JEE Physics 2024 Question with Solution

Given: $V = 100\sin(100t)$ V and $I = 100\sin\left(100t + \frac{\pi}{3}\right)$ mA.

Find: The average power dissipated in one cycle.

Convert current into ampere:

$$I = 0.1\sin\left(100t + \frac{\pi}{3}\right)\, \text{A}$$

For an AC circuit,

$$P_{\text{avg}} = V_{\text{rms}}\, I_{\text{rms}}\cos\phi$$

where $\phi = \frac{\pi}{3}$.

Now,

$$V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}} = \frac{100}{\sqrt{2}} = 50\sqrt{2}\, \text{V}$$ $$I_{\text{rms}} = \frac{I_{\text{peak}}}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.05\sqrt{2}\, \text{A}$$

Also,

$$\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}$$

Substituting,

$$P_{\text{avg}} = (50\sqrt{2})(0.05\sqrt{2})\left(\frac{1}{2}\right)$$ $$P_{\text{avg}} = 2.5\, \text{W}$$

Therefore, the average power dissipated in one cycle is $2.5\, \text{W}$. The correct option is C.

Given: Peak voltage $V_0 = 100\, \text{V}$, peak current $I_0 = 0.1\, \text{A}$, and phase difference $\phi = \frac{\pi}{3}$.

Find: Average power.

Using

$$P_{\text{avg}} = \frac{V_0 I_0}{2}\cos\phi$$

we get

$$P_{\text{avg}} = \frac{100 \times 0.1}{2} \cos\left(\frac{\pi}{3}\right)$$ $$P_{\text{avg}} = 5 \times \frac{1}{2} = 2.5\, \text{W}$$

This works because both RMS factors each contribute a division by $\sqrt{2}$, giving an overall factor of $\frac{1}{2}$ when peak values are used. Hence, the correct option is C.