A series LCR circuit has $L=0.01 \mathrm{H},R=10 \mathrm{\Omega }$ and $C=1 \mu \mathrm{F}$ and it is connected to ac voltage of amplitude $\left(V_m\right) 50 \mathrm{V}$. At frequency $60\%$ lower than resonant frequency, the amplitude of current will be approximately

For an LCR circuit. the resonant angular frequency is given by, ${\omega }_0=\frac{1}{\sqrt{LC}}={10}^4 \mathrm{rad} {\mathrm{s}}^{-1}$

The given frequency is $60\%$ lower than resonant frequency. Therefore,

${\omega }^{\text{'}}=0.4\times {10}^4=4000 \mathrm{rad} {\mathrm{s}}^{-1}$

Reactance of the capacitor at given frequency,

$X_C={\left({\omega }^{\text{'}}C\right)}^{-1}=250 \mathrm{\Omega }$.

Reactance of the inductor at given frequency,

$X_L=\left({\omega }^{\text{'}}L\right)=40 \mathrm{\Omega }$

Now the amplitude of the current in the given circuit will be,

$i_0=\frac{V_0}{\sqrt{R^2+{\left(X_C^{\text{'}}-X_L^{\text{'}}\right)}^2}}=\frac{50}{\sqrt{{10}^2+{\left(250-40\right)}^2}}=238 \mathrm{mA}$