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 isA. 100 VB. 200 VC. 400 VD. 10 V

200 V ac source is fed to series LCR circuit having `X_(L)=50 Omega, X

1.	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 isA. 100 VB. 200 VC. 400 VD. 10 V
Answer» Correct Answer - C Here, `V_("rms")=200 V, 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)=(200 V)/(25 Omega)=8 A` Voltage drop across the inductor is `V_(L)=I_("rms")X_(L)=8 Axx50 Omega = 400 V`

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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 isA. 100 VB. 200 VC. 400 VD. 10 V

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