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A coil has self-inductance L=0.04 H and resistance R=12 Ω. When it is connected to a 220 V, 50 Hz supply, what will be the current flowing through the coil?

Asked in GUJCET 2009 · Current in a series LCR

Answer: (1) 12.7 A

Step-by-step solution

Given: L=0.04 H, R=12 Ω, V=220 V and f=50 Hz.

Idea: the coil is a resistance and an inductance in series, so the current is limited by the impedance Z=√R²+X_L².

X_L=2π fL=2×3.14×50×0.04=12.57 Ω.

Z=√12²+12.57²=√144+158=√302≈17.4 Ω.

I=V/Z=(220)/(17.4)≈12.7 A.

So the current through the coil is 12.7 A.

Why the other options are wrong

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