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Step-by-Step Solution
Step 1: Express the instantaneous current
For a purely resistive AC circuit, the instantaneous current is given by:
$i = i_0 \sin(\omega t)$,
where $i_0$ is the peak (maximum) current, and $\omega$ is the angular frequency given by
$ \omega = 2\pi f$.
Step 2: Determine the time when current is at its maximum value
At maximum current, $i = i_0$. Thus,
$i_0 = i_0 \sin(\omega t_1) \implies \sin(\omega t_1) = 1 \implies \omega t_1 = \frac{\pi}{2}.$
Therefore,
$t_1 = \frac{\pi}{2\omega}.$
Step 3: Determine the time when current is at its RMS value
The RMS value of the current is $i_{\mathrm{rms}} = \frac{i_0}{\sqrt{2}}.$ Hence,
$\frac{i_0}{\sqrt{2}} = i_0 \sin(\omega t_2)
\implies \sin(\omega t_2) = \frac{1}{\sqrt{2}}
\implies \omega t_2 = \frac{\pi}{4}.$
Therefore,
$t_2 = \frac{\pi}{4\omega}.$
Step 4: Calculate the time interval from maximum current to RMS current
The required time interval is:
$(t_1 - t_2) = \frac{\pi}{2\omega} - \frac{\pi}{4\omega}
= \frac{\pi}{4\omega}.$
Step 5: Substitute $\omega = 2\pi f$ and evaluate
Since $\omega = 2\pi f,$ we have
\[
t_1 - t_2
= \frac{\pi}{4 \cdot 2\pi f}
= \frac{1}{8 f}.
\]
Given $f = 50\ \mathrm{Hz},$
\[
t_1 - t_2 = \frac{1}{8 \times 50} = \frac{1}{400} \ \text{second}.
\]
Converting to milliseconds,
\[
\frac{1}{400}\ \mathrm{s} = 2.5\ \mathrm{ms}.
\]
Final Answer
The time taken by the current to change from its maximum value to its RMS value is 2.5 ms.
