Question

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

Answer 1

Resonant frequency ${\omega }_0$=$\frac{1}{\sqrt{LC}}={10}^{4}rad/s$

New frequency,
$\omega =0.4{\omega }_0\dots \left(i\right)=4\cdot {10}^{3}rsd/s$
New $X_L=\omega L=40\Omega$
New $X_C=\frac{1}{\omega C}=250\Omega$
Amplitude of current is:

$\Rightarrow I=\frac{V_0}{Z}=\frac{50}{\sqrt{R^2+{(\omega L-\frac{1}{\omega C})}^2}}\dots \left(ii\right)$
$\Rightarrow I=\frac{50}{\sqrt{{10}^2+{(40-250)}^2}}\dots \left(ii\right)$

$\Rightarrow I=238\text{mA}$
Hence, the correct option is $c$.