find XL of an inductor of l =2 h connected an ac frequency of 50hz ?
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Answer:
3.33 A
The frequency of AC source =f=50Hz
Thus the impedance of the inductor, XL=ωL=(2πf)L=20πΩ
Resistance of the resistor =20Ω
Thus, the net impedance of the circuit =XL2+R2=65.94Ω
Thus, the current in the circuit =XV=65.94Ω220V=3.33A
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