When 100 V dc is applied across a solinoid, 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 Omega and 0.55 H`
B. `100 Omega and 0.86 H`
C. `200 Omega and 0.1 H`
D. `100 Omega and 0.93 H`

Answer

A

Explanation

`R=(E )/(I) (100)/(1) 100 Omega`
for ac : `Z=[R^(2)+(2 pi f L)^(2)]^(1//2)`
`(E_v)/(I_v) =(100)/(0.5) 200 Omega`
or ` 200 =[(100^(2)+(100 pi L)^(2)]^(1//2)`
Solving, we get `L=0.55 H`.