EE MCQ

SSC JE Electrical 2019 with solution SET-1
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If N = linkage flux, then the linkage flux per unit current is defined as:

The inductance L of this coil is defined as the flux linkage per unit current and its unit is Henry.

L =Λ/I

Where

Λ = Flux linkage

I = current

Related Question

SSC JE Electrical 2019 with solution SET-1
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With the current direction marked in the circuit shown, the net voltage applied is

Net voltage = −V2 + V1

Net voltage = −(V2 − V1)

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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The amount of AC content present in the DC output of a rectifier is given by

Ripple factor: The output of the rectifier is of pulsating d.c. type. The amount of a.c. content in the output can be mathematically expressed by a factor called the ripple factor (γ). The ripple factor expresses how much successful the circuit is, in obtaining pure d.c from a.c. input.

Less is the ripple factor, better is the performance of the circuit.

Ripple Factor γ = R.M.S value of a.c component of output ⁄ Average or d.c component of the output

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