PHYSICS 172
WQ 2010
Homework #6
1. Giancoli Chapter 26, Problem 4
See Figure 26-2 for a circuit diagram for this problem. Use Eq. 26-1.
e  Vab 12.0 V  8.4 V
Vab  e  Ir  r 

 0.038 
I
95A
Vab  IR  R 
Vab
I

8.4 V
95A
 0.088 
2. Giancoli Chapter 26, Problem 8
The equivalent resistance of five 100- resistors in parallel is found, and then that resistance
is divided by 10 to find the number of 10- resistors needed.
1
1
1 1 1 1 1
20 
 5 
Req   
 
  
 20   n 10    n 
 2

10 
 100  
 R1 R2 R3 R4 R5 
3. Giancoli Chapter 26, Problem 14
The power delivered to the starter is equal to the square of the current in the circuit
multiplied by the resistance of the starter. Since the resistors in each circuit are in series we
calculate the currents as the battery emf divided by the sum of the resistances.
2
2
P I 2 RS  I   e Req   R0eq   r  RS 

  
 
 

P0 I 02 RS  I 0   e R0eq   Req   r  RS  RC 
2
2
2


0.02   0.15 

  0.40
 0.02   0.15   0.10  
4. Giancoli Chapter 26, Problem 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

4
 R 3R 
Req1  
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
4 
1
11
Req2   
  15 R.
 R 11R 
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
5. Giancoli Chapter 26, Problem 28
Apply Kirchhoff’s loop rule to the circuit starting at the upper left corner of the circuit
diagram, in order to calculate the current. Assume that the current is flowing clockwise.
6V
 I  2.0    18 V  I  6.6    12 V  I 1.0    0  I 
 0.625A
9.6 
The terminal voltage for each battery is found by summing the potential differences across
the internal resistance and EMF from left to right. Note that for the 12 V battery, there is a
voltage gain going across the internal resistance from left to right.
18 V battery: Vterminal   I  2.0    18 V    0.625 A  2.0    18 V  16.75V  17 V
12 V battery: Vterminal  I 1.0    12 V   0.625 A 1.0    12 V  12.625 V  13V
6. Giancoli Chapter 26, Problem 30
(a)
We label each of the currents as shown in the accompanying
figure. Using Kirchhoff’s junction rule and the first three junctions (a-c) we write
equations relating the entering and exiting currents.
I  I1  I 2
[1]
I 2  I3  I 4
[2]
I1  I 4  I 5
[3]
We use Kirchhoff’s loop rule to write equations for loops
abca, abcda, and bdcb.
0   I 2 R  I 4 R  I1 R
[4]
0   I 2 R  I 3R  e
[5]
0   I 3R  I5R  I 4 R
[6]
We have six unknown currents and six equations. We solve these equations by
substitution. First, insert Eq. [3] into [6] to eliminate current I5. Next insert Eq. [2] into
Eqs. [1], [4], and [5] to eliminate I2.
0   I 3 R   I1  I 4  R  I 4 R  0   I 3 R  I1 R  2 I 4 R
[6*]
I  I1  I 3  I 4
[1*]
0    I 3  I 4  R  I 4 R  I1 R  0   I 3 R  2 I 4 R  I 1R
[4*]
0    I 3  I 4  R  I 3R  e  0   I 4 R  2 I 3R  e
[5*]
Next we solve Eq. [4*] for I4 and insert the result into Eqs. [1*], [5*], and [6*].
0   I 3 R  2 I 4 R  I1R  I 4  12 I1  12 I 3
I  I1  I 3  12 I1 - 12 I 3  I  23 I1  12 I 3
0   I 3 R  I1R  2  12 I1 - 12 I 3  R  2 I 3 R  2 I1R  I1  I 3
[1**]
[6**]
0    12 I1  12 I 3  R  2 I 3 R  e  0   12 I1R  23 I 3 R  e
[5**]
Finally we substitute Eq. [6**] into Eq [5**] and solve for I1. We insert this result into
Eq. [1**] to write an equation for the current through the battery in terms of the battery
emf and resistance.
e
e
0   12 I1R  23 I1R  e  I1 
; I  23 I1  12 I1  2 I1  I 
2R
R
(b)
We divide the battery emf by the current to determine the effective resistance.
e
e
Req  
 R
I e R
7. Giancoli Chapter 26, Problem 38
The circuit diagram has been labeled with six different currents. We apply the junction rule to
junctions a, b, and c. We apply the loop rule to the three loops labeled in the diagram.
1 I  I1  I 2 ; 2  I1  I 3  I 5 ; 3 I 3  I 4  I
4   I1R1  I 5 R5  I 2 R2  0 ; 5  I 3 R3  I 4 R4  I 5 R5  0
6  e  I 2 R2  I 4 R4  0
Eliminate I using equations 1) and 3).
1 I 3  I 4  I1  I 2 ; 2  I1  I 3  I 5
b
R1
a
R3
I1
I5
1
4   I1R1  I 5 R5  I 2 R2  0 ; 5  I 3 R3  I 4 R4  I 5 R5  0
6  e  I 2 R2  I 4 R4  0
R5
R4
Eliminate I 1 using equation 2.
1 I 3  I 4  I 3  I 5  I 2  I 4  I 5  I 2
4    I 3  I 5  R1  I 5 R5  I 2 R2  0   I 3 R1  I 5  R1  R5   I 2 R2  0
5  I 3 R3  I 4 R4  I 5 R5  0
6  e  I 2 R2  I 4 R4  0
Eliminate I 4 using equation 1.
d
3
I
2
I4
I2
R2
I3
å
+
–
c
4   I 3 R1  I 5  R1  R5   I 2 R2  0
5  I 3 R3   I 5  I 2  R4  I 5 R5  0   I 3 R3  I 5  R4  R5   I 2 R4  0
6  e  I 2 R2   I 5  I 2  R4  0  e  I 2  R2  R4   I 5 R4  0
Eliminate I 2 using equation 4: I 2 
5  I 3 R3  I 5  R4  R5  
1
R2
1
R2
I R
3
1
I R
3
1
 I 5  R1  R5  .
 I 5  R1  R5   R4  0 
I 3  R1 R4  R2 R3   I 5  R2 R4  R2 R5  R1 R4  R5 R4   0
6 e 
1
R2
I R
3
1
 I 5  R1  R5    R2  R4   I 5 R4  0 
e R2  I 3 R1  R2  R4   I 5  R1R2  R1R4  R5 R2  R5 R4  R2 R4   0
Eliminate I 3 using equation 5: I 3   I 5
 R2 R4  R2 R5  R1R4  R5 R4 
 R1R4  R2 R3 
 R2 R4  R2 R5  R1R4  R5 R4  
 R1  R2  R4   I 5  R1R2  R1R4  R5 R2  R5 R4  R2 R4   0
R
R

