An a.c generator supplying voltage E=10sinωtV to a circuit which consists of a resistor of resistance 3Ω and an inductor of reactance 4Ω as shown in the figure. The voltage across the inductor at t=πω is
A
2V
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B
10V
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C
Zero
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D
4.8V
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Solution
The correct option is D4.8V Impedance of the circuit Z=√R2+X2L =√32+42=5Ω
Power factor of the circuit cosϕ=RZ=35 ⇒ϕ=53∘
Therefore, current in the circuit is ∵i=E0Zsin(ωt−53∘)=2sin(ωt−53∘)
Voltage across the inductor VL=Ldidt=2ωLcos(ωt−53∘) At t=πω, VL=8cos(π−53∘) (∵ωL=4) VL=−8cos53∘=−8×35 ∴VL=−4.8V i.e. potential across inductor lags behind.