The alternating current is given by $\mathrm{i}=\left\{\sqrt{42}\sin \left(\frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{t}\right)+10\right\}\mathrm{A}$. The rms value of this current is ________ A.

Given equation of alternating current
$\mathrm{i}=\left[\sqrt{42}\sin \left(\frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{t}\right)+10\right]\mathrm{A}$
From given equation we get 
i=i₁+i₂
Where, ${\mathrm{i}}_1=\sqrt{42}\sin \left(\frac{2\mathrm{\pi }}{\mathrm{T}}\mathrm{t}\right)\mathrm{A}\text{ and }{\mathrm{i}}_2=10\mathrm{A}$
Now, i₁ is oscillating current, whereas i₂ is direct current and its value does not change with time.
$\begin{array}{l}{\left({\mathrm{i}}_1\right)}_{\mathrm{rms}}=\frac{\sqrt{42}}{\sqrt{2}}=\sqrt{21}\mathrm{A}\\ {\left({\mathrm{i}}_2\right)}_{\mathrm{rms}}=10\mathrm{A}\end{array}$
We know that
${\mathrm{i}}_{\mathrm{rms}}^2={\left({\mathrm{i}}_1\right)}_{\mathrm{rms}}^2+{\left({\mathrm{i}}_2\right)}_{\mathrm{rms}}^2$
Substituting the values we get
$\begin{array}{l}{\mathrm{i}}_{\mathrm{rms}}=\sqrt{(\sqrt{21}{)}^2+{10}^2}=\sqrt{121}\\ \Rightarrow {\mathrm{i}}_{\mathrm{rms}}=11\mathrm{A}\end{array}$