Question 345: AC fundamentals

Question

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

Accepted Answer

The power dissipated in an AC circuit can be calculated using the formula for average power:
$$ P_{	ext{avg}} = VI \cos \phi $$
where $ V $ and $ I $ are the RMS (root mean square) values of voltage and current, and $ \phi $ is the phase difference between them.
Given:
Voltage $ E(t) = 200 \sin (50t) $ volts
Current $ I(t) = 100 \sin \left(50t + \frac{\pi}{3}ight) $ mA
Convert the peak values to RMS values:
For a sinusoidal function, the RMS value is given by:
$$ V_{	ext{RMS}} = \frac{V_{	ext{peak}}}{\sqrt{2}} $$
$$ I_{	ext{RMS}} = \frac{I_{	ext{peak}}}{\sqrt{2}} $$
Therefore:
$$ V_{	ext{RMS}} = \frac{200}{\sqrt{2}} $$
$$ I_{	ext{RMS}} = \frac{100}{\sqrt{2}} $$
The phase difference $ \phi = \frac{\pi}{3} $.
Calculate the power factor:
The power factor is given by $ \cos \phi $.
Given $ \phi = \frac{\pi}{3} $, we use:
$$ \cos \frac{\pi}{3} = \frac{1}{2} $$
Calculate the average power:
Substituting the values into the power formula:
$$ P_{	ext{avg}} = \left(\frac{200}{\sqrt{2}}ight)\left(\frac{100}{\sqrt{2}}ight)\left(\frac{1}{2}ight) $$
$$ P_{	ext{avg}} = \left(\frac{200 	imes 100}{2}ight) 	imes \frac{1}{2} $$
$$ P_{	ext{avg}} = 10000 	imes \frac{1}{2} $$
$$ P_{	ext{avg}} = 5000 \, 	ext{mW} = 5 \, 	ext{W} $$
Thus, the power dissipated in the circuit is 5 watts.
Option D: 5 watt

Answer

20 watt

Answer

15 watt

Answer

10 watt

Answer

5 watt

AC fundamentals

In an A C circuit E=200 \sin (50 t) volt and \mathrm{I}=100 \sin \left(50 \mathrm{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}

Choose the correct answer: