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

Options

A. `0 V, 0.47 A`
B. `1.68 V, 0.47 A`
C. `4 sqrt2 V, sqrt2 A`
D. `2 sqrt2 A, o A`

Answer

C

Explanation

Here, `R = 6 + 4 = 10 Omega , omega = 2000 rad s^(-1)`
`X_(L) = omega L = 2000 xx 5 xx 10^(-3) Omega = 10 Omega`
`X_(C ) = (1)/(omega C) = (1)/(2000 xx 50 xx 10^(-6)) = 10 Omega`
`Z = sqrt(R^(2) + (X_(L) - X_(C ))^(2))`
`= sqrt(10^(2) + (10 - 10)^(2)) = 10 Omega`
`I_(0) = (E_(0))/(Z) = (20)/(10) = 2 A`
`I_(upsilon) = (I_(0))/(Z) = (2)/(sqrt2) = sqrt2 A`
This is the ammeter reading. Combined resistance
of `L` and `C = sqrt(r^(2) + (X_(L) - X_(C ))^(2))`
`Z' = sqrt(4^(2) + (10 - 10)^(2)) = 4 ohm`
`:.` Voltmeter reading ` = I_(upsilon) xx Z'`
`= sqrt2 xx 4 = 4 sqrt2 V`