Solutions to Electric Circuit
1. A battery with an emf of 12 V and an internal resistance of 1  is used to charge
a battery with an emf of 10 V and an internal resistance of 1 . The current in the
circuit is:
A) 1 A
B) 2 A
C) 4 A
D) 11 A
E) 22 A
Solution:
All elements are in series:
2I  10  12

I  1A
Ans: A
2. Two 110-V light bulbs, one "25 W" and the other "100 W", are connected in
series to a 110 V source. Then:
A) the current in the 100-W bulb is greater than that in the 25-W bulb
B) the current in the 100-W bulb is less than that in the 25-W bulb
C) both bulbs will light with equal brightness
D) each bulb will have a potential difference of 55 V
E) none of the above
Series  same I.
Solution:
Bulb ratings:
Series circuit:
V2
: different P , same V  different R .
P
P  I2 R :
R
different R , same I  different P  different brightness.
different R , same I  different V.
V  IR:
Ans: E
3. The positive terminals of two batteries with emf's of ε1 and ε2, respectively, are
connected together. Here ε1 < ε2. The circuit is completed by connecting the negative
terminals. If each battery has an internal resistance of r, the rate in watts at which
electrical energy is converted to chemical energy in the smaller battery is:
A)
B)

2
1
/r
12 / 2r
C)
 2  1  1 / r
D)
 2  1  1 / 2r
E)

2
2
/ 2r
Solution:
All elements are in series:
I
 2  1
2r
P  I 1 
( into battery 1 ).
 2  1
2r
1 (elect. to chem.)
Ans: D
4. The resistance of resistor 1 is twice the resistance of resistor 2. The two are
connected in series and a potential difference is maintained across the combination.
Then:
A) the current in 1 is twice that in 2
B) the current in 1 is half that in 2
C) the potential difference across 1 is twice that across 2
D) the potential difference across 1 is half that across 2
E) none of the above are true
Solution:
V1  IR1  2 IR2  2V2
Ans: C
5. A battery of emf 24 V is connected to a 6- resistor. As a result, current of 3 A
exists in the resistor. The terminal potential difference of the battery is:
A) 0
B)
C)
D)
E)
6V
12 V
18 V
24 V
Solution:
Ans: D
V  3  6  18V
6. The current in the 5.0- resistor in the circuit shown is:
A)
B)
C)
D)
0.42 A
0.67 A
1.5 A
2.4 A
E) 3.0 A
Solution:
6  12
 4  8
6  12
RL  3  5  8
RU 
Upper branch:
Lower branch:
Total:
I 5 
R
1
 8  4
2
I
12
 3A
4
I
 1.5 A
2
Ans: C
7. In the diagrams, all light bulbs are identical and all emf devices are identical. In
which circuit (I, II, III, IV, V) will the bulbs be dimmest?
A) I
B) II
C) III
D)
IV
E) V
Solution:
V2
:
P
R
1  3 
9 2 9
PI  

 p
2 2R
4 R 4
2
1  
PIII  
p
2 R/2
2
1  2 
4
PII  
 p
3 3R
9
2
1  
1
PIV  
 p
2 2R 4
2
1  2 
PV  
 4p
2 R/2
2
Ans: D
8. A 120-V power line is protected by a 15-A fuse. What is the maximum number of
"120 V, 500 W" light bulbs that can be operated at full brightness from this line?
A)
B)
C)
D)
E)
1
2
3
4
5
Solution:
Pmax  120  15  1800 W
P 
n  Floor  max   3
 150 
Ans: C
9. In the figure, voltmeter V1 reads 600 V, voltmeter V2 reads 580 V, and ammeter A
reads 100 A. The power wasted in the transmission line connecting the power house
to the consumer is:
A)
B)
C)
D)
E)
1 kW
2 kW
58 kW
59 kW
60 kW
P  I V1  V2   100   600  580  2kW
Solution:
Ans: B
10. A certain voltmeter has an internal resistance of 10,000  and a range from 0 to
12 V. To extend its range to 120 V, use a series resistance of:
A)
B)
C)
D)
E)
1,111 
90,000 
100,000 
108,000 
120,000 
Solution:
I max
internal resistance is in parallel to ideal voltmeter:
12
120
 R  10,000  100,000


10,000 R  10,000

R  90,000 
Ans: B
11. Four circuits have the form shown in the diagram.
uncharged and the switch S is open.
The capacitor is initially
The values of the emf , resistance and R, and capacitance C for each for the circuits
are
circuit 1:
circuit 2:
 = 18 V, R = 3 , C = 1 F
 = 18 V, R = 6 , C = 9 F
circuit 3:  = 12 V, R = 1 , C = 7 F
circuit 4: = 10 V, R = 5 , C = 7 F
Rank the circuits according to the current just after switch S is closed least to greatest.
A) 1, 2, 3, 4
B) 4, 3, 2, 1
C) 4, 2, 3, 1
D) 4, 2, 1, 3
E) 3, 1, 2, 4
Solution: Since capacitor is initially uncharged, V  0 across it when S is closed.

R
I1  6 A , I 2  3 A , I 3  12 A , I 4  2 A
Rank, least to greatest: 4, 2,1,3
Ans: D

12.
I
In the circuit shown, the capacitor is initially uncharged.
At time t = 0, switch
S is closed. If  denotes the time constant, the approximate current through the 3 
resistor when t = /10 is:
A)
B)
C)
D)
E)
0.38 A
0.50 A
0.75 A
1.0 A
1.5 A
I  I 0 e  t /  I 0e 1/10
Solution:
I0 
10
10

63 9
I
10 0.1
 e  1.0 A
9
Ans: D
13. A charged capacitor is being discharged through a resistor. At the end of one
time constant the charge has been reduced by (1 – 1/e) = 63% of its initial value. At
the end of two time constants the charge has been reduced by what percent of its
initial value?
A) 82%
B) 86%
C) 100%
D) between 90% and 100%
E) need to know more data to answer the question
Solution:
(1 – 1/e) = 0.63

1/ e  0.37
I
2
 1  e 2  1   0.37   86%
I0
Ans: B
14. A certain capacitor, in series with a resistor, is being charged. At the end of 10
ms its charge is half the final value. The time constant for the process is about:
A) 0.43 ms
B) 2.3 ms
C) 6.9 ms
D) 10 ms
E) 14 ms
Solution:
Ans: E
I 1
  e 0.01/
I0 2


0.01
 0.014 s
ln 2