When 100V. DC is applied across a solenoid a current of 1A flows in it. When 100V, AC is applied across the same coil, the current drops to 0.5A. The frequency of the AC is 50Hz. The impendence 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`.