Q131 / 150 · MHT CET 2024 (16 May Shift 1) · Physics

In LCR series circuit, an alternating e.m.f. 'e' and current ' $i$ ' are given by equations $\mathrm{e}=160 \sin (100 \mathrm{t})$ Volt and $\mathrm{i}=250 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \mathrm{mA}$. The average power dissipated in the circuit is

  • A
    $\quad 2.5 \mathrm{~W}$
  • B
    4.0 W
  • Correct answer
    C
    10 W
  • D
    100 W

Solution

$e=1.60 \sin (100 t)$ volt and $\mathrm{i}=250 \sin \left(100 \mathrm{t}+\frac{\pi}{3}\right) \quad \mathrm{mA}$ Comparing given equations with the standard forms, $e=e_0 \sin \omega t$ and $i=i_0 \sin (\omega t+\phi)$ we get, $\begin{aligned} & \mathrm{e}_0=160 \mathrm{~V}, \mathrm{I}_0=250 \mathrm{~mA} \\ & \therefore \quad \text { Power }=\frac{\mathrm{e}_0}{\sqrt{2}} \cdot \frac{\mathrm{I}_0}{\sqrt{2}} \cos \phi \\ &=\frac{160 \times 250 \times 10^{-3}}{2} \times \cos \left(\frac{\pi}{3}\right) \\ &=\frac{40}{2} \times \frac{1}{2} \\ &=10 \mathrm{~W} \end{aligned}$

Hence, the correct answer is option: 'C'.

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