A pure inductive coil having inductance 0.1henry is connected across 200V,50Hz supply. Write the equation for voltage and current
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Answer:
The frequency of AC source =f=50Hz
Thus the impedance of the inductor, X
L
=ωL=(2πf)L=20πΩ
Resistance of the resistor =20Ω
Thus, the net impedance of the circuit =
X
L
2
+R
2
=65.94Ω
Thus, the current in the circuit =
X
V
=
65.94Ω
220V
=3.33A
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