Question

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

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

Answer 1

The correct option is B $\sqrt{\frac{I_1^2+I_2^2}{2}}$

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

$\Rightarrow I=I_1\sin \omega t+I_2\sin ({90}^{\circ }+\omega t)$

Now,
$I_o=\sqrt{I_1^2+I_2^2+2I_1{I}_2\cos \theta }$

$\Rightarrow I_o=\sqrt{I_1^2+I_2^2+2I_1{I}_2\cos {90}^{\circ }}$

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

Further,

$I_{\text{RMS}}=\frac{I_o}{\sqrt{2}}$

​​​​​​​$\Rightarrow I_{\text{RMS}}=\frac{\sqrt{I_1^2+I_2^2}}{\sqrt{2}}=\sqrt{\frac{I_1^2+I_2^2}{2}}$