1.

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

Answer»

250 V
`4xx10^(-3)` V
`2.5xx10^(-10)` V
40 V

Solution :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


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