In a given AC circuit which has maximum voltage V0 and frequency 50 Hz. Find the time instant where the current

Answer: (b) 2.5 ms (the options are written "m/s" by mistake; they are milliseconds)

Take the current as \(i = I_0\sin\omega t\), starting from zero at \(t = 0\). The RMS value is \(I_{\text{rms}} = I_0/\sqrt2\). Setting \(i = I_{\text{rms}}\):

\[I_0\sin\omega t = \frac{I_0}{\sqrt2}\ \Rightarrow\ \sin\omega t = \frac{1}{\sqrt2}\ \Rightarrow\ \omega t = \frac{\pi}{4}\]

With \(\omega = 2\pi f = 100\pi\ \mathrm{rad/s}\):

\[t = \frac{\pi/4}{100\pi} = \frac{1}{400}\ \mathrm{s} = 2.5\ \mathrm{ms}\]

This is the first instant; in general it is one-eighth of the period, \(T/8 = 20\ \mathrm{ms}/8\).