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Question

An a.c generator supplying voltage E=10sinωt V to a circuit which consists of a resistor of resistance 3 Ω and an inductor of reactance 4 Ω as shown in the figure. The voltage across the inductor at t=πω is

A
2 V
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B
10 V
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C
Zero
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D
4.8 V
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Solution

The correct option is D 4.8 V
Impedance of the circuit Z=R2+X2L
=32+42=5 Ω

Power factor of the circuit cosϕ=RZ=35
ϕ=53

Therefore, current in the circuit is
i=E0Zsin(ωt53)=2sin(ωt53)

Voltage across the inductor
VL=Ldidt=2ωLcos(ωt53)
At t=πω, VL=8cos(π53)
(ωL=4)
VL=8cos53=8×35
VL=4.8 V
i.e. potential across inductor lags behind.

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