Question

An alternating current is given by
$i=i_{1}cos\omega t+i_{2}sin\omega t$
The rms is given by

• A. $\frac{i_1+i_2}{\sqrt{2}}$
• B. $\frac{|i_1+i_2|}{\sqrt{2}}$
• C. $\sqrt{\frac{i_1^2+i_2^2}{2}}$
• D. $\sqrt{\frac{i_1^2+i_2^2}{\sqrt{2}}}$

Answer 1

The correct option is C $\sqrt{\frac{i_1^2+i_2^2}{2}}$

$i=i_{1}cos\omega t+i_{2}sin\omega t$
divide and multiply RHS by $\sqrt{{i_1}^2+{i_2}^2}$
$i=\sqrt{{i_1}^2+{i_2}^2}(i_1\frac{cos\omega t}{\sqrt{{i_1}^2+{i_2}^2}}+i_2\frac{sin\omega t}{\sqrt{{i_1}^2+{i_2}^2}})$
Assuming $\frac{i_1}{\sqrt{{i_1}^2+{i_2}^2}}=sin\varphi ⟹\frac{i_2}{\sqrt{{i_1}^2+{i_2}^2}}=cos\varphi$
$i=\sqrt{{i_1}^2+{i_2}^2}(cos\omega t.sin\varphi +sin\omega t.cos\varphi )$
$i=\sqrt{{i_1}^2+{i_2}^2}sin(\omega t+\varphi )--\left(1\right)$
now this is in the form of $i=i_{0}sin(\omega t+\varphi )$
and here i RMS be $\frac{i_0}{\sqrt{2}}$
so similarly i RMS of equation $\left(1\right)$ = $\frac{\sqrt{{i_1}^2+{i_2}^2}}{\sqrt{2}}$