A resistance of $$40$$ $$\Omega$$ is connected to a source of alternating current rated $$220$$ V, $$50$$ Hz. Find the time taken by the current to change from its maximum value to the rms value :

We are given: resistance $$R = 40\ \Omega$$, AC source rated $$220$$ V, $$50$$ Hz. We need to find the time taken by the current to change from its maximum value to its rms value.

The current is given by $$i = I_0 \sin(\omega t)$$, where $$\omega = 2\pi f = 2\pi \times 50 = 100\pi$$ rad/s.

The maximum current occurs when $$\sin(\omega t_1) = 1$$, i.e., $$\omega t_1 = \frac{\pi}{2}$$. The rms value is $$I_{rms} = \frac{I_0}{\sqrt{2}}$$, so setting $$\frac{I_0}{\sqrt{2}} = I_0 \sin(\omega t_2)$$ gives $$\sin(\omega t_2) = \frac{1}{\sqrt{2}}$$. After the maximum (at $$\omega t = \frac{\pi}{2}$$), the next time this occurs is at $$\omega t_2 = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$.

The difference in phase is $$\Delta(\omega t) = \frac{3\pi}{4} - \frac{\pi}{2} = \frac{\pi}{4}$$, so the time difference is $$\Delta t = \frac{\pi}{4\omega} = \frac{\pi}{4 \times 100\pi} = \frac{1}{400} \text{ s} = 2.5 \times 10^{-3} \text{ s} = 2.5 \text{ ms}$$.

Therefore, the correct answer is Option A.