Discussion
The r.m.s current of an alternating current given by, \(i = 4\sqrt{2}\sin\omega t + 3\sqrt{2}\cos\omega t\) is
Answer: A
The r.m.s. current is 5 A.
The two terms combine into one sinusoid with peak value
\[ \begin{aligned} I_0 &= \sqrt{(4\sqrt{2})^{2} + (3\sqrt{2})^{2}} = \sqrt{32 + 18} \\ &= \sqrt{50} = 5\sqrt{2}\ \text{A} \\ I_{rms} &= \frac{I_0}{\sqrt{2}} = 5\ \text{A} \end{aligned} \]
\(5\sqrt{2}\) A is the peak value, not the r.m.s. value, and \(7\sqrt{2}\) A simply adds the two amplitudes.
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