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

The Correct Option is C

To determine the average power dissipated over a cycle in an AC circuit, we have to consider the voltage and current waveforms and their phase difference.

Given the equations: 

• \(V = 100 \sin (100 \, t)\) V
• \(I = 100 \sin \left(100 \, t + \frac{\pi}{3}\right)\) mA

We need to convert the current from milliamps to amps for consistency in units:

• \(I = 0.1 \sin \left(100 \, t + \frac{\pi}{3}\right)\) A

The formula for the average power in an AC circuit is given by:

• \(P_{\text{avg}} = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi)\)

Where:

• \(V_{\text{rms}} = \frac{V_{\text{peak}}}{\sqrt{2}}\)
• \(I_{\text{rms}} = \frac{I_{\text{peak}}}{\sqrt{2}}\)
• \(\phi\) is the phase difference between voltage and current.

Calculating the RMS values:

• \(V_{\text{rms}} = \frac{100}{\sqrt{2}} = 70.71\) V
• \(I_{\text{rms}} = \frac{0.1}{\sqrt{2}} = 0.07071\) A

Given that the phase difference \(\phi = \frac{\pi}{3}\), we find:

• \(\cos\left(\frac{\pi}{3}\right) = 0.5\)

Now, substitute these values into the average power formula:

• \(P_{\text{avg}} = 70.71 \times 0.07071 \times 0.5 = 2.5\) W

Therefore, the average power dissipated in one cycle is 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