Homework 5
Ch19: Q 1; P 1, 3, 17, 23, 25, 37, 45, 51, 59
Questions:
1. Explain why birds can sit on power lines safely, whereas leaning a metal ladder up against a
power line to fetch a stuck kite is extremely dangerous.
Solution
The birds are safe because they are not grounded. Both of their legs are essentially at the same
voltage (the only difference being due to the small resistance of the wire between their feet), and
so there is no current flow through their bodies since the potential difference across their legs is
very small. If you lean a metal ladder against the power line, you are making essentially a short
circuit from the high potential wire to the low potential ground. A large current will flow at least
momentarily, and that large current will be very dangerous to anybody touching the ladder.
Problems:
1.
(I) Calculate the terminal voltage for a battery with an internal resistance of 0.900  and an
emf of 8.50 V when the battery is connected in series with (a) an 81.0- resistor, and (b) an
810- resistor.
a
I
R
ε
r
b
Solution
Using the same analysis as in Example 19-1, the current in the circuit is I 
(a)
Vab  IR 
R
Rr
Vab  E
R
Rr
Rr
.
, or
R
81.0 
 E  E R  r E r
E
  8.50 V 
 8.41V
r 
Rr
Rr
81.0  0.900 
 Rr 
Vab  E  Ir  E  
(b)
E
  8.50 V 
810 
 810  0.900  
 8.49 V
3.
(II) What is the internal resistance of a 12.0-V car battery whose terminal voltage drops
to 8.4 V when the starter draws 75 A? What is the resistance of the starter?
Solution
See Figure from the problem (2) for a circuit diagram for this problem. Use Eq. 19-1.
E  Vab 12.0 V  8.4 V
Vab  E  Ir  r 

 0.048 
I
Vab  IR  R 
17.
Vab
I

8.4 V
75 A
75 A
 0.11 
(II) Determine (a) the equivalent resistance of the circuit shown in Fig. 19–39, and
(b) the voltage across each resistor.
Solution
(a)
The equivalent resistance is found by combining the 820  and 680  resistors in
parallel, and then adding the 470  resistor in series with that parallel combination.
1
 1
1 
Req  

  470   372   470   842   840 
 820  680  
(b)
The current delivered by the battery is I 
V
Req

12.0 V
842 
 1.425  102 A .
This is the current in the 470  resistor. The voltage across that resistor can be found by Ohm’s
law.


V470  IR  1.425  102 A  470    6.7 V
Thus the voltage across the parallel combination must be
12.0 V  6.7 V  5.3 V .
This is the voltage across both the 820  and 680  resistors, since parallel resistors have the
same voltage across them. Note that this voltage value could also be found as follows.


Vparallel  IRparallel  1.425  102 A  372    5.3 V
23. (I) Calculate the current in the circuit of Fig. 19–43 and show that the sum of all the voltage
changes around the circuit is zero.
Solution
All of the resistors are in series, so the
equivalent resistance is just the sum of the
resistors. Use Ohm’s law then to find the
current, and show all voltage changes
starting at the negative pole of the battery
and going counterclockwise.
I
E

9.0 V
 0.409 A  0.41A
 8.0  12.0  2.0  
 voltages  9.0 V  8.0   0.409 A   12.0   0.409 A    2.0   0.409 A 
Req
 9.0 V  3.27 V  4.91V  0.82 V  0.00 V
25. (II) (a) What is the potential difference between points a and d in Fig. 19–45 (same circuit
as Fig. 19–13, Example 19–8), and (b) what is the terminal voltage of each battery?
Solution
From Example 19-8, we have
I1  0.87 A , I 2  2.6 A , I 3  1.7 A .
If another significant figure had been kept,
the values would be
I1  0.858 A , I 2  2.58 A , I 3  1.73A .
We use those results.
(a)
To find the potential difference between points a and d, start at point a and add each
individual potential difference until reaching point d. The simplest way to do this is along the
top branch.
Vad  Vd  Va   I1  30      0.858 A  30    25.7 V
Slight differences will be obtained in the final answer depending on the branch used, due to
rounding. For example, using the bottom branch, we get the following.
Vad  Vd  Va  E1  I 2  21   80 V   2.58 A  21    25.8 V
(b) For the 80-V battery, the terminal voltage is the potential difference from point g to point e.
For the 45-V battery, the terminal voltage is the potential difference from point d to point b.
80 V battery: Vterminal  E1  I 2 r  80 V   2.58 A 1.0    77.4 V
45 V battery: Vterminal  E2  I 3 r  45 V  1.73 A 1.0    43.3 V
37. (I) A 3.00-F and a 4.00-F capacitor are connected in series, and this combination is
connected in parallel with a 2.00-F capacitor (see Fig. 19–52). What is the net capacitance?
Solution
The series capacitors add reciprocally, and then the parallel combination is found by adding
linearly.
1
1
1 1 
1
1


6
6
Ceq      C3  

  2.00  10 F  3.71 10 F  3.71  F
6
6
 3.00  10 F 4.00 10 F 
 C1 C2 
45.
(II) A 0.40-F and a 0.60-F capacitor are connected in series to a 9.0-V battery.
Calculate (a) the potential difference across each capacitor, and (b) the charge on each.
(c) Repeat parts (a) and (b) assuming the two capacitors are in parallel.
Solution
When the capacitors are connected in series, they each have the same charge as the net
capacitance.
1
1 1 
1
1


Q1  Q2  Qeq  CeqV     V  


6
6
 0.40  10  F 0.60 10  F 
 C1 C2 
(a)
1
 9.0 V 
 2.16  106 C
V1 
Q1
C1
(b)

2.16  106 C
0.40  106 F
 5.4 V
V2 
Q2
C2

2.16  106 C
0.60  106 F
 3.6 V
Q1  Q2  Qeq  2.16 106 C  2.2 106 C
When the capacitors are connected in parallel, they each have the full potential difference.
(c) V1  9.0 V
V2  9.0 V



Q1  C1V1  0.40  106 F  9.0 V   3.6  106 C

Q2  C2V2  0.60  106 F  9.0 V   5.4 106 C
51.
(II) The RC circuit of Fig. 19–57 (same as Fig. 19–21a) has R  6.7 k and C  3.0 F. The
capacitor is at voltage V0 at t  0, when the switch is closed. How long does it take the capacitor
to discharge to 1.0% of its initial voltage?
Solution
The voltage of the discharging capacitor is given by VC  V0 e t RC . The capacitor voltage is to be
0.010V0 .
VC  V0 e t RC  0.010 V0  V0 e t RC  0.010  e t RC  ln  0.010   


t
RC


t   RC ln  0.010    6.7 10  3.0 10 6 F ln  0.010   9.3 10 2 s
*59. (II) An ammeter whose internal resistance is 63  reads 5.25 mA when connected in a
circuit containing a battery and two resistors in series whose values are 750  and 480 .
What is the actual current when the ammeter is absent?
Solution
The total resistance with the ammeter present is Req  1293  . The voltage supplied by the battery
is found from Ohm’s law to be Vbattery  IReq   5.25  103 A  1293    6.788 V . When the ammeter
is removed, we assume that the battery voltage does not change. The equivalent resistance
changes to Req  1230  , and the new current is again found from Ohm’s law.
I
Vbattery
Req

6.788 V
1230 
 5.52  10 3 A