An ac current is represented as $$i = 5\sqrt{2} + 10\cos\left(650\pi t + \frac{\pi}{6}\right)$$ Amp. The r.m.s value of the current is :

The given current has two parts:
a constant (dc) part : $$I_{dc}=5\sqrt{2}$$ Amp
a sinusoidal (ac) part : $$i_{ac}=10\cos\!\left(650\pi t+\frac{\pi}{6}\right)$$ Amp

For any quantity that is the sum of a dc value and a sinusoidal value, the mean-square over one full cycle is the sum of the individual mean-squares, because the average of the product of the dc term with the ac term over a complete cycle is zero.

Therefore, the r.m.s. value of the total current is given by
$$I_{rms}=\sqrt{I_{dc}^{\,2}+I_{ac,rms}^{\,2}}$$

First find the r.m.s. of the ac component. If the peak (amplitude) of the cosine term is $$I_0$$, then
$$I_{ac,rms}=\frac{I_0}{\sqrt{2}}$$

Here $$I_0=10$$ Amp, so
$$I_{ac,rms}=\frac{10}{\sqrt{2}}=5\sqrt{2}\text{ Amp}$$

Now substitute into the overall r.m.s. formula:
$$I_{rms}=\sqrt{\left(5\sqrt{2}\right)^{2}+\left(5\sqrt{2}\right)^{2}}$$

Calculate the squares:
$$\left(5\sqrt{2}\right)^{2}=25\times2=50$$

Hence
$$I_{rms}=\sqrt{50+50}= \sqrt{100}=10\text{ Amp}$$

So, the r.m.s. value of the current is 10 Amp.

Answer: Option C (10 Amp)