A series AC circuit containing an inductor (20 mH), a capacitor (120 $$\mu$$F) and a resistor (60 $$\Omega$$) is driven by an AC source of 24 V/50 Hz. The energy dissipated in the circuit in 60 s is:

We begin by recalling that in an AC circuit the average (real) power dissipated as heat in the resistor is given by the well-known relation

$$P = I_{\text{rms}}^{2}\,R,$$

where $$I_{\text{rms}}$$ is the root-mean-square current in the circuit and $$R$$ is the resistance. The total electrical energy converted into heat in a time interval $$t$$ is therefore

$$E = P\,t = I_{\text{rms}}^{2}\,R\,t.$$

So our main task is to find $$I_{\text{rms}}$$ for the given series $$RLC$$ circuit.

For a series combination we may write the impedance $$Z$$ as

$$Z = \sqrt{R^{2} + \left(X_{L}-X_{C}\right)^{2}},$$

where

$$X_{L} = 2\pi f L \qquad\text{(inductive reactance)},$$

$$X_{C} = \frac{1}{2\pi f C} \qquad\text{(capacitive reactance)}.$$

Substituting the numerical data:

$$L = 20\ \text{mH} = 20 \times 10^{-3}\ \text{H},$$ $$C = 120\ \mu\text{F} = 120 \times 10^{-6}\ \text{F},$$ $$f = 50\ \text{Hz},$$ $$R = 60\ \Omega,\quad V_{\text{rms}} = 24\ \text{V}.$$

First we calculate $$X_{L}$$:

$$X_{L} = 2\pi f L = 2\pi \times 50 \times 20\times 10^{-3}$$

$$\;\;= 100\pi \times 0.02$$

$$\;\;= 6.283\ \Omega \;(\text{approximately}).$$

Next we calculate $$X_{C}$$:

$$X_{C} = \frac{1}{2\pi f C} = \frac{1}{2\pi \times 50 \times 120 \times 10^{-6}}$$

$$\;\;= \frac{1}{100\pi \times 120 \times 10^{-6}}$$

$$\;\;= \frac{1}{0.037699}$$

$$\;\;= 26.53\ \Omega \;(\text{approximately}).$$

The net reactance is therefore

$$X = X_{L} - X_{C} = 6.283 - 26.53 = -20.25\ \Omega.$$

(The negative sign merely tells us that the circuit is overall capacitive; we shall use the magnitude in the impedance expression.)

Thus the impedance magnitude is

$$Z = \sqrt{R^{2} + X^{2}} = \sqrt{60^{2} + (-20.25)^{2}}$$

$$\;\;= \sqrt{3600 + 410.06}$$

$$\;\;= \sqrt{4010.06}$$

$$\;\;= 63.36\ \Omega \;(\text{approximately}).$$

Now, Ohm’s law for AC says

$$I_{\text{rms}} = \frac{V_{\text{rms}}}{Z} = \frac{24}{63.36} = 0.3788\ \text{A}\;(\text{approximately}).$$

We can now find the average power dissipated in the resistor:

$$P = I_{\text{rms}}^{2}\,R = (0.3788)^{2} \times 60$$

$$\;\;= 0.1435 \times 60$$

$$\;\;= 8.61\ \text{W}\;(\text{approximately}).$$

The problem asks for the energy dissipated in $$t = 60\ \text{s}$$, so

$$E = P\,t = 8.61 \times 60$$

$$\;\;= 516.6\ \text{J}\;(\text{approximately}).$$

Expressed in scientific notation this is

$$E \approx 5.17 \times 10^{2}\ \text{J}.$$

Hence, the correct answer is Option A.