Question

In an L-R-C series circuit the source is operated at its
resonant angular frequency. At this frequency, the reactance XC
of the capacitor is 200 Ω and the voltage amplitude across the
capacitor is 600 V. The circuit has R = 300 Ω. What is the voltage
amplitude of the source?

Short answer

The Voltage amplitude of the source is 900V.

Step-by-step solution

Step-1: Formulas used  

${\mathit{I}}_{\mathbf{r}\mathbf{m}\mathbf{s}}$represents the d.c. current that dissipates the same amount of power as the average power dissipated by the alternating current/voltage.

$I_{\mathrm{rms}}=\frac{V_{rms}}{Z}$

$V_{rms}$and$I_{rms}$ are the rms voltage and current respectively.

Z is defined as the impedance of the circuit which is the effective resistance of an electric circuit or component to alternating current, arising from the combined effects of ohmic resistance and reactance.

$Z\sqrt{R^2+{\left(X_L-X_{C)}\right)}^2}$, where Z is the impedance

The equivalent Ohm’s law relation to get the amplitude voltage V in the circuit.

V=IZ

Similarly, the amplitude voltage across the resistor, capacitor and inductor is found by the relation

V=IX

Where X is the reactance which is equal to${\mathbf{X}}_{\mathbf{L}}\mathbf{=}\mathbf{\omega L}$ for inductor and${\mathbf{X}}_{\mathbf{C}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{\omega C}}$ for capacitor and Rfor resistor.

Also ${\mathbf{P}}_{\mathbf{av}}\mathbf{=}{\mathbf{I}}^{\mathbf{2}}\mathbf{Z}$, where Z is the impedance.

The process of peaking the current at a particular frequency represents the resonance and that frequency is known as resonant frequency.

$\mathbf{\omega }\mathbf{=}\frac{\mathbf{1}}{\sqrt{\mathbf{LC}}}$,whereis the resonant angular frequency.

At resonance, ${\mathbf{X}}_{\mathbf{L}}\mathbf{=}{\mathbf{X}}_{\mathbf{C}}$.

At resonance, the impedance is minimum and this minimum value is same as that of resistor.

Step-2: Calculations for Voltage amplitude of the source

${\mathbf{V}}_{\mathbf{C}}\mathbf{=}{\mathbf{IX}}_{\mathbf{C}}$

$$\begin{gathered} {\mathrm{V}}_{\mathrm{C}}=600\mathrm{V} \\ {\mathrm{X}}_{\mathrm{C}}=300\mathrm{\Omega } \end{gathered}$$

$$\begin{gathered} \mathrm{I}=\frac{600\mathrm{V}}{200\mathrm{\Omega }} \\ =3\mathrm{A} \end{gathered}$$

At resonance,

role="math" localid="1664180680979" $$\begin{gathered} \mathrm{Z}=\mathrm{R} \\ \mathrm{R}=300\mathrm{\Omega } \\ \mathrm{So},\mathrm{Z}=300\mathrm{\Omega } \end{gathered}$$

Now plug the values for I and Z to get V

$$\begin{gathered} V=\left(3A\right)\left(300\Omega \right) \\ =900V \end{gathered}$$

Therefore, the voltage amplitude of the source is 900V.