Question

The current in an AC circuit is given by I = Io
sin (50π)t, where t is in second. The time interval
in which it will reach from zero to its peak value is:

(1) 10 ms
(2) 5 ms
(3) 20 ms
(4) 15 ms​

Answer 1

Alternating current , I = I₀ sin(50π)t

Peak value of current = I₀

Let time taken to reach peak value from zero be t

$\to \sf I_o=I_osin(50 \pi)t\\\\\to \sf 1=sin(50\pi)t\\\\\to \sf sin(50\pi)t=1\\\\\to \sf sin(50\pi)t=sin\dfrac{\pi}{2}\\\\\to \sf 50\pi t=\dfrac{\pi}{2}\\\\\to \sf 50t=\dfrac{1}{2}\\\\\to \sf t=\dfrac{1}{2\times 50}\\\\\to \sf t=\dfrac{1}{100}\ s\\\\\to \sf t=\dfrac{10}{1000}s\\\\\to \sf t=10\times 10^{-3}\ s\\\\\to \sf t=10\ ms$

Option (1)