An AC generator 10 V ( rms ) at ( Rms ) at 200 rad //s is connected in series with a 50 `Omega ` Resistor , a `400 mH` inductor and a `200 muF ` capacitor . The rms voltage across the inductor is

Options

A. 2.5V
B. 3.4V
C. 6.7V
D. 10.8V

Answer

D

Explanation

Answer (4) V = 10V, `omega` = 200 rad/s,
R = `50 Omega, L = 400 mH, C = 200 muF` .
Impedance, z = `sqrt(R^2 + (X_L - X_C)^2)` .
`X_L = omegaL = (200)(`400 xx 10^-3) = 80.
`X_C = 1/omegaC = 1/(200 xx 200 xx 10^-6 = 25` .
`therfore Z = sqrt((50)^2 + (80 - 25)^2) = 74.3 Omega` .
RMS voltage across inductor, .
`(V_L)_rms = I_rms . X_L` .
=` V/Z X_L` = `10/74.3 xx 80` .
Therefore `(V_L)_rms = 10.76V cong 10.8V`