1. Write the current equation

For a purely resistive AC circuit, $i = I_0 \sin(\omega t)$ where:

• $I_0$ = maximum current
• $\omega = 2\pi f$

Given frequency: $f = 50\ \text{Hz}$ so, $\omega = 2\pi(50) = 100\pi\ \text{rad/s}$

2. Current at maximum value

The current is maximum when $\sin(\omega t_1)=1$ which gives $\omega t_1 = \frac{\pi}{2}$

3. Current at rms value

The rms value of current is $I_{\text{rms}} = \frac{I_0}{\sqrt{2}}$

So we need the next instant when $i = \frac{I_0}{\sqrt{2}}$ Thus, $I_0\sin(\omega t_2)=\frac{I_0}{\sqrt{2}}$ $\sin(\omega t_2)=\frac{1}{\sqrt{2}}$ After the maximum point, this occurs at $\omega t_2 = \frac{3\pi}{4}$

4. Find the time interval

Therefore, $\Delta t = t_2 - t_1 = \frac{1}{\omega}\left(\frac{3\pi}{4}-\frac{\pi}{2}\right)$ $\Delta t = \frac{1}{\omega}\left(\frac{\pi}{4}\right)$ Substitute $\omega = 100\pi$: $\Delta t = \frac{\pi/4}{100\pi} = \frac{1}{400}\ \text{s}$ $\Delta t = 0.0025\ \text{s} = 2.5\ \text{ms}$

5. Check options

• A: $2.5\ \text{ms}$ ✅
• B: $25\ \text{ms}$ ❌
• C: $2.5\ \text{s}$ ❌
• D: $0.25\ \text{ms}$ ❌

Hence, the correct option is A.