In the circuit of fig, the source freqency is `omega=2000 rad s^(-1)`. The current in the will be

Options

A. (A) `2A`
B. (B) `3.3A`
C. (C) `2 //sqrt(5)A`
D. (D) `sqrt(5) A`

Answer

A

Explanation

`(X_L) = L omega = 5 xx 10 ^(-3) xx 2000 = 10 Omega`
`(X_C) = (1)/(C omega)=(1)/(50 xx 10^(-6) xx 2000) =(100)/(10) Omega = 10 Omega`
since `(X_L)=(X_C)` therefore,
`Z=R = 5.9 + 0.10 +4 = 10 Omega`
`(I_V) =(E_V)/(Z)=(20)/(10) A =2A`.