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Question

A current of 4A flows in a coil when connected to a 12 V dc source . If the same coil is connected to a 12V,50 rad/s a.c source, a current of 2.4 A flows in the circuit. Determine the inductance of the coil.

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Solution

Resistance of coil R=VdcIdc=124=3Ω
Rms value of AC source Vrms=12 V
Rms value of alternating current Irms=2.4 A
Impedance of circuit Z=VrmsIrms=122.4=5Ω
Using Z2=R2+X2L
We get 52=32+X2L
XL=42=4Ω
Frequency of AC source f=50 Hz
Thus inductance of circuit L=XL2πf=42π×50=0.0127 H=12.7 mH

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