Question

The instantaneous values of alternating current and voltages in a circuit are \(I = \frac{3}{\sqrt{2}} \sin(200\pi t)\) ampere and \(V = \frac{3}{\sqrt{2}} \sin\left(200\pi t + \frac{\pi}{6}\right)\) volt. What is the average power consumed in the circuit in watt?

• A. \(\frac{\sqrt{3}}{8}\)
• B. \(\frac{6\sqrt{3}}{8}\)
• C. \(\frac{8\sqrt{2}}{9}\)
• D. \(\frac{9\sqrt{3}}{8}\)

Hint:

Always convert peak values to RMS before using power formula: \(P = V_{rms} I_{rms} \cos\phi\).

Answer 1

The Correct Option is D

Step 1: Identify peak values.
Given:
\[ I_0 = \frac{3}{\sqrt{2}}, \quad V_0 = \frac{3}{\sqrt{2}} \]

Step 2: Convert to RMS values.
\[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{3}{2} \]
\[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{3}{2} \]

Step 3: Identify phase difference.
From given expressions:
\[ \phi = \frac{\pi}{6} \]

Step 4: Use average power formula.
\[ P = V_{rms} \cdot I_{rms} \cdot \cos\phi \]

Step 5: Substitute values.
\[ P = \frac{3}{2} \times \frac{3}{2} \times \cos\left(\frac{\pi}{6}\right) \]

Step 6: Use value of cosine.
\[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
\[ P = \frac{9}{4} \times \frac{\sqrt{3}}{2} \]
\[ P = \frac{9\sqrt{3}}{8} \]

Step 7: Final answer.
\[ \boxed{\frac{9\sqrt{3}}{8}} \]