Homework 6 Solutions
PHYS 212
Dr. Amir
Chapter 25:
19. (II) A 100-W lightbulb has a resistance of about 12 Ω when cold (20°C) and 140 Ω when on
(hot). Estimate the temperature of the filament when hot assuming an average temperature
coefficient of resistivity α= 0.0042(C°)-1 .
19.
Use Eq. 25-5 multiplied by l A so that it expresses resistances instead of resistivity.
R  R0 1   T  T0  
T  T0 
1 R

1
 140  
 1  20C 
 1  2390C  2400C

1 
  R0 
0.0045  C   12 

25. (II) A length of wire is cut in half and the two lengths are wrapped together side by side to
make a thicker wire. How does the resistance of this new combination compare to the
resistance of the original wire?
25. The resistance depends on the length and area as R   l A. Cutting the wire and running the wires
side by side will halve the length and double the area.
  12 l  1  l
R2 
4
 14 R1
2A
A
31. (I) What is the maximum power consumption of a 3.0-V portable CD player that draws a
maximum of 270 mA of current?
31.
Use Eq. 25-6 to find the power from the voltage and the current.
P  IV   0.27 A  3.0 V   0.81W
Chapter 26
18. (II) (a) Determine the equivalent resistance of the “ladder” of equal 125-V resistors shown in
Fig. 26–40. In other words, what resistance would an ohmmeter read if connected between
points A and B? (b) What is the current through each of the three resistors on the left if a
50.0-V battery is connected between points A and B?
Homework 6 Solutions
PHYS 212
Dr. Amir
18.
(a)
The three resistors on the far right are in series, so their
equivalent
resistance is 3R. That combination is in parallel with the next
resistor to the left, as shown in the dashed box in the second figure.
The equivalent resistance of the dashed box is found as follows.
1
1 1   3R
Req1   

4
 R 3R 
This equivalent resistance of 43 R is in series with the next two
resistors, as shown in the dashed box in the third figure (on the next page). The equivalent
resistance of that dashed box is Req2  2 R  43 R  114 R. This 114 R is in parallel with the next
resistor to the left, as shown in the fourth figure. The equivalent
resistance of that dashed box is found as follows.
1
 1  4   11 R.

15
 R 11R 
Req2  
This is in series with the last two resistors, the ones connected
directly to A and B. The final equivalent resistance is given below.
11
Req  2 R  15
R  1541 R  1541 125    341.67   342 
(b) The current flowing from the battery is found from Ohm’s law.
V
50.0 V
I total 

 0.1463A  0.146 A
Req 341.67 
This is the current in the top and bottom resistors. There will be less current in the next resistor
because the current splits, with some current passing through the resistor in question, and the
rest of the current passing through the equivalent resistance of 114 R , as shown in the last figure.
The voltage across R and across 114 R must be the same, since they are in parallel. Use this to
find the desired current.
VR  V R  I R R  I R  114 R    I total  I R   114 R  
11
4
11
4
11
11
I R  15
I total  15
 0.1463A  I total  0.107 A
32. (II) Calculate the currents in each resistor of Fig. 26–50.
32. There are three currents involved, and so there must be three
independent equations to determine those three currents. One
comes from Kirchhoff’s junction rule applied to the junction of
the three branches at the top center of the circuit.
I1  I 2  I 3
I1
25
58V
120
3.0 V
I2
64
82
I3
110
Homework 6 Solutions
PHYS 212
Dr. Amir
Another equation comes from Kirchhoff’s loop rule applied to the left loop, starting at the negative
terminal of the battery and progressing counterclockwise.
58V  I1 120   I1 82   I 2  64   0  58  202 I1  64 I 2
The final equation comes from Kirchhoff’s loop rule applied to the right loop, starting at the negative
terminal of the battery and progressing counterclockwise.
3.0 V  I 3  25   I 2  64   I 3 110   0  3  64 I 2  135I 3
Substitute I1  I 2  I 3 into the left loop equation, so that there are two equations with two unknowns.
58  202  I 2  I 3   64 I 2  266 I 2  202 I 3
Solve the right loop equation for I 2 and substitute into the left loop equation, resulting in an
equation with only one unknown, which can be solved.
135I 3  3
 135I 3  3   202 I 
3  64 I 2  135 I 3  I 2 
; 58  266 I 2  202 I 3  266 

