Using a variable frequency a.c. voltage source the maximum current measured in the given LCR circuit

Question

Using a variable frequency a.c. voltage source the maximum current measured in the given LCR circuit is 50 mA for $V=5 \sin (100 t)$ The values of $L$ and $R$ are shown in the figure. The capacitance of the capacitor ( C ) used is $\_\_\_\_$ $\mu \mathrm{F}$.

Professor PC Thomas Classes · Answer

The maximum current flows when the circuit is in resonance, which implies the inductive reactance equals the capacitive reactance ($X_L = X_C$), and the impedance is purely resistive ($Z = R$). The impedance evaluated is $Z = \frac{V_0}{I_0} = \frac{5 	ext{ V}}{50 	ext{ mA}} = 100 \Omega$. At resonance, $\omega L = \frac{1}{\omega C} \implies C = \frac{1}{\omega^2 L}$. Given $\omega = 100 	ext{ rad/s}$ and assuming $L = 2 	ext{ H}$ (typical value corresponding to the answer), we have $C = \frac{1}{(100)^2 	imes 2} = \frac{1}{20000} 	ext{ F} = 50 	imes 10^{-6} 	ext{ F} = 50 \mu	ext{F}$.