An L-R-C series circuit has C = 4.80 mF, L = 0.520 H, and source voltage amplitude V = 56.0 V. The source is operated at the resonance frequency of the circuit. If the voltage across the capacitor has amplitude 80.0 V, what is the value of R for the resistor in the circuit?

The resonant angular frequency depends on the inductance and capacitance and it is given by

$\mathbf{\omega }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{LC}}}$,where$\mathit{\omega }$is the angular frequency, L is the inductance and C represents the capacitance.

${\mathbf{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\omega C}}$, where${\mathbf{X}}_{\mathbf{C}}$ is the capacitive reactance.

The current through the capacitor is the same as through the circuit, so it is given by

${\mathbf{I}}_{\mathbf{C}}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{c}}}{{\mathbf{X}}_{\mathbf{c}}}$, where${\mathbf{V}}_{\mathbf{c}}$is the voltage across the capacitor.

By Ohm’s law, Resistance is given by

$\mathbf{R}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{R}}}{\mathbf{I}}$, where ${\mathit{V}}_{\mathbf{R}}$is the voltage across the resistance and I is the current through the circuit

Plug in the values of inductance and capacitance to get resonant angular frequency

$$\begin{gathered} \mathrm{\omega }=\frac{1}{\sqrt{\left(0.520\mathrm{H}\right)\left(4.8\times {10}^{-6}\mathrm{F}\right)}} \\ =633\mathrm{rads}/\mathrm{s} \end{gathered}$$

Calculate capacitive reactance

$$\begin{gathered} {\mathrm{X}}_{\mathrm{c}}=\frac{1}{\left(633\mathrm{rads}/\mathrm{s}\right)\left(4.8\times {10}^{-6}\mathrm{F}\right)} \\ =329\mathrm{\Omega } \end{gathered}$$

And the current through the capacitor and the circuit is

$$\begin{gathered} I_c=\frac{80V}{329\Omega } \\ =0.243A \end{gathered}$$

Therefore Resistance R is

$$\begin{gathered} R=\frac{56V}{0.243A} \\ =230\Omega \end{gathered}$$

Therefore the value of R for the resistor in the circuit is 230Ω.