Q159 / 200 · Manipal (MET) 2011 · Physics
In an AC circuit the potential difference and current are represented respectively by $V=100 \sin (100 t)$ volt,
$I=100 \sin \left(100 t+\frac{\pi}{3}\right)$ milliampere. The power in the circuit is
- Correct answerA$2.5 W$
- B$5 W$
- C$10 W$
- D$10^4 \mathrm{~W}$
Solution
Given, $V=100 \sin 100 t$
$i=100 \sin \left(100 t+\frac{\pi}{3}\right)$
Comparing with standard forms
$\begin{aligned} & V=V_0 \sin \omega t \\ & I=I_0 \sin (\omega t+\phi)\end{aligned}$
$\therefore \quad V_0=100 \mathrm{~V}$
$\begin{aligned} & I_0=100 \mathrm{~mA}=100 \times 10^{-3} \mathrm{~A} \\ & \phi=\frac{\pi}{3}\end{aligned}$
Power $P=\frac{V_0 I_0}{2} \cos \phi$ $\therefore \quad P=\frac{100 \times 100 \times 10^{-3}}{2} \cos \frac{\pi}{3}=2.5 \mathrm{~W}$
Power $P=\frac{V_0 I_0}{2} \cos \phi$ $\therefore \quad P=\frac{100 \times 100 \times 10^{-3}}{2} \cos \frac{\pi}{3}=2.5 \mathrm{~W}$
Hence, the correct answer is option: 'A'.