Question

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

• A. 5 W
• B. 10 W
• C. 2.5 W
• D. 25 W

Answer 1

Answer: (C)

2.5 W

Answer 2

The formula for the average power dissipated in an AC circuit with sinusoidal voltage and current is:
$P_{\text{avg}} = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi$ where $\phi$ is the phase difference between the voltage and the current.

Step 1. Convert voltage and current to RMS values:
- Given $V = 100\sin(100t)$, the peak voltage $V_0 = 100 \, \text{V}$.

$V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{100}{\sqrt{2}} = 50\sqrt{2} \, \text{V}$  
  

- Given $I = 100\sin(100t + \frac{\pi}{3})$, the peak current $I_0 = 100 \, \text{mA} = 0.1 \, \text{A}$.

$I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{0.1}{\sqrt{2}} = 0.05\sqrt{2} \, \text{A}$  

Step 2. Determine the phase difference:
- The phase difference $\phi = \frac{\pi}{3}$.

Step 3. Calculate the average power:

$P_{\text{avg}} = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos \phi$

Substituting the values:

$P_{\text{avg}} = (50\sqrt{2}) \cdot (0.05\sqrt{2}) \cdot \cos \frac{\pi}{3}$

$P_{\text{avg}} = 50 \cdot 0.05 \cdot \cos \frac{\pi}{3}$

$P_{\text{avg}} = 2.5 \, \text{W}$
The Correct Answer is: 2.5 W