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This mock test of Test: Basic Concepts for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam.
This contains 20 Multiple Choice Questions for Electrical Engineering (EE) Test: Basic Concepts (mcq) to study with solutions a complete question bank.
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students definitely take this Test: Basic Concepts exercise for a better result in the exam. You can find other Test: Basic Concepts extra questions,
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QUESTION: 1

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

Solution:

QUESTION: 2

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: 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:

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:

Solution:

**W = QV**- W = 120 * 3 = 360 J

QUESTION: 5

Find the value of 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 given, 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

**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
- 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 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 =** 6 C**

QUESTION: 10

Find the value of R_{1.}

Solution:

- Voltage across 60 Ω resistor = 30 V
- Current = 30 / 60 = 0.5 A
- Voltage across R
_{1}is = (70 - 20) V = 50 V **R**_{1}= 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

Solution:

The current i will be distributed in the cube branches symmetrically

QUESTION: 12

Find the value of v_{1.}

Solution:

If we go from +ve side of 1 kΩ through 7 V, 6 V and 5 V

We get, v_{1} = 7 + 6 - 5 = **8 V**

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

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?

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 _________ ?

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 10 ^{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:

12 * C = 2 m x 10

C = 1.67 mF

Hence, Capacitance is equal to **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:

= **3125 t ^{2}**

At t = 4 ms, v

It will be parabolic path at t = 0

t-axis will be tangent.

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