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**Hint:**The average value of the alternating current always tends to zero. Hence the root means square value is used. From the given alternating current, square its value. Taking the square root of the mean value provides the value of the root means square of the alternating current.

**Formula used:**

(1) The formula of the root means square value of the current is given by

${I_{rms}} = \sqrt {{I^2}} $

Where ${I_{rms}}$ is the root means square value of the current and $I$ is the mean value of the current.

(2) The algebraic formula is given by

${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$

(3) The trigonometric formula is given by

$\sin 2\theta = 2\sin \theta \cos \theta $

**Complete step by step solution:**

It is given that the alternating current, $I = {I_1}\cos \omega t + {I_2}\sin \omega t$

For getting the mean value of the current, square the given alternating current,

$\Rightarrow {I^2} = {\left( {{I_1}\cos \omega t + {I_2}\sin \omega t} \right)^2}$

By using the formula of the ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ , we get

$\Rightarrow {I^2} = I_1^2{\cos ^2}\omega t + {I_2}^2{\sin ^2}\omega t + 2{I_1}{I_2}\cos \omega t\sin \omega t$

Since the instantaneous value of the alternating current is ${45^ \circ }$ , substitute this value in the above equation, we get

$\Rightarrow {I^2} = \dfrac{{I_1^2}}{2} + \dfrac{{{I_2}^2}}{2} + 2{I_1}{I_2}\sin 2\omega t$

By simplifying the above step, we get

$\Rightarrow {I^2} = \dfrac{{I_1^2}}{2} + \dfrac{{{I_2}^2}}{2} + 2{I_1}{I_2} \times 0$

By further simplification,

$\Rightarrow {I^2} = \dfrac{{{I_1}^2}}{2} + \dfrac{{{I_2}^2}}{2}$

$\Rightarrow {I^2} = \dfrac{{{I_1}^2 + {I_2}^2}}{2}$

By taking the square root on both sides, in order to neglect the square in the left hand side of the equation.

$\Rightarrow I = \sqrt {\dfrac{{{I_1}^2 + {I_2}^2}}{2}} $

The value of the root means the square value of the current is obtained as $I = \sqrt {\dfrac{{{I_1}^2 + {I_2}^2}}{2}} $ .

**Thus the option (C) is correct.**

**Note:**In the above solution, $\sin \omega t$ and the $\cos \omega t$ is substituted as the $\dfrac{1}{{\sqrt 2 }}$ . Also remember the formula, $\sin 2\theta = 2\sin \theta \cos \theta $ . From the above formula, it is framed as $\cos \theta \sin \theta = 2\sin 2\theta $ . The root means the square value of the sinusoidal wave will produce the same heating effect similar to that of direct current.

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