Physics - EMI/AC Question with Solution | TestHub

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

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

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

Reactance of the capacitor at given frequency,

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

Reactance of the inductor at given frequency,

$X_L=\left({\omega }^{{}^{\prime }}L\right)=40\Omega$

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

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