An AC source of voltage $V=200\sin (100t)$ is connected to a series LCR circuit with $L=2\text{ H}$, $C=50\mu \text{F}$, and $R=10\Omega$. What is the average power dissipated in the circuit?

Solution

1. Given $V=200\sin (100t)$, we identify the peak voltage $V_0=200\text{ V}$ and angular frequency $\omega =100\text{ rad/s}$.
2. Calculate the reactances: $$\begin{array}{rl}{X}_L& =\omega L\\ & =100\times 2\\ & =200\Omega \end{array}$$ and $$\begin{array}{rl}{X}_C& =\frac{1}{\omega C}\\ & =\frac{1}{100\times 50\times 10^{-6}}\\ & =200\Omega \end{array}$$.
3. Since $X_L=X_C$, the circuit is in resonance, meaning the impedance $$\begin{array}{rl}Z& =R\\ & =10\Omega \end{array}$$.
4. Calculate the RMS voltage: $$\begin{array}{rl}{V}_{\text{rms}}& =\frac{V_0}{\sqrt{2}}\\ & =\frac{200}{\sqrt{2}}\text{ V}\end{array}$$.
5. Calculate the average power: $$\begin{array}{rl}P& =\frac{V_{\text{rms}}^2}{R}\\ & =\frac{(200/\sqrt{2}{)}^2}{10}\\ & =\frac{20000}{10}\\ & =2000\text{ W}\end{array}$$.
6. Hence the answer is (A).