When `100 V` DC is applied across a solenoid, a current of `1.0 A` flows in it. When `100 V` AC is applied across the same coil. The current drops to `0.5A`. If the frequency of the ac source is `50 Hz`, the impedance and inductance of the solenoid areA. `200 Omega` and 0.55 HB. `100 Omega` and 0.86 HC. `200 Omega` and 1.0 HD. 1100`Omega` and 0.93 H

When `100 V` DC is applied across a solenoid, a current of `1.0 A` flo

1.	When `100 V` DC is applied across a solenoid, a current of `1.0 A` flows in it. When `100 V` AC is applied across the same coil. The current drops to `0.5A`. If the frequency of the ac source is `50 Hz`, the impedance and inductance of the solenoid areA. `200 Omega` and 0.55 HB. `100 Omega` and 0.86 HC. `200 Omega` and 1.0 HD. 1100`Omega` and 0.93 H
Answer» Correct Answer - A Impedance = `V_AC/I_AC = 100/0.5 = 200 Omega = Z` `R = V_DC/I_DC = 100 / 1 = 100 Omega` `X_L = sqrt Z^(2) - R^(2) = 100sqrt 3Omega = (2pifL)` `L = 100sqrt 3/ 2pifL = 100 xx 1.732 / 2 xx 3.14 xx 50` = 0.55 H

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