In a L-R circuit, the value of L is `((0.4)/(pi))` H and the value of R is `30Omega`. If in the circuit, an alternating emf of 200 V at 50 cycles/s is connected, the impedance of the circuit and current will be

Options

A. `11.4Omega, 17.5A`
B. `30.7Omega, 6.5A`
C. `40.4Omega,5A`
D. `50Omega, 4A`

Answer

D

Explanation

`Z^(2)=R^(2)+(2pifL)^(2)`
`=(30)^(2)+(2pixx50xx(0.4)/(pi))^(2)`
`=(900+1600)=2500`
or `Z=50Omega`
Also, `I=(V)/(Z)=(200)/(50)=4A`