SSC JE Electrical Engineering 2015 Paper

Question : 103 of 199

Marks: +1, -0

The rate of change of current in a 4 H inductor is 2 Amps/sec. Find the voltage across inductor.

Solution:

Voltage across inductor (E)

L {di}/{dt}

Her

L =4H,{dl }/{dt} = 2

Amp/sec = 4 × 2 = 8V

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