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

The Correct Option is D

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:

\[ 50 \, \text{W}. \]