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InterviewSolution
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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 ohm and 0.55 henryB. 100 ohm and 0.86 henryC. 100 ohm and 1.0 henryD. 100 ohm and 0.93 henry. |
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Answer» Correct Answer - A A Solenoid consists of inductance and resistance. When `100 V` dc is applied, `omega = 0 implies Z = R` `Z = (V_("rms"))/(l_("rms")) implies R = (100)/(1) = 100 Omega` When `100 V, 50 Hz` ac is applied, `Z = (V_("rms"))/(l_("rms")) = (100)/(0.5) = 200 Omega` `Z^(2) = R^(2) + X_(L)^(2) implies 200^(2) = (100)^(2) + X_(C )^(2)` `implies X_(L) = - 100 sqrt(3) implies 2 pi f L = 100 sqrt(3)` `implies L = (100 sqrt(3))/(2 xx pi xx 50) = 0.55 H` |
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