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

Options

A. 100 V
B. 200 V
C. 400 V
D. 10 V

Answer

C

Explanation

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`