1. Given data

• Inductance: $L = 20\,\text{mH} = 0.02\,\text{H}$
• Capacitance: $C = 120\,\mu\text{F} = 120 \times 10^{-6}\,\text{F}$
• Resistance: $R = 60\,\Omega$
• AC source: $V = 24\,\text{V},\; f = 50\,\text{Hz}$
• Time: $t = 60\,\text{s}$

We need the energy dissipated in the series AC circuit in 60 s.

2. Angular frequency

$\omega = 2\pi f = 2\pi(50) = 100\pi\,\text{rad/s}$

3. Inductive reactance

$X_L = \omega L = 100\pi \times 0.02 = 2\pi \approx 6.28\,\Omega$

4. Capacitive reactance

$X_C = \frac{1}{\omega C} = \frac{1}{100\pi \times 120\times 10^{-6}}$

$X_C \approx 26.53\,\Omega$

5. Net reactance

$X = X_L - X_C = 6.28 - 26.53 = -20.25\,\Omega$

Magnitude of reactance:

$|X| = 20.25\,\Omega$

6. Impedance of series RLC circuit

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

$Z = \sqrt{3600 + 410.06} = \sqrt{4010.06} \approx 63.3\,\Omega$

7. RMS current

Assuming the given 24 V is the RMS voltage of the AC source,

$I = \frac{V}{Z} = \frac{24}{63.3} \approx 0.379\,\text{A}$

8. Average power dissipated

Only the resistor dissipates energy, so

$P = I^2R = (0.379)^2 \times 60$

$P \approx 8.62\,\text{W}$

9. Energy dissipated in 60 s

$E = Pt = 8.62 \times 60 \approx 517\,\text{J}$

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

10. Checking options

• A: $5.65 \times 10^2\,\text{J}$
• B: $2.26 \times 10^3\,\text{J}$
• C: $5.17 \times 10^2\,\text{J}$
• D: $3.39 \times 10^3\,\text{J}$

The correct option is:

$\boxed{\text{C}}$