A resistance of \(20\Omega\) is connected to a source of alternating potential \(V=220\,\sin (100\pi t)\). The time taken by current to change from its peak value to r.m.s. value is \(\left(\sin \frac{\pi }{4}=\frac{1}{\sqrt{2}}\right)\)
Show Hint
In a resistor the current is in phase with the voltage, so find the phase gap between the peak and the r.m.s. value and divide it by the angular frequency.
Step 1: Find the period
From $\omega=100\pi$ the frequency is $f=50$ Hz, so the period is $T=\dfrac{1}{50}=0.02$ s.
Step 2: Fraction of a cycle
Going from the peak to the r.m.s. value covers a phase of $\dfrac{\pi}{4}$, which is $\dfrac{1}{8}$ of a full cycle $2\pi$.
Step 3: Time taken
$t=\dfrac{T}{8}=\dfrac{0.02}{8}=2.5\times10^{-3}\text{ s}$, which is option (D).
Final Answer:
One eighth of the 0.02 s period gives 2.5 milliseconds.
\[ \boxed{\text{(D) }2.5\times10^{-3}\text{ s}} \]