R
R




1 4
2 3

I    R R  R2 R5  R1R4  R5 R4  
e   5  2 4
R1  R2  R4    R1R2  R1R4  R5 R2  R5 R4  R2 R4  

R2  
 R1R4  R2 R3 



e R2   I 5
   25  14     25  15     22  14    15  14   

 22   25   14   
I5 

 22  14     25  12  




25  

   22   25     22 14   15   25    15  14     25  14  

  I 5  5261    I 5  
6.0 V
5261 
 1.140 mA  upwards 
 R2 R4  R2 R5  R1R4  R5 R4 
 R1R4  R2 R3 
 25  14     25  15     22  14    15  14  
   1.140 mA 
 0.1771A
 22  14     25  12  
I3   I5
I2 
1
R2
I R
3
1
 I 5  R1  R5   
1
25 
 0.1771A  22     0.00114 A  37    0.1542 A
I 4  I 5  I 2  0.00114 A  0.1542 A  0.1531A
I1  I 3  I 5  0.1771A  0.00114 A  0.1760 A
We keep an extra significant figure to show the slight difference in the currents.
I 22   0.176A
I 25   0.154 A
I12   0.177 A
I14   0.153A
I15  0.001A, upwards
8. Giancoli Chapter 26, Problem 44
(a) From Eq. 26-7 the product RC is equal to the time constant.
 24.0  106 s
  RC  C  
 1.60  109 F
3
R 15.0  10 
(b) Since the battery has an EMF of 24.0 V, if the voltage across the resistor is 16.0 V, the
voltage across the capacitor will be 8.0 V as it charges. Use the expression for the
voltage across a charging capacitor.
t
 V 
 V 
VC  e 1  e  t /    e  t /    1  C     ln  1  C  
e 

e 




t   ln  1 
VC 
8.0 V 

6
6
   24.0  10 s ln  1 
  9.73  10 s
e 
24.0
V




9. Giancoli Chapter 26, Problem 47
The capacitance is given by Eq. 24-8 and the resistance by Eq. 25-3. The capacitor plate
separation d is the same as the resistor length l. Calculate the time constant.
  d   K A    K  1.0  1012  m 5.0 8.85  1012 C 2 N m 2  44 s
  RC  

 

 0 
0
d
 A 
10. Giancoli Chapter 26, Problem 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 
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 C 2 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 C 2 .
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.