A 10 $$\Omega$$ resistance is connected across 220 V - 50 Hz AC supply. The time taken by the current to change from its maximum value to the rms value is:

For a purely resistive circuit the alternating current has the same waveform as the alternating voltage, only scaled by the resistance. Thus, if the instantaneous voltage is written as

$$v(t)=V_{\text m}\sin(\omega t),$$

then the instantaneous current is

$$i(t)=\dfrac{v(t)}{R}=I_{\text m}\sin(\omega t),\qquad\text{where }I_{\text m}= \dfrac{V_{\text m}}{R}.$$

First we express the maximum (peak) and rms values.

Formula: $$V_{\text m}=\sqrt{2}\,V_{\text{rms}},\qquad I_{\text{rms}}=\dfrac{I_{\text m}}{\sqrt{2}}.$$

The supply gives $$V_{\text{rms}}=220\ \text V,$$ so

$$V_{\text m}=\sqrt{2}\times220\ \text V=220\sqrt{2}\ \text V.$$

With the resistance $$R=10\ \Omega$$ we have

$$I_{\text m}=\dfrac{V_{\text m}}{R}=\dfrac{220\sqrt{2}}{10}=22\sqrt{2}\ \text A,$$

and correspondingly

$$I_{\text{rms}}=\dfrac{I_{\text m}}{\sqrt{2}}=22\ \text A.$$

Now we look at how the current changes with time. We write

$$i(t)=I_{\text m}\sin(\omega t).$$

The maximum value occurs when $$\sin(\omega t)=1.$$ The smallest positive angle giving this is $$\omega t=\dfrac{\pi}{2},$$ so the time at that instant is

$$t_1=\dfrac{\pi/2}{\omega}.$$

Next we want the time when the current has fallen to its rms value. We set

$$i(t_2)=I_{\text{rms}}=\dfrac{I_{\text m}}{\sqrt{2}}.$$

Substituting into the waveform,

$$I_{\text m}\sin(\omega t_2)=\dfrac{I_{\text m}}{\sqrt{2}} \;\Longrightarrow\; \sin(\omega t_2)=\dfrac{1}{\sqrt{2}}.$$

The principal angles whose sine equals $$1/\sqrt{2}$$ are $$\omega t_2=\dfrac{\pi}{4}\ \text{or}\ \dfrac{3\pi}{4}.$$

Since the current is decreasing after reaching its maximum at $$\omega t=\pi/2$$, we select the later angle

$$\omega t_2=\dfrac{3\pi}{4}.$$

The angular frequency is determined by the supply frequency

Formula: $$\omega = 2\pi f.$$

With $$f = 50\ \text{Hz},$$

$$\omega = 2\pi \times 50 = 100\pi\ \text{rad s}^{-1}.$$

Hence,

$$t_1=\dfrac{\pi/2}{100\pi}=\dfrac{1}{200}\ \text s=0.005\ \text s,$$

$$t_2=\dfrac{3\pi/4}{100\pi}=\dfrac{3}{400}\ \text s=0.0075\ \text s.$$

The required time interval for the current to drop from its maximum to its rms value is

$$\Delta t = t_2 - t_1 = \left(\dfrac{3}{400}-\dfrac{2}{400}\right)\text s=\dfrac{1}{400}\ \text s=0.0025\ \text s.$$

Converting to milliseconds,

$$\Delta t = 0.0025\ \text s = 2.5\ \text{ms}.$$

Hence, the correct answer is Option A.