An AC current is given by I = i₁ coswt + i₂sinwt The rms current is given by:

The correct answer is:

$\frac{1}{\sqrt{2}}{i_1^2+i_2^2}^{\frac{1}{2}}$

Step-by-Step Explanation:

To find the root-mean-square (rms) value of the alternating current given by:

$I=i_1\cos \left(\omega t\right)+i_2\sin \left(\omega t\right)$

We use the definition of the root-mean-square value of a function over a time period $T$:

$I_{\mathrm{rms}}=\sqrt{⟨I^2⟩}$

where $⟨I^2⟩$ denotes the average (mean) value of $I^2$ over one complete cycle.

First, let's square the expression for the current $I$:

$I^2={(i_1\cos \omega t+i_2\sin \omega t)}^2$

Expanding the squared term using the identity ${(a+b)}^2=a^2+b^2+2ab$:

$I^2=i_1^2{\cos }^2\omega t+i_2^2{\sin }^2\omega t+2i_1{i}_2\sin \omega t\cos \omega t$

Now, we take the average of each term over one full cycle of duration $T$:

1. The average value of ${\cos }^2\omega t$ over a full cycle is $\frac{1}{2}$.
2. The average value of ${\sin }^2\omega t$ over a full cycle is $\frac{1}{2}$.
3. The term $2\sin \omega t\cos \omega t$ is equal to $\sin \left(2\omega t\right)$. The average value of a pure sinusoidal wave $\sin \left(2\omega t\right)$ over a full period is $0$.

Substituting these average values, we obtain the mean value $⟨I^2⟩$:

$⟨I^2⟩=i_1^2\left(\frac{1}{2}\right)+i_2^2\left(\frac{1}{2}\right)+0$

$⟨I^2⟩=\frac{1}{2}(i_1^2+i_2^2)$

Finally, we take the square root of the mean squared current to find $I_{\mathrm{rms}}$:

$I_{\mathrm{rms}}=\sqrt{\frac{1}{2}(i_1^2+i_2^2)}$

$I_{\mathrm{rms}}=\frac{1}{\sqrt{2}}{(i_1^2+i_2^2)}^{\frac{1}{2}}$