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**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