Test: Basic Concepts

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:

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
Charge on sphere will be positive as this electric charge is removed.

QUESTION: 3

A lightning bolt carrying 15,000 A lasts for 100 μs. If the lightning strikes an airplane flying at 2 km, the charge deposited on the plane is:

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

Solution:

ΔQ = i x Δ t = 15000 x 100μ = 1.5 C

QUESTION: 4

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

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

Solution:

W = QV
W = 120 * 3 = 360 J

QUESTION: 5

Find the value of i.

A.
1 A
B.
2 A
C.
3 A
D.
4 A

Solution:

Based on kirchoffs current law, we know that incoming currents at a node is equals to outgoing currents.
So, based on this 4 and 3 will get canceled with 5 and 2.
So, finally i = 1.

QUESTION: 6

In the circuit given, a charge of 600 C is delivered to the 100 V source in a 1 minute. The value of v₁ must be:

A.
240 V
B.
120 V
C.
60 V
D.
30 V

Solution:

In order for 600 C charge to be delivered to the 100 V source, the current must be anticlockwise.
Applying KVL we get:
v₁ + 60 - 100 = 10 x 20
v₁ = 240 V

QUESTION: 7

In the circuit of the fig the value of the voltage source E is:

A.
-16 V
B.
4 V
C.
- 6 V
D.
16 V

Solution:

Going from 10 V to 0 V:

10 + 5 + E + 1 = 0
E = -16V
Minus sign indicates that the polarity of battery should be reversed.

QUESTION: 8

Consider the circuit graph shown in fig. 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 + v₄ - v₃- 65 = 0
⇒ v₄ = 25 V
v₄ + 15 - 55 + v₁ = 0
⇒ v₁ = 15 V

QUESTION: 9

What quantity of charge must be delivered by a battery with a Potential Difference of 110 V to do 660J of Work?

A.
0.6 C
B.
6 C
C.
60 C
D.
600 C

Solution:

► V = W / Q
► Q = W / V = 660 / 110 = 6 C

QUESTION: 10

Find the value of R_1.

A.
25 Ω
B.
50 Ω
C.
100 Ω
D.
2000 Ω

Solution:

Voltage across 60 Ω resistor = 30 V
Current = 30 / 60 = 0.5 A
Voltage across R₁ is = (70 - 20) V = 50 V
R₁ = 50 / 0.5 = 100 Ω

QUESTION: 11

Twelve 6 Ω resistor are used as edge to form a cube.The resistance between two diagonally opposite corner of the cube is

A.
5/6 Ω
B.
6/5 Ω
C.
5 Ω
D.
6 Ω

Solution:

The current i will be distributed in the cube branches symmetrically

QUESTION: 12

Find the value of v_1.

A.
-11 V
B.
5 V
C.
8 V
D.
18 V

Solution:

If we go from +ve side of 1 kΩ through 7 V, 6 V and 5 V
We get, v₁ = 7 + 6 - 5 = 8 V

QUESTION: 13

The voltage v_o in fig is always equal to:

A.
Cannot be determined
B.
5 V
C.
19 V
D.
9 V

Solution:

From above figure
V – 4 – 5 = 0
or
V = 9 V
Hence alternative (D) is the correct choice.

QUESTION: 14

In a AC circuit, resistive and total impedance are 10 and 20 ohms respectively. What is the phase angle difference between the voltage and current?

A.
0 degree
B.
30 degree
C.
45 degree
D.
60 degree

Solution:

Cosɸ = R/Z; where R= 10 ohms and Z = 20 ohms
Cosɸ = 1/2 means phase angle is 60 degree.

QUESTION: 15

The quantity of a charge that will be transferred by a current flow of 10 A over 1 hour period is _________ ?

A.
10 C
B.
3.6 x 10⁴ C
C.
2.4 x 10³ C
D.
1.6 x 10² C

Solution:

We know that,
I=Q / t or Q = I x t = 10 x 1 hours
{since unit of current is Cs^-1, therefore time should be in seconds}
∴ Q = 10 x 60 x 60
= 36000 C
= 3.6 x 10⁴C