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:
- \( 21.6 \, \text{W} \)
- \( 200 \, \text{W} \)
- \( 173.2 \, \text{W} \)
- \( 50 \, \text{W} \)
The Correct Option is D
Approach Solution - 1
To determine the average power dissipated in the circuit, we start by identifying the expressions for the AC voltage and current provided:
- Voltage: \(V = 20 \sin 200 \pi t\)
- Current: \(I = 10 \sin \left(200 \pi t + \frac{\pi}{3}\right)\)
The average power dissipated in an AC circuit can be expressed as:
\(P_{\text{avg}} = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi\)
Where:
- \(V_{\text{rms}}\) is the root mean square (RMS) voltage.
- \(I_{\text{rms}\) is the root mean square (RMS) current.
- \(\phi\) is the phase angle between the voltage and current.
Step 1: Calculate RMS Values
For a sinusoidal function, the RMS value can be calculated as \(\frac{1}{\sqrt{2}}\) times its maximum (peak) value:
- \(V_{\text{rms}} = \frac{20}{\sqrt{2}} = 10\sqrt{2} \, \text{V}\)
- \(I_{\text{rms}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \, \text{A}\)
Step 2: Determine the Phase Angle (\(\phi\))
The phase difference given is \(\frac{\pi}{3}\), hence \(\phi = \frac{\pi}{3}\).
Step 3: Calculate Cosine of Phase Angle
\(\cos \phi = \cos \left( \frac{\pi}{3} \right) = \frac{1}{2}\)
Step 4: Calculate Average Power
Substitute these values into the average power formula:
\(P_{\text{avg}} = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi = (10\sqrt{2}) \cdot (5\sqrt{2}) \cdot \frac{1}{2}\)
The simplification is as follows:
\(P_{\text{avg}} = 10 \cdot 5 \cdot 2 \cdot \frac{1}{2} = 50 \, \text{W}\)
Conclusion:
The average power dissipated in the circuit is \(50 \, \text{W}\). Thus, the correct answer is:
- \(50 \, \text{W}\)
Approach Solution -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:
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