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.

We have the instantaneous current as

$$i(t)=\sqrt{42}\,\sin\!\left(\frac{2\pi}{T}\,t\right)+10.$$

To find its r.m.s. (root-mean-square) value, we begin with the basic definition: for a periodic current of period $$T$$, the r.m.s. value is

$$I_{\text{rms}}=\sqrt{\frac1T\int_0^T i^2(t)\,dt}.$$

So we must first square the given current:

$$i^2(t)=\Bigl(\sqrt{42}\,\sin\!\left(\tfrac{2\pi}{T}t\right)+10\Bigr)^2.$$

Expanding the square term-by-term, we obtain

$$i^2(t)=\left(\sqrt{42}\,\sin\!\left(\tfrac{2\pi}{T}t\right)\right)^2 +2\left(\sqrt{42}\,\sin\!\left(\tfrac{2\pi}{T}t\right)\right)(10) +10^2.$$

Simplifying each part separately gives

$$i^2(t)=42\,\sin^2\!\left(\tfrac{2\pi}{T}t\right) +20\sqrt{42}\,\sin\!\left(\tfrac{2\pi}{T}t\right) +100.$$

We now take the average of this expression over one complete cycle. To do so, we recall two standard time-averages for a pure sine wave over its period:

$$\frac1T\int_0^T\sin\!\left(\tfrac{2\pi}{T}t\right)\,dt=0,$$

$$\frac1T\int_0^T\sin^2\!\left(\tfrac{2\pi}{T}t\right)\,dt=\frac12.$$

Using these results, the mean (average) of each term is

$$\frac1T\int_0^T 42\,\sin^2\!\left(\tfrac{2\pi}{T}t\right)\,dt =42\left(\frac12\right)=21,$$

$$\frac1T\int_0^T 20\sqrt{42}\,\sin\!\left(\tfrac{2\pi}{T}t\right)\,dt =20\sqrt{42}\,(0)=0,$$

$$\frac1T\int_0^T 100\,dt=100.$$

Adding these average values we have the mean square of the current:

$$\langle i^2(t)\rangle =21+0+100=121.$$

Finally, the r.m.s. value is the square root of this mean square:

$$I_{\text{rms}}=\sqrt{121}=11\ \text{A}.$$

So, the answer is $$11\text{ A}.$$