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This mock test of Basic & Electric Circuits - 2 for Electrical Engineering (EE) helps you for every Electrical Engineering (EE) entrance exam.
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QUESTION: 1

A current of 3 A flows through a resistor of 20 ohms. The energy dissipated in the resistor per minute is

Solution:

Power dissipated by the resistor is

P = I^{2} R = 3^{2} x 20 = 180 W

Now, energy dissipated by the resistor in one hour (or 60 min)

= [180 x 60] watt-min

∴ Energy dissipated by the resistor in one minutes

QUESTION: 2

A long uniform coil of a inductance L henries and associated resistance R ohms is physically cut into two exact halves which are then rewound in parallel. The resistance and inductance of the combination are

Solution:

We know that,

and

when coil is cut into two equal halves,

(Area of cross - section = constant)

So,

Also,

So,

Hence, new values of resistance and inductance which are reconnected in parallel is

and

QUESTION: 3

In the circuit shown, the power dissipated in the resistor R is 1 W when only source ‘1’ is present and ‘2’ is replaced by short circuit. The power dissipated in the same resistor R is 4 W when only source '2’ is present and '1' is replaced by a short circuit When both the sources ‘1’ and ‘2’ are present, the power dissipated in R will be

Solution:

We have;

or,

or,

When both the sources are present, net current through R will be

I = (I_{2 }- I_{1})

[as polarity of |V_{1}| is reverse]

So, power loss in R is

QUESTION: 4

If v, w, q stand for voltage, energy and charge, then v can be expressed as:

Solution:

QUESTION: 5

In the figure shown below, φ = power-factor angle, W = watts, VA = volt ampere and VAr = volt-ampere .reactive for an ac circuit. The correct figure is

Solution:

QUESTION: 6

A circuit possesses resistance R and inductive reactance X_{L} in series, its susceptance is given by

Solution:

Susceptance is the imaginary part of admittance,

or,

Here, G = Conductance

and S = Susceptance

QUESTION: 7

If V = a + jb and I = c + jd, then the power is given by

Solution:

P = VI* = (a + jb).(c + jd)*

= (a + jb) (c - jd)

= (ac + bd) + j(bc - ad)

= P + jQ

Here, P = Active power

and Q = Reactive power

QUESTION: 8

A 230 V, 100 W bulb has resistance R_{A} and a 230 V, 200 W bulb has resistance R_{B}. Here,

1. R_{A} > R_{B}

2. R_{B} > R_{A}

3. R_{A} = 2 R_{B}

4. R_{B} = 2 R_{A}

5. R_{A} = 4 R_{B}

From these, the correct answer is

Solution:

Power,

Since voltage ratings of both bulbs are same, therefore

PR = constant

or, P_{A} R_{A} = P_{B} R_{B}

or, 100 x R_{A} = 200 x R_{B}

or, R_{A} = 2 R_{B}

Thus, R_{A} > R_{B}

QUESTION: 9

The voltage phapor of a circuit is 10∠15° V and the current phasor is 2∠- 45° A. The active and the reactive powers in the circuit are

Solution:

S = VI* = (10∠15°) (2∠ - 45°)*

= 20∠15 + 45° = 20∠60°

= 20 (cos 60° + j sin 60°)

= (10 + j10√3) = (10 + j17.32)

= P+ jQ

QUESTION: 10

In the circuit shown below, if the power consumed by the 5 Ω resistor is 10 W, then power factor or the circuit will be

Solution:

Given, power consumed is

P = I^{2 }R_{eq}

or,

Also,

or, |Z| = 25Ω

or,

or,

Hence, p.f. of given circuit is

QUESTION: 11

The minimum requirements for causing flow of current are.

Solution:

QUESTION: 12

Ohm’s law is applicable to

Solution:

Carbon resistor and semiconductors have nonlinear relationship between V and I. Hence, Ohm’s law is not applicable. Also, these are not bilateral.

QUESTION: 13

For a fixed supply voltage, the current flowing through a conductor will increase when its

Solution:

When l is reduced. I will be increased and vice-versa.

QUESTION: 14

Two registors R_{1}, and R_{2} give combined resistance of 4.5 Ω when in series and 1 Ω when in parallel. The resistances are

Solution:

R_{1 }+ R_{2} = 4.5 ...(i)

and

∴ (R_{1}-R_{2})^{2} = (R_{1} + R_{2})^{2 }- 4R_{1}R_{2}

or, ...(ii)

On solving equations (i) and (ii), we get

R_{1} = 3 Ω and R_{2} = 1.5 Ω

or, R_{1} = 1.5 Ω and R_{2 }= 3Ω.

QUESTION: 15

Which of the following is not equivalent to watts?

Solution:

QUESTION: 16

Two heaters, rated at 1000 W, 250 V each are connected in series across a 250 V, 50 Hz ac mains. The total power drawn from the supply would be

Solution:

For series connection,

R_{eq} = R_{1} + R_{2}

or,

or,

Given,

P_{1} = P_{2 }= 1000 W

∴ P_{eq }= 500 Watt

QUESTION: 17

A 100 watt light bulb burns on an average of 10 hours a day for one week. The weekly consumption of energy will be

Solution:

Energy consumption per week is

W= 100 x 10 x 7

= 7000 kW - hr

= 7 units

QUESTION: 18

Assertion (A): The direction of flow of conventional current is taken opposite to that of electrons.

Reason (R): Electrons have negative charge.

Solution:

QUESTION: 19

Match List-I (Materials) with List-lI (Range of resistivity) and select the correct answer using the codes given below the lists:

List-I

A. Conducting materials

B. Semiconductor materials

C. insulating material

List-II

1. 10^{0} to 10^{2} Ω-m

2. 10^{-8} to 10^{-6} Ω-m

3. 10^{12}to 10^{18 }Ω-m

4. 10^{20} to 10^{30} Ω-m

Codes:

Solution:

QUESTION: 20

A constant current source supplies a current of 200 mA to a load of 2 kΩ When the load is changed to 100 Ω, the load current is

Solution:

For a constant C.S., current will remain constant for all values of loads.

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