AC source represented as V = `V_(0) sin omega`t is applied to a circuit and current I = `I_(0) sin (omega t + pi//2)` is found to flow through the circuit . What will be average power consumed in the circuit ?

Options

A. `V_(0) I_(0)`
B. `(V_(0)I_(0))/(2)`
C. `(V_(0)I_(0))/(sqrt(2))`
D. 0

Explanation

To find the average power consumed in the circuit where the voltage and current are given by:

- Voltage: \( V = V_0 \sin(\omega t) \)
- Current: \( I = I_0 \sin(\omega t + \frac{\pi}{2}) \)

we can follow these steps:

### Step 1: Identify the phase difference
The current is given as \( I = I_0 \sin(\omega t + \frac{\pi}{2}) \). This indicates that the current leads the voltage by \( \frac{\pi}{2} \) radians (or 90 degrees).

### Step 2: Calculate the phase angle \( \phi \)
From the relationship between voltage and current, the phase angle \( \phi \) can be determined:
- Since the current leads the voltage by \( \frac{\pi}{2} \), we have:
\[ \phi = -\frac{\pi}{2} \]

### Step 3: Calculate the average power formula
The average power \( P_{\text{avg}} \) in an AC circuit is given by the formula:
\[ P_{\text{avg}} = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi) \]
where:
- \( V_{\text{rms}} = \frac{V_0}{\sqrt{2}} \)
- \( I_{\text{rms}} = \frac{I_0}{\sqrt{2}} \)

### Step 4: Substitute the values into the power formula
Substituting \( V_{\text{rms}} \) and \( I_{\text{rms}} \) into the power formula, we get:
\[ P_{\text{avg}} = \left(\frac{V_0}{\sqrt{2}}\right) \cdot \left(\frac{I_0}{\sqrt{2}}\right) \cdot \cos\left(-\frac{\pi}{2}\right) \]

### Step 5: Calculate \( \cos(-\frac{\pi}{2}) \)
We know that:
\[ \cos\left(-\frac{\pi}{2}\right) = 0 \]
Thus, substituting this value into the power equation gives:
\[ P_{\text{avg}} = \left(\frac{V_0}{\sqrt{2}}\right) \cdot \left(\frac{I_0}{\sqrt{2}}\right) \cdot 0 = 0 \]

### Conclusion
The average power consumed in the circuit is:
\[ P_{\text{avg}} = 0 \]