A coil has resistance `30 Omega` and inductive reactance `20 Omega` at 50 Hz frequency. If an ac source of 200V, 100 Hz is connected across the coil, the current in the coil will be

Options

A. 2.0A
B. 4.0A
C. 8.0A
D. `20/sqrt13A`

Answer

B

Explanation

In this, case, `R=30Omega`and `X_L=20Omega`
`thereforeX_L=omegaL=2pifL=`[f=frequency]
`thereforeX_L/(X'_L)=f/(f')`or,`X'_L=X_Ltimes((f')/f)=20times(100/50)=40Omega`
`thereforeZ=sqrt(R^2+X'_L^2)=sqrt((30)^2+(40)^2)=50`
`I=V/Z=200/50=4A`