Question

The peak value of an alternating current $5A$ and its frequency is $60Hz$. How long will the current take to reach the peak value starting from zero ?

• A. $\frac{1}{120}s$
• B. $\frac{3}{240}s$
• C. $\frac{1}{140}s$
• D. $\frac{1}{340}s$

Answer 1

The correct option is A $\frac{1}{120}s$

An $ac$ current can be represented as $I=i_0\sin \omega t$

(where $i_0$ is the peak value of ac current and $\omega$ is the angular frequency of current. $I=i_0\sin \omega t$)

We are given peak value $i_0$ to be $5$ and $f=60H_z$.

So, $i=5\sin 120\pi t$.

Now, RMS value $\frac{i_0}{\sqrt{2}}=\frac{5}{\sqrt{2}}$

Time taken to reach peak value from $0$ is $\frac{T}{4}=\frac{1}{4f}$

$\frac{T}{4}=\frac{1}{120}=4.16\times {10}^{-3}s$.