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Q. $200\,V$ 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

COMEDKCOMEDK 2008Alternating Current

Solution:

Here, $V_{rms} = 200\,V$,
$X_L=50\,\Omega$,
$X_C=50\,\Omega$,
$R=25\,\Omega$
Impedance of the circuit,
$Z=\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}$
$=\sqrt{25^{2}+\left(50-50\right)^{2}}$
$=25\,\Omega$
Current in the circuit, $I_{rms}=\frac{V_{rms}}{Z}$
$=\frac{200\,V}{25\,\Omega}$
$=8\,A$
Voltage drop across the inductor is
$V_{L}=I_{rms}X_{L}$
$=8\,A\times50\,\Omega$
$=400\,V$