Question

A series AC circuit containing an inductor $$\left(20mH\right)$$, a capacitor $$\left(120\mu F\right)$$ and a resistor $\left(60\mathrm{\Omega }\right)$ is driven by an AC source of $24V/50Hz$. The energy dissipated in the circuit in $60s$ is:
$\begin{array}{rl}& \text{A}\text{. }2.06\times {10}^{3}J\\ & \text{B}\text{. 3}\text{.39}\times {10}^{3}J\\ & \text{C}\text{. 5}\text{.56}\times {10}^{2}J\\ & \text{D}\text{. 5}\text{.17}\times {10}^{2}J\end{array}$

Answer 1

Hint: Energy dissipated or power dissipated in an AC circuit is given by the product of root mean square voltage, root mean square current and the power factor. Then calculate the power factor of the given AC circuit using the formula for power factor. As root mean square voltage value is given, root mean square current can be calculated using the relation between them. After calculating the root mean square voltage, root mean square current and the power factor we can multiply to get the power or energy dissipated in one second. To calculate the energy dissipated in $60s$, we have to multiply $60s$ with the calculated result to get our answer.

Formulas used:
The average power dissipated in a series AC circuit is given by
$P_{av}=V_{rms}.i_{rms}.\cos \varphi$.
$i_{rms}={\displaystyle \frac{V_{rms}}{Z}}$
$\cos \varphi ={\displaystyle \frac{R}{Z}}$
The impedance of series LCR circuit is given by
$Z=\sqrt{R^2+X^2}$
Where $X=X_C-X_L={\displaystyle \frac{1}{\omega C}}-\omega L$

Complete answer:
The average power dissipated in a series AC circuit is given by
$P_{av}=V_{rms}.i_{rms}.\cos \varphi$
Where $V_{rms}$ is the root mean square voltage and $i_{rms}$ is the root mean square current and $\cos \varphi$is the power factor of series AC circuit.
The root mean square current is given by
$i_{rms}={\displaystyle \frac{V_{rms}}{Z}}$
If circuit containing resistance $\left(R\right)$, inductance $\left(L\right)$, and capacitance $\left(C\right)$ connected in series then its power factor is given by
$\cos \varphi ={\displaystyle \frac{R}{Z}}$
Where $Z$ is the net impedance of the circuit
Now the average power is given by
$P_{av}=V_{rms}\times {\displaystyle \frac{V_{rms}}{Z}}\times {\displaystyle \frac{R}{Z}}={\displaystyle \frac{R{V}_{rms}^2}{Z^2}}$
The impedance of series LCR circuit is given by
$Z=\sqrt{R^2+X^2}$
Where $X=X_C-X_L={\displaystyle \frac{1}{\omega C}}-\omega L$
Given $\omega =2\pi f=2\pi \times 50=100\pi$,$C=120\mu F$and $L=20mH$
So
$\begin{array}{rl}& X={\displaystyle \frac{1}{100\pi \times 120\times {10}^{-6}}}-100\pi \times 20\times {10}^{-3}={\displaystyle \frac{1}{3.14\times 1.2\times {10}^4\times {10}^{-6}}}-3.14\times 2\times {10}^3\times {10}^{-3}\\ & \Rightarrow X=26.54-6.28=20.26\mathrm{\Omega }\end{array}$
So now $Z=\sqrt{R^2+X^2}=\sqrt{{60}^2+{20.26}^2}=\sqrt{4010.46}=63.32\mathrm{\Omega }$
Now the average power is
$P_{av}={\displaystyle \frac{R{V}_{rms}^2}{Z^2}}={\displaystyle \frac{60\times {\left(24\right)}^2}{{\left(63.32\right)}^2}}={\displaystyle \frac{34560}{4009.42}}=8.61watt$
This is average power in one second. So the average power in $60s$or the energy dissipated in $60s$ is
$P=8.61\times 60=516.6J=5.166\times {10}^{2}J\simeq 5.17\times {10}^{2}J$

So the correct option is D.

Note:
A circuit containing components capacitor and inductor will help the circuit to store energy. The inductor stores energy in the form of a magnetic field and the capacitor stores the energy in the form of an electric field. Both of these components, for half time, keep on accumulating energy and then provide energy back to the circuit. Therefore a capacitor and an inductor do not consume any power if averaged over full cycle.