resistor. The 10  resistor contributes nothing to the circuit. So the total circuit is equivalent to a 2  resistor in
series with the 2  equivalent resistance in series with the final 3  resistor. The equivalent resistance of these
three series resistors is
Rab  2   2   3   7 
31.57. Model: The batteries are ideal, the connecting wires are ideal, and the ammeter has a negligibly small
resistance.
Visualize: Please refer to Figure P31.57.
Solve: Kirchhoff’s junction law tells us that the current flowing through the 2.0  resistance in the middle
branch is I1  I2  3.0 A. We can therefore determine I1 by applying Kirchhoff’s loop law to the left loop. Starting
clockwise from the lower left corner,
9.0 V – I1(3.0 ) – (3.0 A)(2.0 )  0 V  I1  1.0 A  I2  (3.0 A – I1)  (3.0 A – 1.0 A)  2.0 A
Finally, to determine the emf , we apply Kirchhoff’s loop law to the right loop and start counterclockwise from
the lower right corner of the loop:
  I2(4.5 )  (3.0 A)(2.0 )  0 V   – (2.0 A)(4.5 ) – 6.0 V  0 V    15.0 V
31.65. Model: Assume ideal connecting wires.
Visualize: Please refer to Figure P31.65. Because the ammeter we have shows a full-scale deflection with a
current of 500 A  0.500 mA, we must not allow a current more than 0.500 mA to pass through the ammeter.
Since we wish to measure a maximum current of 50 mA, we must split the current in such a way that 0.500 mA
flows through the ammeter and 49.500 mA flows through the resistor R.
Solve: (a) The potential difference across the ammeter and the resistor is the same. Thus,
VR  Vammeter  (49.500  103 A)R  (0.500  103 A)(50 )  R  0.505 
(b) Effective resistance is
1
1
1


 R  0.50 
Req 0.505  50 
31.69. Model: The battery and the connecting wires are ideal.
Visualize:
The figure shows how to simplify the circuit in Figure P31.69 using the laws of series and parallel resistances.
We will reverse the procedure and “build up” the circuit using the loop law and junction law to find the current
and potential difference of each resistor.
Solve: Having found Req  12 , the current from the battery is I  (24 V)/(12 )  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 of the above figure, the 12  resistor is returned to 4  and 8  resistors in series. Both resistors must
have the same 2.0 A as the 12  resistor. We use Ohm’s law to find V4  8 V and V8  16 V. As a check, 8 V
 16 V  24 V, which was V of the 12  resistor. In Step 2, the 8  resistor is returned to the 12  and 24 
resistors in parallel. Both resistors must have the same V  16 V as the 8  resistor. From Ohm’s law,
I12  16 V 12   43 A and I24  23 A . As a check, I12  I24  2.0 A, which was the current I of the 8 
resistor. In Step 3, the 12  resistor is returned to the two 6  resistors in series. Both resistors must have the
same 43 A as the 12  resistor. We use Ohm’s law to find V6  8 V and V6  8 V. As a check, 8 V  8 V  16
V, which was V of the 12  resistor. Finally, in Step 4, the 6  resistor is returned to the 8  and 24  resistors
in parallel. Both resistors must have the same V  8 V as the 6  resistor. From Ohm’s law, I8  (8 V)/(8 )  1
A and I24  13 A . As a check, I8  I24  43 A , which was the current I of the 6  resistor.
Resistor
4
6
8
Bottom 24 
Right 24 
Potential difference (V)
8
8
8
8
16
Current (A)
2
4
3
1
1
3
2
3
31.73. Model: The capacitor discharges through the resistor, and the wires are ideal.
Solve: In a RC circuit, the charge at a given time is related to the original charge as Q  Q0et/. For a capacitor
Q  CV, so V  V0et/. From the Figure P 71.73, we note that V0  30 V and V  10 V at t  2 ms. So,
10 V   30 V  e

2 ms R 50106 F
2  103 s
  ln  10 V    2  103 s  R  
 36.4 
 30 V 
R  50  106 F 


 50  106 F  ln  13 
31.77. Model: Assume the battery and the connecting wires are ideal.
Visualize: Please refer to Figure P31.77.
Solve: (a) If the switch has been closed for a long time, the capacitor is fully charged and there is no current
flowing through the right branch that contains the capacitor. So, a voltage of 60 V appears across the 60 
resistor and a voltage of 40 V appears across the 40  resistor. That is, maximum voltage across the capacitor is
40 V. Thus, the charge on the capacitor is
Q0  C  (40 V)(2.0  106 F)  80 C
(b) Once the switch is opened, the battery is disconnected from the capacitor. The capacitor C has two
resistances (10  and 40 ) in series and discharges according to Q  Q0et/RC. For Q  0.10 Q0,
t
 0.10 
0.10 Q0  Q0et/(50 )(2 F)  ln 
   50  2  F

 
 1 
 t    50  2 F  ln  0.10   0.23 ms