Question

In an AC circuit the potential difference and current are represented respectively by \( V = 100 \sin (100t) \) volt, \( I = 100 \sin \left( 100t + \frac{\pi}{3} \right) \) milliampere. The power in the circuit is:

• A. 2.5 W
• B. 5 W
• C. 10 W
• D. \( 10^4 \, \text{W} \)

Hint:

The average power in an AC circuit is calculated using the maximum voltage, maximum current, and the cosine of the phase difference between them.

Answer 1

The Correct Option is A

Step 1: Power formula for an AC circuit.
The average power \( P_{\text{avg}} \) in an AC circuit is given by: \[ P_{\text{avg}} = \frac{1}{2} V_{\text{max}} I_{\text{max}} \cos(\phi) \] where \( V_{\text{max}} \) is the maximum voltage, \( I_{\text{max}} \) is the maximum current, and \( \phi \) is the phase difference between voltage and current.

Step 2: Determine the maximum values and phase difference.
From the given equations: - \( V = 100 \sin(100t) \), so \( V_{\text{max}} = 100 \, \text{V} \). - \( I = 100 \sin\left( 100t + \frac{\pi}{3} \right) \), so \( I_{\text{max}} = 100 \, \text{mA} = 0.1 \, \text{A} \). The phase difference \( \phi = \frac{\pi}{3} \).

Step 3: Calculate the power.
Now, using the formula: \[ P_{\text{avg}} = \frac{1}{2} \times 100 \times 0.1 \times \cos\left( \frac{\pi}{3} \right) \] Since \( \cos\left( \frac{\pi}{3} \right) = \frac{1}{2} \): \[ P_{\text{avg}} = \frac{1}{2} \times 100 \times 0.1 \times \frac{1}{2} = 2.5 \, \text{W} \]