Question

In a given $AC$ circuit through a specific branch, the current varies as a function of time given as
$i=\{\begin{array}{cc}& i_0{\sin }^2\omega t,0\le \omega t<\pi \\ & i_0\sin \omega t,\pi \le \omega t<2\pi \end{array}$
Calculate the average current per cycle of this $\text{AC}$.

• A. $\frac{i_0}{\pi }$
• B. $i_0(\frac{1}{4}-\frac{1}{\pi })$
• C. $i_0(\frac{1}{2}-\frac{1}{\pi })$
• D. $\frac{i_0}{\sqrt{2}}(\pi -1)$

Answer 1

The correct option is B $i_0(\frac{1}{4}-\frac{1}{\pi })$

Given:

Current as a function of time $t$

$i=\{\begin{array}{cc}& i_0{\sin }^2\omega t,0\le \omega t<\pi \\ & i_0\sin \omega t,\pi \le \omega t<2\pi \end{array}$

As, $0\le \omega t<\pi$

$\Rightarrow 0\le t<\frac{\pi }{\omega }$

Also, $T=\frac{2\pi }{\omega }$, where $T=$ Time period

$\therefore 0\le t<\frac{T}{2}$

Similarly, $\pi \le \omega t<2\pi$

$\Rightarrow \frac{T}{2}\le t<T$

So, current as a function of time $t$ can be rewritten as,

$i=\{\begin{array}{cc}& i_0{\sin }^2\omega t,0\le t<\frac{T}{2}\\ & i_0\sin \omega t,\frac{T}{2}\le t<T\end{array}$

The average value of the given $AC$ current can be calculated as

$\Rightarrow I_{avg}=\frac{1}{T}\left({\int }_0^{T/2}\left(i_0{\sin }^2\omega t\right)dt+{\int }_{T/2}^T\left(i_0\sin \omega t\right)dt\right)$

$=\frac{1}{T}\left(\frac{i_0}{2}{\int }_0^{T/2}(1-\cos 2\omega t)dt+i_0{[-\frac{\cos \omega t}{\omega }]}_{T/2}^T\right)$

$=\frac{1}{T}(\frac{i_0}{2}{[t-\frac{1}{2\omega }\sin 2\omega t]}_0^{T/2}-\frac{i_0}{\omega }[1-(-1)\left]\right)$

$=\frac{1}{T}(\frac{i_0}{2}\left(\frac{T}{2}\right)-\frac{2i_0}{\omega })$

$=i_0(\frac{1}{4}-\frac{1}{\pi })$

Hence, $\left(B\right)$ is the correct option.