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

Options

A. 5 watt
B. 10 watt
C. 2.5 watt
D. 0

Answer

To solve the problem, we need to find the average power in the AC circuit given the potential difference \( V \) and the current \( I \).

### Step-by-Step Solution:

1. **Identify the given equations:**
- The potential difference is given by:
\[ V = 5 \cos(\omega t) \text{ volts} \]
- The current is given by:
\[ I = 2 \sin(\omega t) \text{ amps} \]

2. **Convert the current equation to cosine form:**
- We can express the sine function in terms of cosine:
\[ I = 2 \sin(\omega t) = 2 \cos\left(\omega t - \frac{\pi}{2}\right) \]
- This shows that the current lags the voltage by \( \frac{\pi}{2} \) radians (or 90 degrees).

3. **Determine the phase difference:**
- The phase difference \( \phi \) between the voltage and the current is:
\[ \phi = \frac{\pi}{2} \]

4. **Calculate the maximum voltage and current:**
- The maximum voltage \( V_m \) is 5 volts.
- The maximum current \( I_m \) is 2 amps.

5. **Use the formula for average power in an AC circuit:**
- The average power \( P \) in an AC circuit is given by:
\[ P = V_m I_m \cos(\phi) \]
- Substituting the values we have:
\[ P = 5 \times 2 \times \cos\left(\frac{\pi}{2}\right) \]

6. **Calculate \( \cos\left(\frac{\pi}{2}\right) \):**
- We know that:
\[ \cos\left(\frac{\pi}{2}\right) = 0 \]

7. **Final calculation of average power:**
- Therefore, the average power \( P \) becomes:
\[ P = 5 \times 2 \times 0 = 0 \text{ watts} \]

### Conclusion:
The average power developed in the circuit is \( 0 \) watts.