Question

In an AC circuit, the rms value of the current, $I_{rms}$, is related to the peak current $I_0$ as

• A. $I_{rms}=\frac{1}{\pi }{I}_0$
• B. $I_{rms}=\frac{1}{\sqrt{2}}{I}_0$
• C. $I_{rms}=\sqrt{2}{I}_0$
• D. $I_{rms}=\pi I_0$

Answer 1

Answer: (B)

The alternating current is represented as
$I=I_0\sin (\omega t)$ ................(1)
where, $I_0$ is the peak value of current
Now, formula for a continuous function (or waveform) $f(t)$ defined over the interval $t_1\le t\le t_2$ is
$f_{rms}=\sqrt{\frac{1}{t_2-t_1}\underset{t_1}{\overset{t_2}{\int }}f(t{)}^{2}dt}$
Therefore,
$I_{rms}=\sqrt{\frac{1}{t_2-t_1}\underset{t_1}{\overset{t_2}{\int }}{\left(I_0\sin (\omega t)\right)}^{2}dt}$
since, $I_0$ is positive, we can write
$I_{rms}=I_0\sqrt{\frac{1}{t_2-t_1}\underset{t_1}{\overset{t_2}{\int }}\left({\sin }^2(\omega t)\right)dt}$
Using a trigonometric identity to eliminate squaring of trigonometric function
$I_{rms}=I_0\sqrt{\frac{1}{t_2-t_1}\underset{t_1}{\overset{t_2}{\int }}\frac{1}{2}(1-\cos (2\omega t))dt}$
$I_{rms}=I_0\sqrt{\frac{1}{t_2-t_1}{\left[\frac{t}{2}-\frac{\sin (2\omega t)}{4\omega }\right]}_{t_1}^{t_2}}$
but since the interval is a whole number of complete cycles (as per definition of RMS), the $\$\$\backslash \sin \$\$$ terms will cancel out, leaving
$I_{rms}=I_0\sqrt{\frac{1}{t_2-t_1}{\left[\frac{t}{2}\right]}_{t_1}^{t_2}}$
$I_{rms}=I_0\sqrt{\frac{1}{t_2-t_1}{\left[\frac{t}{2}\right]}_{t_1}^{t_2}}$
$I_{rms}=I_0\sqrt{\frac{1}{t_2-t_1}\frac{t_2-t_1}{2}}$
$I_{rms}=\frac{I_0}{\sqrt{2}}$