Question

An AC current is given by$I=I_1\sin \omega t+I_2\cos \omega t$. A hot wire ammeter will give a reading :

• A. $\begin{array}{l}\frac{I_1+I_2}{\sqrt{2}}\end{array}$
• B. $\begin{array}{l}\sqrt{\frac{I_1^2+I_2^2}{2}}\end{array}$
• C. $\begin{array}{l}\sqrt{\frac{I_1^2-I_2^2}{2}}\end{array}$
• D. $\begin{array}{l}\frac{I_1+I_2}{2\sqrt{2}}\end{array}$

Answer 1

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:

$I=I_1\sin \omega t+I_2\cos \omega t$

Let us substitute $I_1=I_0\sin \left(\varphi \right)$………$\left(1\right)$

$I_2=I_0\cos \left(\varphi \right)$………$\left(2\right)$

Step 2: Finding the Root Mean Square(RMS) value of the current:

Squaring and adding equations $\left(1\right)$ & $\left(2\right)$

we get, ${I_1}^2+{I_2}^2={I_0}^2[{\sin }^2\left(\varphi \right)+{\cos }^2\left(\varphi \right)]$

$\Rightarrow \sqrt{{I_1}^2+{I_2}^2}=I_0$

Using, ${\mathrm{I}}_{\mathrm{RMS}}=\frac{{\mathrm{I}}_0}{\sqrt{2}}$

$\Rightarrow {\mathrm{I}}_{\mathrm{RMS}}=\frac{\sqrt{{I_1}^2+{I_2}^2}}{\sqrt{2}}$

$\Rightarrow {\mathrm{I}}_{\mathrm{RMS}}=\sqrt{\frac{{I_1}^2+{I_2}^2}{2}}$

Hence, option B is correct.