Intermediate Exam II: Problem #1 (Spring ’05)
The circuit of capacitors connected to a battery is at equilibrium.
(a) Find the equivalent capacitance C eq
.
(b) Find the voltage V
3 across capacitor C
3
.
(c) Find the the charge Q
2 on capacitor C
2
.
C
3
= 3
µ
F
8V C
1
= 1
µ
F
Solution:
(a) C
12
= C
1
+ C
2
= 3 µ F , C eq
=
„ 1
C
12
+
1
C
3
« −
1
= 1 .
5 µ F .
(b) Q
3
= Q
12
= Q eq
= C eq
(8V) = 12 µ C
Q
3
⇒ V
3
= =
12 µ C
= 4V .
C
3
3 µ F
(c) Q
2
= V
2
C
2
= 8 µ C .
C
2
= 2
µ
F tsl336 – p.1/4

Intermediate Exam II: Problem #2 (Spring ’05)
Consider the electrical circuit shown.
(a) Find the equivalent resistance R eq
.
(b) Find the current I
3 through resistor R
3
.
2Ω
12V
3Ω
Solution:
(a) R
36
=
„ 1
R
3
+
1
R
6
« −
1
= 2Ω , R eq
= R
2
+ R
36
= 4Ω .
(b) I
2
= I
36
=
12V
R eq
= 3A
⇒ V
3
= V
36
= I
36
R
36
= 6V ⇒ I
3
=
V
3
R
3
= 2A .
6Ω tsl337 – p.2/4

Intermediate Exam II: Problem #3 (Spring ’05)
This RC circuit has been running for a long time.
(a) Find the current I
2 through the resistor R
2
.
(b) Find the voltage V
C across the capacitor.
Solution:
E
(a) I
C
= 0 , I
2
=
R
1
+ R
2
=
12V
6Ω
= 2A .
(b) V
C
= V
2
= I
2
R
2
= (2A)(4Ω) = 8V .
R =
2Ω
12V
C = 7nF
4Ω tsl338 – p.3/4

Intermediate Exam II: Problem #4 (Spring ’05)
Consider a charged particle moving in a uniform magnetic field as shown. The velocity is in y -direction and the magnetic field in the yz -plane at 30 ◦ from the y -direction.
(a) Find the direction of the magnetic force acting on the particle.
(b) Find the magnitude of the magnetic force acting on the particle.
z
B = 4mT
+ q = 5nC
30 o y v = 3m/s
Solution: x
(a) Use the right-hand rule: positive x -direction (front, out of page).
(b) F = qvB sin 30 ◦ = (5 × 10 −
9 C)(3m / s)(4 × 10 −
3 T)(0 .
5) = 3 × 10 −
11 N .
tsl339 – p.4/4