In an AC circuit $\mathrm{E}=200 \sin (50 \mathrm{t})$ volt and $I=100 \sin \left(50 t+\frac{\pi}{3}\right)…

In an AC circuit $\mathrm{E}=200 \sin (50 \mathrm{t})$ volt and $I=100 \sin \left(50 t+\frac{\pi}{3}\right) \mathrm{mA}$. The power dissipated in the circuit is $\binom{\sin 30^{\circ}=\cos 60^{\circ}=0.5}{\sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3} / 2}$
  1. 20 watt
  2. 15 watt
  3. 10 watt
  4. 5 watt

Solution

$e=200 \sin (50 t) \text { and } I=100 \sin \left(50 t+\frac{\pi}{3}\right)$
Comparing these equations with the standard forms $\mathrm{e}=\mathrm{e}_0 \sin \omega \mathrm{t}$ and $\mathrm{I}=\mathrm{I}_0 \sin \omega \mathrm{t}$, we get, $\mathrm{e}_0=200 \mathrm{~V}, \mathrm{I}_0=100 \times 10^{-3} \mathrm{~A}, \phi=$ phase difference $=60^{\circ}$ $\begin{aligned} \therefore \quad P=e_{r m s} I_{r m s} \cos \phi & =\frac{e_0}{\sqrt{2}} \times \frac{I_0}{\sqrt{2}} \times \cos 60^{\circ} \\ & =\frac{200}{\sqrt{2}} \times \frac{100 \times 10^{-3}}{\sqrt{2}} \times \frac{1}{2}=5 \mathrm{~W} \end{aligned}$

Asked in: MHT CET 2024 (03 May Shift 1)

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