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(A) The given equation is $i = i_{1} \sin \omega t + i_{2} \cos \omega t$.We can rewrite $\cos \omega t$ as $\sin(\omega t + 90^{\circ})$.So,$i = i_{1

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$\frac{1}{\sqrt{2}}(i_{1}^{2} + i_{2}^{2})^{1/2}$

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(A) The given equation is $i = i_{1} \sin \omega t + i_{2} \cos \omega t$.We can rewrite $\cos \omega t$ as $\sin(\omega t + 90^{\circ})$.So,$i = i_{1} \sin \omega t + i_{2} \sin(\omega t + 90^{\circ})$.This represents the superposition of two sinusoidal currents with a phase difference of $90^{\circ}$.The resultant peak current $i_{0}$ is given by $i_{0} = \sqrt{i_{1}^{2} + i_{2}^{2} + 2i_{1}i_{2} \cos(90^{\circ})}$.Since $\cos(90^{\circ}) = 0$,we get $i_{0} = \sqrt{i_{1}^{2} + i_{2}^{2}}$.The root mean square (rms) current is defined as $i_{rms} = \frac{i_{0}}{\sqrt{2}}$.Therefore,$i_{rms} = \frac{\sqrt{i_{1}^{2} + i_{2}^{2}}}{\sqrt{2}} = \frac{1}{\sqrt{2}}(i_{1}^{2} + i_{2}^{2})^{1/2}$.

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

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