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

SSC JE Electrical 2019 with solution SET-1
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Two decimal coils A and B of 1000 turns each lies in the parallel plane such that 80% of the flux produced by one coil links with the other. If a current of 5A flowing in A produces a flux of 0.05 mWb, then the flux linking with coil B is:

Flux in coil A = 0.05 mWb = 5 × 10−5 wb = φA

No. of turns NA = NB = 1000

Flux linkage in a coil with B = Flux linkage in coil A × 80/100

= 0.8 × 5 × 10−5

= 4 × 10−5 wb =0.04 mwb

SSC JE Electrical 2019 with solution SET-1
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Which of the following fuel has the least calorific value?

In the above option, the wood has the lowest calorific value. Its value is around 17000 – 20000 kilojoule/kg.

The calorific value of LPG is 55000 kilojoule/kg.

The calorific value of Kerosene oil is 45000 kilojoule/kg.

The calorific value of hydrogen oil is 150000 kilojoule/kg.

SSC JE Electrical 2019 with solution SET-1
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Which of the following is the correct relation regarding coupled coils?

Mutual inductance between two coupled coils is given by the relation

M = K√L1L2

Where

L1L2 = Inductance of the coil

K = Coefficient of coupling

When K = 1 coils are said to be tightly coupled and if K is a fraction the coils are said to be loosely coupled.

Here

Kmax =1

Kmin =0

∴ m ≤ (L1 L2 )0.5