The alternating current
Question:

The alternating current is given by $i=\left\{\sqrt{42} \sin \left(\frac{2 \pi}{T} t\right)+10\right\} A$

The r.m.s. value of this current is A.

Solution:

$f_{\mathrm{rms}}^{2}=\mathrm{f}_{1 \mathrm{rms}}^{2}+\mathrm{f}_{2 \mathrm{rms}}^{2}$

$=\left(\frac{\sqrt{42}}{\sqrt{2}}\right)^{2}+10^{2}$

$=121 \Rightarrow \mathrm{f}_{\mathrm{ms}}=11 \mathrm{~A}$

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