When 100 V DC is applied across a solenoid , a current of 1 A flows in it. When 100 V AC is applied across the same solenoid 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`
B. `50 Omega`
C. `100 Omega`
D. `50 sqrt(3) Omega`

Answer

A

Explanation

when dc is applied , `R=V/I =(100)/(1) = 100 Omega`
and when ac of 50 Hz is applied
`I=V/Z i.e., Z=V/I = 100/0.5 = 200 Omega`.
`Z=sqrt(R^(2)+omega^(2)L^(2))`
i.e., `omega^(2)L^(2)=Z^(2)-R^(2)`
i.e., `(2 pi fL)^(2)=(200^2)-(100^2)=3xx10^(4)` (`as o,ega = 2 pi f`)
so,` L=(sqrt(3)xx10^(2))/(2 pi xx 50 ) =(sqrt(3))/(pi)H=0.55H`.