EE MCQ

SSC JE Electrical 2019 with solution SET-1
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In a transmission line, the voltage regulation is negative whenever receiving end voltage Vs

Voltage regulation:- When a transmission line is carrying current, there is a voltage drop in the line due to resistance and inductance of the line. The result is that the receiving end voltage (VR) of the line is generally less than the sending end voltage (Vs). This voltage drop (Vs − VR) in the line is expressed as a percentage of receiving end voltage VR and is called voltage regulation.

The difference in voltage at the receiving end of a transmission line between conditions of no-load and full-load is called voltage regulation and is expressed as a percentage of the receiving end voltage. Mathematically,

% Voltage regulation = (Vs − VR) ⁄ VR × 100

Obviously, it is desirable that the voltage regulation of a transmission line should be low i.e., the increase in load current should make very little difference in the receiving end voltage. In the transmission line, the voltage regulation is negative whenever the receiving end voltage VR is Greater than the sending end voltage.

The regulation will depend upon the power factor of the load. If the power factor is lagging, the voltage at the sending end is more than that at the receiving end. Hence, voltage regulation is positive. On the other hand, if the power factor is leading, the voltage at the sending end will be somewhat less than that at the receiving end. In that case, the regulation is negative.

Related Question

SSC JE Electrical 2019 with solution SET-1
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The output of the logic circuit given below represents the gate.

In the given diagram all are NOR Gate . The final output is shown in the figure.

At stage 1 the output will be \overline A \& \overline B

At stage 2 the output will be \overline {\overline A + \overline B } = A.B

And the final output will be \overline {A.B}

Hence for input A & B the output is \overline {AB} in case of Nand gate.

SSC JE Electrical 2019 with solution SET-1
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If the copper loss of a transformer at half full load is 400 W, then the copper loss corresponding to full load is

Copper loss is proportional to the square of load current. At half load, load current becomes half as voltage remains the same, so the copper loss will become (1/2)2 i.e 1/4 times of full load copper loss.

At full load copper Loss = I2R

At half load copper Loss = (I/2)2 × R = I2/4 × R

400 = I2/4 × R

I2R = 4 × 400

I2R = Full load copper Loss = 1600 W

SSC JE Electrical 2019 with solution SET-1
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What would the total loss of the 2kVA transformer corresponding to maximum efficiency be, provided the transformer has the iron of 150 W and full-load copper loss of 250W?

When variable loss becomes equal to the constant loss, efficiency is maximum.

Losses = Pi + Pc

Since copper loss is a variable loss therefore

Losses = Pi + Pi = 2pi

Thus at a maximum efficiency of this transformer total loss

= 150 x 2 = 300 W

SSC JE Electrical 2019 with solution SET-1
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What is the reason behind using a centrifugal switch in a single-phase induction motor?

The centrifugal switch is used to disconnect the starting winding of the motor once the motor approaches its normal operating speed i.e 50% – 70% speed.