In the circuit shown in figure the `AC` source gives a voltage `V=20cos(2000t)`. Neglecting source resistance, the voltmeter and and ammeter readings will be
.

Options

A. 0V, 0.47A
B. 1.68V, 0.47A
C. 0V, 1.4 A
D. 5.6V, 1.4 A

Answer

D

Explanation

`V = 20 cos (2000 t)`
`X_(L) = omega L = 2000 xx 5 xx 10 ^(-3) = 10 Omega `
`X_(C) = (1)/(omega C) = (1)/(2000 xx 5 xx 10^(-3) = 10 Omega `
`X_(C) = (1)/(omegaC) = (1)/( 2000 xx 50 xx 10^(-6)) = 10 Omega `
Since ` X_(L) = X_(C) and Z = 10 Omega = R `
`i_(max) = (V_(0))/(Z) = (20)/(10) = 2A , i_(max) = (I_(max))/(sqrt(2)) = (2)/( sqrt(2)) = 1.4 A `
and `V_(rms) = 4 xx 1.41 = 5.64 ` Volt