Question

An AC voltage $V = 20 \sin 200 \pi t$ is applied to a series LCR circuit which drives a current $I = 10 \sin \left( 200 \pi t + \frac{\pi}{3} \right)$. The average power dissipated is:

• A. $21.6 \, \text{W}$
• B. $200 \, \text{W}$
• C. $173.2 \, \text{W}$
• D. $50 \, \text{W}$

Answer 1

Answer: (D)

$50 \, \text{W}$

Answer 2

The average power dissipated $\langle P \rangle$ in an AC circuit with voltage $V$ and current $I$ is given by:

$\langle P \rangle = V_{\text{rms}} I_{\text{rms}} \cos \phi,$

where $V_{\text{rms}}$ is the root mean square (RMS) value of the voltage, $I_{\text{rms}}$ is the RMS value of the current, and $\phi$ is the phase difference between the voltage and current.

The given voltage is $V = 20 \sin 200 \pi t$, so the peak voltage $V_0$ is 20 V. The RMS value of the voltage $V_{\text{rms}}$ is:

$V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, \text{V}.$

Similarly, the given current is $I = 10 \sin (200 \pi t + \frac{\pi}{3})$, so the peak current $I_0$ is 10 A. The RMS value of the current $I_{\text{rms}}$ is:

$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, \text{A}.$

The current leads the voltage by $\frac{\pi}{3}$, meaning the phase difference $\phi$ is:

$\phi = \frac{\pi}{3} = 60^\circ.$

Since $\phi = 60^\circ$:

$\cos \phi = \cos 60^\circ = \frac{1}{2}.$

Now, substitute the values of $V_{\text{rms}}$, $I_{\text{rms}}$, and $\cos \phi$ into the formula for average power:

$\langle P \rangle = V_{\text{rms}} I_{\text{rms}} \cos \phi = (10\sqrt{2})(5\sqrt{2}) \cdot \frac{1}{2}.$

Simplify the expression:

$\langle P \rangle = (10 \cdot 5 \cdot 2) \cdot \frac{1}{2} = 50 \, \text{W}.$

The average power dissipated in the circuit is:

$50 \, \text{W}.$