Animated Solution for Physics - Alternating Current: In an AC circuit, the instantaneous emf and current are given by $e=100\sin 30t$, $i=20\sin \left(30t-\frac{\pi }{4}\right)$ In one cycle of AC, the average power consumed by the circuit and the wattless current are, respectively

Have you ever wondered why sometimes electricity flows through a circuit, but no actual 'work' gets done? It sounds like a paradox, right? Welcome to the fascinating world of Alternating Current (AC), where phase differences create magical phenomena like 'wattless current'. Let's dive into this classic JEE problem to unravel the mystery of power dissipation in AC circuits.

We are given the instantaneous equations for electromotive force (voltage) and current in an AC circuit:

By comparing these with the standard forms $e=V_0\sin (\omega t)$ and $i=I_0\sin (\omega t-\varphi )$, we can immediately extract the vital statistics of our circuit. The peak voltage $V_0$ is $100\text{ V}$, and the peak current $I_0$ is $20\text{ A}$.

More importantly, we see a phase difference $\varphi =\frac{\pi }{4}$. The negative sign in the current equation tells us that the current lags behind the voltage by $45^{\circ }$. This phase difference is the key to everything that follows.

Active Component

In a DC circuit, power is simply voltage times current. But in an AC circuit, because the voltage and current are constantly changing and are out of sync, we must use their RMS (Root Mean Square) values and account for the phase difference. The average power consumed over one complete cycle is given by:

The term $\cos \varphi$ is known as the power factor. It represents the fraction of the total apparent power that is actually doing real work. Geometrically, $I_{rms}\cos \varphi$ is the component of the current phasor that is perfectly aligned with the voltage phasor.

Let's plug in our numbers. Remember that $V_{rms}=\frac{V_0}{\sqrt{2}}$ and $I_{rms}=\frac{I_0}{\sqrt{2}}$:

Since $\cos \left(\frac{\pi }{4}\right)=\frac{1}{\sqrt{2}}$, the calculation simplifies beautifully:

Wattless Current

Now, what about the rest of the current? If $I_{rms}\cos \varphi$ is doing the work, what is the perpendicular component, $I_{rms}\sin \varphi$, doing?

This component is perfectly out of phase ($90^{\circ }$) with the voltage. It surges back and forth, temporarily storing energy in magnetic fields (inductors) or electric fields (capacitors) and then returning it to the source. Because it doesn't result in any net energy dissipation (heat), it is famously called the wattless current.

Substituting our values once more:

Since $\sin \left(\frac{\pi }{4}\right)$ is also $\frac{1}{\sqrt{2}}$, we get:

Final Conclusion: The circuit consumes an average power of $\frac{1000}{\sqrt{2}}\text{ W}$ while sustaining a wattless current of $10\text{ A}$. This perfectly matches option (b). Understanding the phasor breakdown of current into active and wattless components is a superpower for mastering AC circuits!