EE MCQ

SSC JE Electrical 2019 with solution SET-1
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If admittance Y= a + jb, then a =?

Admittance (Y) is the reciprocal of the impedance of a circuit. Admittance of an AC circuit is analogous to the conductance of a DC circuit. The unit of Admittance is Simen or MHO

Admittance = 1/Z simen

Y = Conductance ± J Susceptance

Or the Admittance can be written as

Y = (G ± J B) Simen

Now comparing the above equation by the given equation in the question i.e Y= a + jb

∴ a = G = Conductance

Related Question

SSC JE Electrical 2019 with solution SET-1
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The trade name of the Nickel-Copper alloy, that is used as a heating element is:

Constantin‘, also known as ‘Eureka wire‘, is the trade-name for a copper-nickel alloy (approx. 60:40 ratio) formulated in the late 1800s by Edward Weston.

SSC JE Electrical 2019 with solution SET-1
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In the circuit shown, find the equivalent resistance between A and B.

The resistance R3 & R4 are connected in series

= (30 + 70) = 100Ω

Now three resistor i.e 100Ω, 100Ω & 50Ω is connected in parallel

1/Rp = 1/100 + 1/100 + 1/50

Rp = 100/4 = 25 Ω

Resistance 50Ω and 25Ω are connected in series.

Rtotal = (50 + 25)Ω

Rtotal = 75Ω

SSC JE Electrical 2019 with solution SET-1
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Hysteresis loss is NOT a function of

Hysteresis Loss = Kh × BM1.67 × f × v watts

where

Kh = Hysteresis constant depends upon the material

Bm = Maximum flux density

f = frequency

v = Volume of the core

Hence the hysteresis loss does not depend upon the ambient temperature.