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 determined by: \[P_{\text{avg}} = V_{\text{rms}} I_{\text{rms}} \cos(\Delta \phi),\] where: - \( V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \) represents the root-mean-square voltage. - \( I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \) represents the root-mean-square current. - \( \Delta \phi \) is the phase difference between the voltage and current. Given values: \[V_0 = 100 \, \text{V}, \quad I_0 = 100 \times 10^{-3} \, \text{A}, \quad \Delta \phi = \frac{\pi}{3}.\] Step 1: Calculate \( V_{\text{rms}} \) and \( I_{\text{rms}} \)} \[V_{\text{rms}} = \frac{100}{\sqrt{2}}, \quad I_{\text{rms}} = \frac{100 \times 10^{-3}}{\sqrt{2}}.\] Step 2: Substitute these values 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: \[\boxed{2.5 \, \text{W}}\]