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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.
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QUESTION: 1

A solid copper sphere, 10 cm in diameter is deprived of 10^{20} electrons by a charging scheme. The charge on the sphere is

Solution:

n =10^{20}, Q = ne = e*10^{20} = -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 v_{1} 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

v_{1} + 60 - 100 = 10 x 20 or v_{1} = 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 v_{1} is

Solution:

100 = 65 + V_{2} ⇒ v_{2 }= 35 V

v_{3} - 30 = v_{2} ⇒ v_{3} = 65 V

105 + v_{4} - v_{3 }- 65 = 0 ⇒ v_{4} = 25 V

v_{4} + 15 - 55 + v_{1} = 0 ⇒ v_{1} = 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

R_{1} = ?

Solution:

Voltage across 60 Ω resistor = 30 V Current = 30/60 =0.5 A

Voltage across R_{1} is =70 - 20 = 50 V

R_{1} = 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

v_{1} = ?

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 v_{o} in fig is always equal to

Solution:

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

QUESTION: 14

R_{eq} = ?

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=3.6x10^{4 }C

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:

= 3125t^{2}

At t = 4 ms, v_{c} = 0.05 V

It will be parabolic path. at t = 0 t-axis will be tangent.

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