The figure shows a series $LCR$ circuit connected to a variable frequency $230 \; V$ source. Given $L = 5.0 \; H$,$C = 80 \; \mu F$,and $R = 40 \; \Omega$.
$(a)$ Determine the source frequency which drives the circuit in resonance.
$(b)$ Obtain the impedance of the circuit and the amplitude of current at the resonating frequency.
$(c)$ Determine the $rms$ potential drops across the three elements of the circuit. Show that the potential drop across the $LC$ combination is zero at the resonating frequency.

(N/A) Given: $V_{rms} = 230 \; V$,$L = 5.0 \; H$,$C = 80 \; \mu F = 80 \times 10^{-6} \; F$,$R = 40 \; \Omega$.
$(a)$ The resonant angular frequency is given by $\omega_0 = \frac{1}{\sqrt{LC}}$.
$\omega_0 = \frac{1}{\sqrt{5.0 \times 80 \times 10^{-6}}} = \frac{1}{\sqrt{400 \times 10^{-6}}} = \frac{1}{20 \times 10^{-3}} = 50 \; rad/s$.
$(b)$ At resonance,the impedance $Z = R = 40 \; \Omega$.
The $rms$ current is $I_{rms} = \frac{V_{rms}}{Z} = \frac{230}{40} = 5.75 \; A$.
The amplitude of current $I_0 = \sqrt{2} \times I_{rms} = 1.414 \times 5.75 \approx 8.13 \; A$.
$(c)$ The $rms$ potential drops are:
$V_R = I_{rms} R = 5.75 \times 40 = 230 \; V$.
$X_L = \omega_0 L = 50 \times 5 = 250 \; \Omega$.
$X_C = \frac{1}{\omega_0 C} = \frac{1}{50 \times 80 \times 10^{-6}} = \frac{1}{4000 \times 10^{-6}} = 250 \; \Omega$.
$V_L = I_{rms} X_L = 5.75 \times 250 = 1437.5 \; V$.
$V_C = I_{rms} X_C = 5.75 \times 250 = 1437.5 \; V$.
For the $LC$ combination,$V_{LC} = I_{rms} |X_L - X_C| = 5.75 \times |250 - 250| = 0 \; V$. Thus,the potential drop across the $LC$ combination is zero at resonance.