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# Basic Concepts - MCQ Test

## 20 Questions MCQ Test Mock Test Series for Electrical Engineering (EE) GATE 2020 | Basic Concepts - MCQ Test

Description
This mock test of Basic Concepts - MCQ Test for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam. This contains 20 Multiple Choice Questions for Electrical Engineering (EE) Basic Concepts - MCQ Test (mcq) to study with solutions a complete question bank. The solved questions answers in this Basic Concepts - MCQ Test quiz give you a good mix of easy questions and tough questions. Electrical Engineering (EE) students definitely take this Basic Concepts - MCQ Test exercise for a better result in the exam. You can find other Basic Concepts - MCQ Test extra questions, long questions & short questions for Electrical Engineering (EE) on EduRev as well by searching above.
QUESTION: 1

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

Solution:

n =1020, Q = ne = e*1020 = -16.02 C.
Charge on sphere will be positive as this electric charge is removed

QUESTION: 2

### 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

Solution:

ΔQ = i x Δ t = 15000 x 100μ = 1.5 C

QUESTION: 3

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

Solution: QUESTION: 4

The energy required to move 120 coulomb through 3 V is

Solution:

W = Qv = 360 J

QUESTION: 5

i = ? 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 of fig P1.1.6 a charge of 600 C is delivered to the 100 V source in a 1 minute. The value of v1 must be 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 or v1 = 240 V

QUESTION: 7

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

Going from 10 V to 0 V 10+5+E+1 = 0

QUESTION: 8

Consider the circuit graph shown in fig. Each branch of circuit graph represent a circuit element. The value of voltage v1 is Solution:

100 = 65 + V2 ⇒ v2 = 35 V v3 - 30 = v2  ⇒ v3 = 65 V
105 + v4 - v- 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?

Solution:

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

QUESTION: 10

R1 = ? Solution:

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

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

Solution:

The current i will be distributed in the cube branches symmetrically  QUESTION: 12

v1 = ? Solution:

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

QUESTION: 13

The voltage vo in fig is always equal to Solution:

It is not possible to determine the voltage across 1 A source.

QUESTION: 14

Req =  ? Solution:

We infer from the given diagram that the same pattern is followed after every 10ohm resistor. The infinite pattern in parallel as a whole is considered as to be Rx.
Thus, Rx = R + (R||Rx)
Solving for Rx, we get Rx = 1.62R, where R=10ohm.
So we have Rx = (1.62ohm)*(10ohm), which gives Rx=16.2ohm
Therefore, Req = 5 + (10||16.2)
=> 5 + [(10*16.2)/(10+16.2)] => Req = 5 + (162/26.2) which gives Req=11.18 ohm.

QUESTION: 15

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

Solution:

We know that, I=Q/t or Q=I x t= 10x1 hours
{since unit of current is Cs-1, therefore time should be in seconds}
Therefore, Q=10x60x60=36000=3.6x10C

QUESTION: 16

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

Solution:  12C  =2m x 10  C ⇒ 1.67 mF

QUESTION: 17

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

Solution:  QUESTION: 18

The current in a 100 μF capacitor is shown in fig. If capacitor is initially uncharged, then the waveform for the voltage across it is Solution:  This 0.2 V increases linearly from 0 to 0.2 V. Then current is zero. So capacitor hold this voltage.

QUESTION: 19

The voltage across a 100 μF capacitor is shown in fig. The waveform for the current in the capacitor is Solution:   = 600 mA
For 1 ms < t < 2 ms,  QUESTION: 20

The waveform for the current in a 200 μF capacitor is shown in fig. The waveform for the capacitor voltage is Solution:   = 3125t2
At t = 4 ms, vc = 0.05 V
It will be parabolic path. at t = 0 t-axis will be tangent.