Question

For an AC Circuit, the potential difference and current are given by $V=10\sqrt{2}\sin \omega t$ (inV) and $I=2\sqrt{2}\cos \omega t$ (in A) respectively.The power dissipated in the instrument is:

• A. $20W$
• B. $40W$
• C. $40\sqrt{2}W$
• D. Zero

Answer 1

The correct option is D Zero

Given,

$V=10\sqrt{2}sin\omega t$

$I=2\sqrt{2}cos\omega t=2\sqrt{2}sin(\omega t+\frac{\pi }{2})$

The power dissipated in the instrument is,

$P=V_{rms}.I_{rms}.cos\varphi$

$P=\frac{V_0}{\sqrt{2}}.\frac{I_0}{\sqrt{2}}.cos\varphi$

$P=\frac{10\sqrt{2}}{\sqrt{2}}\times \frac{2\sqrt{2}}{\sqrt{2}}\times cos\frac{\pi }{2}$

$P=10\times 2\times 0=0W$

The correct option is D.