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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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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,
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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**

W = 120 * 3

W = 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 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

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

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

= 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) 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

v_{1} = ?

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

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 10 ohm resistor. The infinite pattern in parallel as a whole is considered as to be R
_{x}.

Thus, R_{x }= R + (R || R_{x})

Solving for R_{x}, we get R_{x }= 1.62 R, where R = 10 ohm.

So, we have R_{x }= (1.62 ohm)*(10 ohm)

This gives R_{x }= 16.2 ohm

∴ R_{eq }= 5 + (10 || 16.2)

⇒ 5 + [(10 * 16.2) / (10 + 16.2)]

⇒ R_{eq }= 5 + (162 / 26.2)

which gives R_{eq }= 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 = 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:

= 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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