PHYS 2421 Exam 2 Spring 2008 NAME___________________________________
Problem 1 (10 points).
Select True or false
1.
Resistors connected in parallel suffer the same voltage drop.
2.
Resistors connected in parallel carry similar currents. .
.
.
3.
Kirchhof’s laws are based on conservation of charge and energy. .
4.
An ideal voltmeter would have a very low resistance. . .
.
.
.
.
.
.
.
.
.
.
.
.
5.
The drift velocity of charges in a wire are smaller when I = 1 A than when I = 2 A. .
6.
The capacitance of a capacitor depends on the voltage of the battery it gets connected to .
7.
The voltage provided by a Duracell 1.5 V battery varies depending on the current drawn. .
8.
Dielectrics increase the electric field inside a capacitor and thus reduce its capacitance. .
9.
Capacitors connected in series get charged to the same value of charge. .
10.
Capacitors connected in parallel get charged to different voltages. . .
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Problem 2 (10 points).
Determine: A) The equivalent capacitance of the capacitors shown:
B) The charge on the 9 µ F capacitor:
C) The potential difference across the 11 µ F capacitor:
Solution:
Q = Q
Eq
= CV = (
) ( 24 V ) = 144
C
C
Eq


1

 3 11


1 1
= + + + +
9 12 6


− 1 



− 1
=
V
11 µ F
= V − V
9 µ F
Another way:
= V − Q / C
9 µ F
= −
C
= 8 V
V
11 µ F
= V
Lower 4
=
Q
Lower 4
C
Lower 4
, C
Lower 4
3 11
 1 1
= + + +
12 6


− 1
= Q
Lower 4
= Q
⇒
V
11 µ F
=
144 µ C
=8 V

Problem 3 (10 points).
The parallel plate capacitor shown has a capacitance of C
O
=5 x10
-12
F when there is air between the plates. A dielectric of K=2.70 is then inserted completely filling the volume between the plates, and the capacitor is connected to a power supply which charges the capacitor until the electric field in the dielectric reaches a value of 3x10
4
V/m. What is the charge on the plates of the capacitor?
Solution.
C = KC
0
=
Q
=
Q
⇒
=
V Ed
Q KC Ed
0
= 2.7 5 pF × 30000 V/m 0.0015m
= × − 10
C

Problem 4 (10 points).
In the circuit shown, the resistor R
1
is dissipating 20 W of energy. Determine
A) the magnitudes of R
1,
B) the magnitude of the EMF E .
C) the magnitude of the current through the 10 Ω resistor.
Solution . A) P
1
=
2
I R
1 1
⇒
R
1
= P I
1 1
2
= 20 W ( A ) 2
= 5 Ω
B)
ε
I R
1
( 2 A )( 5 )
C)
ε
= I R
10 10
⇒
I
10
=
ε
10 V
/ R
10
= 10 V/10 Ω = 1 A

Problem 5 (10 points).
Find the EMFs E
1
and E
2
and the potential difference between points a and b.
Solution .
• Assuming I as shown, Kirkhoff’s law at node a yields:
I + 1 A = 2 A ⇒
I = 1 A
• Kirkhoff’s law of outer loop starting at node a:
− ( 2 A )( 1 )
2
− ( 2 )( ) ( 1 )( ) ( 20 V ) ( 1 A )( 1 ) 0
⇒
2
= − − − + − = V
• Kirkhoff’s law of upper loop starting at node a:
( 1 )( ) ( 1 )( )
1
− ( 1 )( ) ( 20 V ) ( 1 A )( 1 ) 0
⇒
2
4 1 6 20 1 18 V

Problem 6 (10 points).
For the circuit shown determine,
A) the current and direction of the current
B) the terminal voltage Vab of the 16 V battery
C) the power output of the 8 V EMF
Solution:
A) 16 V − I (1.6
Ω + 5 Ω + 1.4
Ω + 9 ) 8 V = 0
⇒
I = 8 V /17 Ω = 0.47
A , counterclockwise
B) V = ab
16 V − 0.47
A × 1.6
Ω = 0 = 015.24
V
C) P
I ( 8 V ) ( 0.47
A )
3.76
W

EXTRA Problem 7 (10 EXTRA points).
What length of a copper wire of diameter 0.462 mm has a resistance of 1 Ω ? The resistivity of copper is ρ =
1.72x10
-8
Ω m
Solution.