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 are

Options

A. 200 ohm and 0.55 henry
B. 100 ohm and 0.86 henry
C. 100 ohm and 1.0 henry
D. 100 ohm and 0.93 henry.

Answer

A

Explanation

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`