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

Answer

A

Explanation

`R=100/1=100 Omega`
`Z=100/0.5 =200 Omega`
`200=sqrt(100^(2)+x_(L)^(2))`
`X_(L)=100sqrt(3)`
`L=(sqrt(3))/(pi)=0.55H`