Question 62: AC fundamentals

Question

An alternating current is given by $\mathrm{I}=\mathrm{I}_{\mathrm{A}} \sin \omega \mathrm{t}+\mathrm{I}_{\mathrm{B}} \cos \omega \mathrm{t}$. The r.m.s current will be

Accepted Answer

$$ I_{	ext{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}} $$
To determine the root mean square (r.m.s) value of the alternating current given by
$$ I(t) = I_A \sin (\omega t) + I_B \cos (\omega t), $$
we start by squaring the current:
$$ I(t)^2 = I_A^2 \sin^2 (\omega t) + I_B^2 \cos^2 (\omega t) + 2 I_A I_B \sin (\omega t) \cos (\omega t). $$
The r.m.s value is defined as the square root of the average of this squared current over one complete period. When averaging over a full cycle, we utilize the known averages:
$$ \langle \sin^2 (\omega t) angle = \frac{1}{2}, \quad \langle \cos^2 (\omega t) angle = \frac{1}{2}, \quad \langle \sin (\omega t) \cos (\omega t) angle = 0. $$
Thus, the time-averaged square of the current becomes:
$$ \langle I(t)^2 angle = I_A^2 \frac{1}{2} + I_B^2 \frac{1}{2} = \frac{I_A^2 + I_B^2}{2}. $$
Taking the square root gives the r.m.s current:
$$ I_{	ext{rms}} = \sqrt{\frac{I_A^2 + I_B^2}{2}}. $$
This corresponds to Option B.

Answer

$\frac{\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}}{2}$

Answer

$\sqrt{\frac{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}{2}}$

Answer

$\frac{\left|\mathrm{I}_{\mathrm{A}}+\mathrm{I}_{\mathrm{B}}\right|}{\sqrt{2}}$

Answer

$\sqrt{\mathrm{I}_{\mathrm{A}}^2+\mathrm{I}_{\mathrm{B}}^2}$

AC fundamentals

An alternating current is given by \mathrm{I}=\mathrm{I}_{\mathrm{A}} \sin \omega \mathrm{t}+\mathrm{I}_{\mathrm{B}} \cos \omega \mathrm{t}. The r.m.s current will be

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