In an L-R-C series circuit, R = 400 Ω, L = 0.350 H, and C = 0.0120 mF. (a) What is the resonance angular frequency of the circuit? (b) The capacitor can withstand a peak voltage of 670 V. If the voltage source operates at the resonance frequency, what maximum voltage amplitude can it have if the maximum capacitor voltage is not exceeded?

The angular frequency depends on the inductance and the capacitance and it is given by $\mathbf{\omega }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{LC}}}$.

Capacitive reactance is given by${\mathbf{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\omega C}}$

When the angular frequency is varied, the current amplitude will change and the maximum current is produced at the minimum impedance. The process of peaking the current represents the resonance where the initial angular frequency is the same for the final angular frequency.

Plug the values of L and C into equation

$$\begin{gathered} \mathrm{\omega }=\frac{1}{\sqrt{\mathrm{LC}}} \\ =\frac{1}{\sqrt{\left(0.350\mathrm{H}\right)\left(0.0120\times {10}^{-6}\mathrm{F}\right)}} \\ =1.54\times {10}^4\mathrm{rad}/\mathrm{s} \end{gathered}$$

First calculate the capacitive reactance

$$\begin{gathered} X_C=\frac{1}{\left(1.54\times {10}^{4}rad/s\right)\left(0.0120\times {10}^{-6}F\right)} \\ =5410\Omega \end{gathered}$$

The current through the capacitor is given by

$$\begin{gathered} I=\frac{V_c}{X_c} \\ =\frac{550V}{5410\Omega } \\ =0.101A \end{gathered}$$

Now, get the amplitude voltage by

$$\begin{gathered} \mathrm{V}=\left(0.101\mathrm{A}\right)\left(400\mathrm{\Omega }\right) \\ =40.7\mathrm{V} \end{gathered}$$

Hence, (a)The resonance angular frequency of the circuit is$\mathrm{\omega }=1.54\times {10}^4\mathrm{rad}/\mathrm{s}$ and (b) Maximum voltage amplitude it can have if the maximum capacitor voltage is not exceeded is 40.7V.