Test: Basic Concepts : PN Junction

QUESTION: 1

If 120 C of charge passes through an electric conductor in 60 sec, the current in the conductor is:

A.
0.5 A
B.
2 A
C.
3.33 mA
D.
0.3 mA

Solution:

i = dQ/dt = 120/60 = 2A.

QUESTION: 2

A solid copper sphere, 10 cm in diameter is deprived of 10²⁰ electrons by a charging scheme. The charge on the sphere is:

A.
160.2 C
B.
-160.2 C
C.
16.02 C
D.
-16.02 C

Solution:

n = 10²⁰
Q = ne = e * 10²⁰ = 16.02 C.
The charge on the sphere will be positive.

QUESTION: 3

The energy required to move 120 coulomb through 3 V is:

A.
25 mJ
B.
360 J
C.
40 J
D.
2.78 mJ

Solution:

W = QV = 360 J.

QUESTION: 4

A lightning bolt carrying 30,000 A lasts for 50 microseconds. If the lightning strikes an airplane flying at 20,000 feet, what is the charge deposited on the plane?

A.
13.33 C
B.
75 C
C.
1500 C
D.
1.5 C

Solution:

QUESTION: 5

Consider the circuit graph shown in figure below. Each branch of circuit graph represent a circuit element. The value of voltage V1 is:

A.
30 V
B.
25 V
C.
20 V
D.
15 V

Solution:

100 = 65 + V2 ⇒ V2 = 35 V
V3 – 30 = V2 ⇒ V3 = 65 V
105 – V3 + V4 – 65 = 0 ⇒ V4 = 25 V
V4 + 15 – 55 + V1 = 0 ⇒ V1 = 15 V.

QUESTION: 6

What is the value of R_eq = ?

A.
11.86 ohm
B.
10 ohm
C.
25 ohm
D.
11.18 ohm

Solution:

Req – 5 = 10(Req + 5)/(10 + 5 +Req).
Solving for Req we have
Req = 11.18 ohm.

QUESTION: 7

The superposition theorem requires as many circuits to be solved as there are:

A.
Sources, nodes and meshes
B.
Sources and nodes
C.
Nodes
D.
Sources

Solution:

QUESTION: 8

Twelve 6 resistor are used as edge to form a cube. The resistance between two diagonally opposite corner of the cube is: (in ohm)

A.
5/6
B.
6/5
C.
5
D.
6

Solution:

QUESTION: 9

The energy required to charge a 10 µF capacitor to 100 V is:

A.
0.1 J
B.
0.05 J
C.
5 x 10^(-9) J
D.
10 x 10^(-9) J

Solution:

Energy provided is equal to 0.5 CV²= 0.5x10^-6 x 100 x 100 = 0.05 J

QUESTION: 10

A capacitor is charged by a constant current of 2 mA and results in a voltage increase of 12 V in a 10 sec interval. The value of capacitance is ________.

A.
0.75 mF
B.
1.33 mF
C.
0.6 mF
D.
1.67 mF

Solution:

The capacitor current is given as i=C*(dv/dt), where dv/dt is the derivative of voltage, dt=t2-t1 given as 10 sec and dv is the change in voltage which is given as 12 V.
So, we have C = i / (dv/dt)
C = 2mA/(12/10) = 2mA/(1.2).
Hence C = 1.67mF