Question

The frequency of an alternating current is 50 Hz. What is the minimum time taken by current to reach its peak value from rms value?

• A. $0.02s$
• B. $5\times 10^{-3}$ s
• C. $10\times 10^{-3}$ s
• D. $2.5\times 10^{-3}$ s

Answer 1

Answer: (D)

We know that $V_{rms}=V_{0}sin\omega t$
$\frac{V_0}{\sqrt{2}}=V_{0}sin\omega t$
$\therefore sin\omega t=\left(\frac{1}{\sqrt{2}}\right)$
$i.e.,\omega t=\frac{\pi }{4}\Rightarrow \frac{2\pi }{T}.t=\frac{\pi }{4}$
$\therefore t=\frac{T}{8}$
Time for current to reach from rms value to peak value is i.e.,$l_{rms}\to l_0$
$\frac{T}{4}-\frac{T}{8}=\frac{T}{8}\left(\frac{I_0}{\sqrt{2}}\to I_0\right)$
$f=50\therefore t=\frac{I}{8}$
$=\frac{1}{50\times 8}=\frac{1}{400}=0.25\times 10^{-2}$
$=2.5\times 10^{-3}sec\therefore \left[D\right]iscorrect.$