Question

# An AC current is given by$I={I}_{1}\mathrm{sin}\omega t+{I}_{2}\mathrm{cos}\omega t$. A hot wire ammeter will give a reading :

A

$\begin{array}{l}\frac{{I}_{1}+{I}_{2}}{\sqrt{2}}\end{array}$

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B

$\begin{array}{l}\sqrt{\frac{{I}_{1}^{2}+{I}_{2}^{2}}{2}}\end{array}$

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C

$\begin{array}{l}\sqrt{\frac{{I}_{1}^{2}-{I}_{2}^{2}}{2}}\end{array}$

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D

$\begin{array}{l}\frac{{I}_{1}+{I}_{2}}{2\sqrt{2}}\end{array}$

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Solution

## The correct option is B $\begin{array}{l}\sqrt{\frac{{I}_{1}^{2}+{I}_{2}^{2}}{2}}\end{array}$Explanation for the correct option:Step 1: Given data:$I={I}_{1}\mathrm{sin}\omega t+{I}_{2}\mathrm{cos}\omega t$Let us substitute ${I}_{1}={I}_{0}\mathrm{sin}\left(\varphi \right)$………$\left(1\right)$ ${I}_{2}={I}_{0}\mathrm{cos}\left(\varphi \right)$………$\left(2\right)$Step 2: Finding the Root Mean Square(RMS) value of the current:Squaring and adding equations $\left(1\right)$ & $\left(2\right)$we get, ${{I}_{1}}^{2}+{{I}_{2}}^{2}={{I}_{0}}^{2}\left[{\mathrm{sin}}^{2}\left(\varphi \right)+{\mathrm{cos}}^{2}\left(\varphi \right)\right]$ $⇒\sqrt{{{I}_{1}}^{2}+{{I}_{2}}^{2}}={I}_{0}$ Using, ${\mathrm{I}}_{\mathrm{RMS}}=\frac{{\mathrm{I}}_{0}}{\sqrt{2}}$$⇒{\mathrm{I}}_{\mathrm{RMS}}=\frac{\sqrt{{{I}_{1}}^{2}+{{I}_{2}}^{2}}}{\sqrt{2}}$$⇒{\mathrm{I}}_{\mathrm{RMS}}=\sqrt{\frac{{{I}_{1}}^{2}+{{I}_{2}}^{2}}{2}}$Hence, option B is correct.

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