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Question

In an LR circuit, the value of L is 0.4π H and the value R is 30 Ω. If in the circuit, an alternating emf of 200 V rms value at 50 cycles/s is connected, the peak current of the circuit is k2 A .The value of k is (integer only)

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Solution

Impedence for LR circuit is given by,

Z=R2+X2L=R2+ω2L2

As, f=50 Hz ω=2πf=100π

XL=ωL=100π(0.4)π=40 Ω

So, Z=302+402=50 Ω

Irms=VrmsZ=20050=4 A

Ipeak=Irms2=42

Given, Ipeak=k2

k=4

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