JEE Physics 2024 Question with Solution

Given: $V(t) = 220 \sin(100\pi t)$ and the load is purely resistive with $R = 50\,\Omega$.

Find: The time taken for current to rise from half of its peak value to its peak value.

For a purely resistive circuit, current is in phase with voltage, so

$$I(t) = \frac{V(t)}{R} = \frac{220\sin(100\pi t)}{50} = 4.4\sin(100\pi t)$$

Hence, the peak current is $I_{\text{peak}} = 4.4\,\text{A}$.

When the current is half of the peak value,

$$I(t_1) = \frac{I_{\text{peak}}}{2}$$

So,

$$4.4\sin(100\pi t_1) = 2.2$$ $$\sin(100\pi t_1) = \frac{1}{2}$$

For the rising part of the first quadrant,

$$100\pi t_1 = \frac{\pi}{6}$$ $$t_1 = \frac{1}{600}\,\text{s}$$

When the current reaches peak value,

$$I(t_2) = I_{\text{peak}}$$

So,

$$4.4\sin(100\pi t_2) = 4.4$$ $$\sin(100\pi t_2) = 1$$

Thus,

$$100\pi t_2 = \frac{\pi}{2}$$ $$t_2 = \frac{1}{200}\,\text{s}$$

Therefore, the required time interval is

$$\Delta t = t_2 - t_1 = \frac{1}{200} - \frac{1}{600} = \frac{1}{300}\,\text{s}$$ $$\Delta t = 3.33\,\text{ms}$$

Therefore, the time taken is $3.33\,\text{ms}$.

Given: $V(t) = 220\sin(100\pi t)$, so the angular frequency is $\omega = 100\pi\,\text{rad/s}$.

Find: The time interval for current to increase from $\frac{1}{2}I_{\text{peak}}$ to $I_{\text{peak}}$.

In a purely resistive AC circuit, current and voltage are in phase. Therefore, it is enough to track the sine function.

For a sine wave:

• $\sin\theta = \frac{1}{2}$ at $\theta = \frac{\pi}{6}$ in the rising first quadrant.
• $\sin\theta = 1$ at $\theta = \frac{\pi}{2}$.

So the required phase change is

$$\Delta \theta = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}$$

Using $\Delta t = \frac{\Delta \theta}{\omega}$,

$$\Delta t = \frac{\pi/3}{100\pi} = \frac{1}{300}\,\text{s}$$

Converting into milliseconds,

$$\Delta t = \frac{1000}{300}\,\text{ms} = 3.33\,\text{ms}$$

So, the required time interval is $3.33\,\text{ms}$.

The solution also contains a discrepancy: one approach writes $t = \frac{\pi/6}{\omega}$ directly as the required interval, but the detailed working correctly computes the interval as the difference between the two instants, giving $\frac{1}{300}\,\text{s}$.