EE MCQ

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

Related Question

SSC JE Electrical 2019 with solution SET-1
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Find the mutual inductance between two ideally coupled coils of 2H and 8H

Mutual inductance between the two coils is K = √L1.L2

K = √2 × 8

k = √16

k = 4 H

SSC JE Electrical 2019 with solution SET-1
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A coil is wound with 50 turns and a current 8A produces a flux of 200μWb. Calculate the inductance of the coil

The inductance of the coil is given by the relation

L = Nφ/I

Where

N = number of turns = 50

φ = flux = 200μWb

I = current = 8 A

L = 50 × 200 × 10−6 ⁄ 8

L = 1.25 mH

SSC JE Electrical 2019 with solution SET-1
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In the circuit shown, if R = 0, then the phase angle between v(t) and i(t) is

In the given circuit if R = 0 then the circuit becomes purely inductive. So the phase angle between v(t) and i(t) is 90°.

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

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