Question

The AC current in a circuit is given by $3 \sin(\omega t)$ amperes. The time taken by the current to drop from RMS value to zero is

• A. $\dfrac{\pi}{20}$
• B. $\dfrac{\pi}{40}$
• C. $\dfrac{\pi}{60}$
• D. $\dfrac{\pi}{80}$

Hint:

Use $\sin(\omega t) = \dfrac{1}{\sqrt{2}}$ for RMS point and analyze its drop to next zero crossing.

Answer 1

The Correct Option is B

Given $I(t) = 3 \sin(\omega t)$
RMS value $I_{\text{rms}} = \dfrac{I_0}{\sqrt{2}} = \dfrac{3}{\sqrt{2}}$
Set $I(t) = \dfrac{3}{\sqrt{2}}$ to find time $t$ when current equals RMS value:
$3 \sin(\omega t) = \dfrac{3}{\sqrt{2}} \Rightarrow \sin(\omega t) = \dfrac{1}{\sqrt{2}}$
$\Rightarrow \omega t = \dfrac{\pi}{4} \Rightarrow t = \dfrac{\pi}{4\omega}$
Now, to find time taken to drop from this value to zero, solve for when $\sin(\omega t) = 0$ after this point, which occurs at $\omega t = \dfrac{\pi}{2}$
$\Delta t = \dfrac{\pi}{2\omega} - \dfrac{\pi}{4\omega} = \dfrac{\pi}{4\omega}$
If $\omega = 10\pi$, then $\Delta t = \dfrac{\pi}{4 \cdot 10\pi} = \dfrac{1}{40}$ s = $\dfrac{\pi}{40}$ in angular notation