In an AC circuit, \( I=100 \sin 200 \pi t \), the time required for the current to achieve its peak value is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \displaystyle\frac{1}{400}\,\mathrm{s} \)
Comparing with \( I = I_0\sin\omega t \), the peak value \( I_0 = 100 \) is reached for the first time when the sine equals 1, that is when \( \omega t = \pi/2 \). Here \( \omega = 200\pi\,\mathrm{rad\,s^{-1}} \).
\( \displaystyle \sin(200\pi t) = 1 \Rightarrow 200\pi t = \frac{\pi}{2} \)
\( \displaystyle t = \frac{1}{400}\,\mathrm{s} \)
Shortcut: The time period is \( T = 2\pi/\omega = 1/100\,\mathrm{s} \). A sine wave starting from zero reaches its peak after a quarter period, \( T/4 = 1/400\,\mathrm{s} \).