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

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:

• \(50 \, \text{W}\)

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:

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