EE MCQ

SSC JE Electrical 2019 with solution SET-1
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A 200 V d.c machine as Ra = 0.5 Ω and its full-load Ia= 20A. Determine the induced e.m.f when the machine acts as a motor

Given Data

Voltage Va = 200 v

Armature Resistance Ra = 0.5Ω

Armature Current Ia = 20 A

Induced EMF = Ea = ?

The Induced EMF of a DC machine working as a Motor is

Ea = Va − IaRa

Ea = 200 − 20 × 0.5

Ea = 190 V

Related Question

SSC JE Electrical 2019 with solution SET-1
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The laws involved in the expression e = −dφ/dt are

Faraday’s 1st laws of electromagnetic induction tell us about the condition under which an e.m.f. is induced in a conductor or coil a when the magnetic flux linking a conductor or coil changes.

Faraday’s 2nd laws of electromagnetic induction give the magnitude of the induced e.m.f in a conductor or coil and may be stated as:

The magnitude of the e.mf induced in a conductor or coil is directly proportional to the rate of change of magnetic flux linkages.

Suppose a coil has N turns and magnetic flux linking the coil increases (i.e. changes) from φ1 Wb to φ2 Wb in t seconds. Now, magnetic flux linkages mean the product of magnetic flux and the number of turns of the coil.

N = e dφ/dt

Lenz Law:- Lenz’s law states: the direction of the induced e.m.f. is such as to oppose the change producing it. Therefore, the magnitude and direction of induced e.m.f. should be written as :

N = −e dφ/dt

SSC JE Electrical 2019 with solution SET-1
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In a single-phase capacitor start-and-run motor, the minimum number of capacitors to be used in it is:

Capacitor start and capacitor run motor: Two capacitors are used for starting, but one of them is cut out when speed reaches 70 percent of the synchronous speed. The capacitor start-and-run motor starts with a high value and a low-value capacitor connected in parallel with each other but in series with the starting winding. This provides a very high starting torque. The centrifugal switch disconnects the high-value capacitor at 80 percent speed, but the lower value capacitor remains in the circuit.

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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The average value of a sinusoidal wave is:

The average value of the sine wave over one complete cycle is actually zero. Hence, for a sine wave, the average value is defined over half the period. The average value expressed in terms of peak value is given by

Average value = 0.637 × peak value