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200 V ac source is fed to series LCR cir...

200 V ac source is fed to series LCR circuit having `X_(L)=50 Omega, X_(C )=50 Omega` and `R = 25 Omega`. Potential drop across the inductor is

A

100V

B

200V

C

400V

D

10V

Text Solution

Verified by Experts

The correct Answer is:
C
Here , `V_(rms) = 200V`
`X_L = 50 Omega , X_C = 50 Omega , R = 25 Omega`
Impedance of the circuit ,
`Z = sqrt(R^2 + (X_L - X_C)^2) = sqrt(25^2 + (50-50)^2) = 25 Omega`
Current in the circuit , `I_(rms) = (V_(rms))/(Z) = (200V)/(25Omega) = 8A`
Voltage drop across the inductor is
`V_L =I_(rms) X_L = 8A xx 50 Omega = 400 V`
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