In L-C-R series circuit at resonance p.d. between two ends of resistor is 100V. `R = 1kOmega` and C = `2mu`F, if resonant angular frequency is `omega = 200 "rad/s"^(-1)`, p.d. between two ends of inductor of resonance is .....

Options

A. 250 V
B. `4xx10^(-3)` V
C. `2.5xx10^(-10)` V
D. 40 V

Answer

A

Explanation

Resonance angular frequency `omega_0=1/sqrt(LC)`
`therefore L=1/(omega^2C)`
`=1/((200)^2xx2xx10^(-6))`
`=100/8`=12.5 H
p.d. between two ends of inductor,
`V_L=IX_L = I omegaL`
=0.1 x 200 x 12.5
`therefore V_L`= 250 V