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

Options

A. `4.0 A`
B. `8.0 A`
C. `20/(sqrt(13)) A`
D. `2.0 A`

Answer

A

Explanation

If `omega=50xx2pi`, then `omegaL=20 Omega`
Similarly `omega'=-100xx2pi`, then `omega'L=40 Omega`
`i=-200/Z=200/(sqrt(R^(2)+(omega'L)^(2)))`
`200/(sqrt((30)^(2)+(40)^(2)))=4A`