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

The problem asks us to find the total energy dissipated in a series L-C-R circuit over a period of 60 seconds. At first glance, an AC circuit with three different components might seem intimidating, but the key to solving this lies in understanding where the energy actually goes.

We are given a series circuit with three components: A resistor with resistance

$$R=60\Omega$$

An inductor with inductance * A capacitor with capacitance

The circuit is driven by an AC source providing an RMS voltage at a frequency . We need to find the energy dissipated in .

In any ideal L-C-R circuit, the inductor and capacitor are purely reactive components. They store energy in their magnetic and electric fields respectively, and then release it back into the circuit. They do not dissipate any real power. The only component that dissipates energy as heat is the resistor.

Therefore, the average power dissipated in the circuit is simply the power dissipated across the resistor:

The total energy dissipated over time is:

To find this energy, our primary mission is to calculate the RMS current, .

To find the current, we first need the total opposition to current flow, known as the impedance (). But before we find , we must calculate the individual reactances of the inductor and capacitor.

The inductive reactance () is given by:

The capacitive reactance () is given by:

Now, we find the net reactance:

The negative sign indicates that the capacitive reactance dominates, making the circuit overall capacitive. However, since we will square this value to find the impedance, the sign will not affect the magnitude.

The total impedance () of a series L-C-R circuit is the vector sum of the resistance and the net reactance:

With the total impedance known, we can use Ohm's law for AC circuits to find the RMS current:

We finally have all the pieces of the puzzle. Let's plug the RMS current back into our energy equation:

Expressing this in scientific notation to match the options:

This perfectly matches option (d). The beauty of this problem lies in recognizing that despite the presence of an inductor and a capacitor, the energy dissipation is solely governed by the resistor and the RMS current flowing through it.