3
64
 64 
I 3  0.09235A ; I 2 
135 I 3  3
 0.1479 A ; I1  I 2  I 3  0.24025A
64
The current in each resistor is as follows:
120  : 0.24 A
82  : 0.24 A
64  : 0.15A
25  : 0.092 A
110  : 0.092 A
50. (III) Determine the time constant for charging the capacitor in the circuit of Fig. 26–61. [Hint:
Use Kirchhoff’s rules.] (b) What is the maximum charge on the capacitor?
50.
(a)
With the currents and junctions labeled as in the
diagram, we use point a for the junction rule and the right and left loops for
the loop rule. We set current I3 equal to the derivative of the charge on the
capacitor and combine the equations to obtain a single differential equation
in terms of the capacitor charge. Solving this equation yields the charging
time constant.
Q
I1  I 2  I 3 [1] ; e  I1R1  I 2 R2  0 [2] ;   I 2 R2  0 [3]
C
We use Eq. [1] to eliminate I1 in Eq. [2]. Then we use Eq. [3] to eliminate I2 from Eq.
[2].
 Q 
0  e   I 2  I 3  R1  I 2 R2 ; 0  e  I 2  R1  R2   I 3R1 ; 0  e  
  R1  R2   I 3R1
 R2C 
We set I3 as the derivative of the charge on the capacitor and solve the differential equation by
separation of variables.
Homework 6 Solutions
PHYS 212
 Q 
dQ
0= e  
R1 
  R1  R2  
dt
 R2C 


R Ce 
1  e
Q 2
R1  R2 

 R1  R2 
R1R2C
t
t
t R  R 
dQ 
1
2

dt  
0
R1R2C
 R2C e 
Q  

 R1  R2 

 R2C e  
Q  

R1  R2  
 R  R2  t 


 ln
 1
  R2C e  
R1R2C
 
 
  R1  R2  

 R Ce 
 R1  R2  t 
ln Q    2
  
R1R2C
 R1  R2   0

0
Q
Q
0




From the exponential term we obtain the time constant,  
(b)
Dr. Amir
R1R2C
.
R1  R2
We obtain the maximum charge on the capacitor by taking the limit as time goes to
infinity.

R Ce 
1  e
Qmax  lim 2
t  R1  R2 

 R1  R2 
R1 R2C
t

R Ce
 2
 R1  R2

51. (III) Two resistors and two uncharged capacitors are arranged as shown in Fig. 26–62. Then a
potential difference of 24 V is applied across the combination as shown. (a) What is the
potential at point a with switch S open? (Let V  0 at the negative terminal of the source.) (b)
What is the potential at point b with the switch open? (c) When the switch is closed, what is
the final potential of point b? (d) How much charge flows through the switch S after it is
closed?
51. (a) With the switch open, the resistors are in series with each other, and so have the same current.
Apply the loop rule clockwise around the left loop, starting at the negative terminal of the
source, to find the current.
V
24 V
V  IR1  IR2  0  I 

 1.818 A
R1  R2 8.8   4.4 
Homework 6 Solutions
PHYS 212
Dr. Amir
The voltage at point a is the voltage across the 4.4 -resistor.
Va  IR2  1.818 A  4.4    8.0 V
(b) With the switch open, the capacitors are in series with each other. Find the equivalent
capacitance. The charge stored on the equivalent capacitance is the same value as the charge
stored on each capacitor in series.
1
1
1
CC
 0.48 F  0.36 F 


 Ceq  1 2 
 0.2057  F
Ceq C1 C2
C1  C2  0.48 F  0.36 F 
Qeq  VCeq   24.0 V  0.2057  F   4.937 C  Q1  Q2
The voltage at point b is the voltage across the 0.24 F -capacitor.
Q
4.937 C
Vb  2 
 13.7 V  14 V
C2
0.36 F
(c) The switch is now closed. After equilibrium has been reached a long time, there is no current
flowing in the capacitors, and so the resistors are again in series, and the voltage of point a must
be 8.0 V. Point b is connected by a conductor to point a, and so point b must be at the same
potential as point a, 8.0 V . This also means that the voltage across C2 is 8.0 V, and the
voltage across C1 is 16 V.
(d) Find the charge on each of the capacitors, which are no longer in series.
Q1  V1C1  16 V  0.48 F   7.68C
Q2  V2C2   8.0 V  0.36 F   2.88C
When the switch was open, point b had a net charge of 0, because the charge on the negative
plate of C1 had the same magnitude as the charge on the positive plate of C2 . With the switch
closed, these charges are not equal. The net charge at point b is the sum of the charge on the
negative plate of C1 and the charge on the positive plate of C2 .
Qb  Q1  Q2  7.68C  2.88C  4.80C  4.8C
Thus 4.8C of charge has passed through the switch, from right to left.