Question

In an a.c. circuit, voltage and current are given by: \[ V = 100 \sin(100t) \, \text{V} \quad \text{and} \quad I = 100 \sin\left(100t + \frac{\pi}{3}\right) \, \text{mA}. \] The average power dissipated in one cycle is:

• A. \( 10 \, \text{W} \)
• B. \( 2.5 \, \text{W} \)
• C. \( 25 \, \text{W} \)
• D. \( 5 \, \text{W} \)

Hint:

In AC circuits, the average power depends on the product of the rms values of voltage and current and the cosine of the phase angle. Always ensure consistent units and correct trigonometric values for accurate results.

Answer 1

The Correct Option is B

The average power in an AC circuit over one complete cycle is calculated as: \[ P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos(\Delta \phi), \] where: - \( V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \), - \( I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \), - \( \Delta \phi \) is the phase difference between the voltage and current. Given: \[ V_0 = 100 \, \text{V}, \quad I_0 = 100 \times 10^{-3} \, \text{A}, \quad \Delta \phi = \frac{\pi}{3}. \] Step 1: Calculate \( V_{\text{rms}} \) and \( I_{\text{rms}} \)} \[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}}, \quad I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{100 \times 10^{-3}}{\sqrt{2}}. \] Step 2: Substitute into the Power Formula \[ P_{\text{avg}} = \frac{100}{\sqrt{2}} \cdot \frac{100 \times 10^{-3}}{\sqrt{2}} \cdot \cos\left(\frac{\pi}{3}\right). \] Step 3: Simplify the Expression \[ P_{\text{avg}} = \frac{100 \cdot 0.1}{2} \cdot \frac{1}{2} = \frac{10}{4} = 2.5 \, \text{W}. \] Final Answer: \[ \boxed{2.5 \, \text{W}} \]