Physics 112
Homework 5 (Ch18 & Ch19)
Ch18
1. An ac voltage, whose peak value is 180 V, is across a 330- resistor. What are the rms and
peak currents in the resistor?
Solution
Find the peak current from Ohm’s law, and then find the rms current from the relationship
between peak and rms values.
I peak 
Vpeak
R

180 V
330 
 0.54545 A
0.55 A
I rms  I peak
2   0.54545 A 
2  0.39 A
2. The peak value of an alternating current in a 1500-W device is 5.4 A. What is the rms voltage
across it?
Solution
P  I rmsVrms 
I peak
2
Vrms  Vrms 
2P
I peak
2 1500 W 

5.4 A
 3.9  102 V
3. A 0.65-mm-diameter copper wire carries a tiny current of 2.3 A. What is the electron drift
speed in the wire?
Solution
We follow exactly the derivation in Example 18-14, which results in an expression for the drift
velocity.
vd 

I
neA

N 1 mole 
m 1 mole 
I
 D e  12 d 


2
4I m
N  D e d 2

m 1.60  10
4 2.3  10 6 A 63.5  10 3 kg
 6.02 10 8.9 10
23
3
kg
3
19

 
C  0.65 10 m
3

2
 5.1 1010 m s
Physics 112
Homework 5 (Ch18 & Ch19)
Ch19
4. 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.
Solution
The current in the circuit is I 
(a)
Vab  IR 
R
Rr
E
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)
Vab  E
R
Rr
  8.50 V 
810 
 810  0.900  
 8.49 V
5. 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: Use equations for terminal voltage:
E  Vab 12.0 V  8.4 V
Vab  E  Ir  r 

 0.048 
I
Vab  IR  R 
Vab
I

8.4 V
75 A
75 A
 0.11 
6. Calculate the current in the circuit of the figure below 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
Physics 112
Homework 5 (Ch18 & Ch19)
7. What is the potential difference between points a and d in the figure below (the same circuit
as Fig. 19–13, Example 19–8 in the textbook), 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
8. The RC circuit 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?
Physics 112
Homework 5 (Ch18 & Ch19)
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
9. 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