Physics, asked by vbbg, 8 hours ago

find XL of an inductor of l =2 h connected an ac frequency of 50hz ?​

Answers

Answered by vk624419
0

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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