The potential difference `V` across and the current `I` flowing through an instrument in an `AC` circuit are given by what will be the avrage power:
`V=5 cos omega t` volt
`I=2 sin omega t` Amp.

Options

A. zero
B. `5` watt
C. `10` watt
D. `2.5` watt

Explanation

To find the average power in the given AC circuit, we can follow these steps:

### Step 1: Identify the Voltage and Current Equations
The voltage \( V \) and current \( I \) are given as:
\[ V = 5 \cos(\omega t) \quad \text{(1)} \]
\[ I = 2 \sin(\omega t) \quad \text{(2)} \]

### Step 2: Convert Voltage to Sine Form
We can express the cosine function in terms of sine:
\[ V = 5 \cos(\omega t) = 5 \sin\left(\omega t + \frac{\pi}{2}\right) \]
This shows that the voltage leads the current by \( 90^\circ \) (or \( \frac{\pi}{2} \) radians).

### Step 3: Determine the Phase Difference
From the transformation in Step 2, we can see that the phase difference \( \phi \) between the voltage and current is:
\[ \phi = 90^\circ \]

### Step 4: Calculate RMS Values
The root mean square (RMS) values for voltage and current are calculated as follows:
\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{5}{\sqrt{2}} \quad \text{(where } V_0 = 5 \text{)} \]
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{2}{\sqrt{2}} \quad \text{(where } I_0 = 2 \text{)} \]

### Step 5: Calculate Average Power
The average power \( P \) in an AC circuit is given by the formula:
\[ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi) \]
Substituting the values we have:
- \( V_{\text{rms}} = \frac{5}{\sqrt{2}} \)
- \( I_{\text{rms}} = \frac{2}{\sqrt{2}} \)
- \( \cos(90^\circ) = 0 \)

Thus, the average power becomes:
\[ P = \left(\frac{5}{\sqrt{2}}\right) \cdot \left(\frac{2}{\sqrt{2}}\right) \cdot 0 = 0 \]

### Conclusion
The average power in the circuit is:
\[ \boxed{0} \]