The potential differences `V` and the current `i` flowing through an instrument in an `AC` circuit of frequency `f` are given by `V=5 cos omega t` and `I=2 sin omega t` amperes (where `omega=2 pi f`). The power dissipated in the instrument is

Options

A. Zero
B. 10 W
C. 5 W
D. `2.5` W

Answer

To solve the problem, we need to find the power dissipated in the instrument given the voltage and current equations in an AC circuit.

### Step-by-Step Solution:

1. **Identify the given equations:**
- Voltage: \( V = 5 \cos(\omega t) \)
- Current: \( I = 2 \sin(\omega t) \)

2. **Convert the current equation to cosine form:**
- We know that \( \sin(\omega t) = \cos(\omega t - \frac{\pi}{2}) \).
- Therefore, we can rewrite the current as:
\[ I = 2 \sin(\omega t) = 2 \cos\left(\omega t - \frac{\pi}{2}\right) \]

3. **Determine the phase difference (\( \phi \)):**
- The voltage is in cosine form, while the current is in cosine form with a phase shift of \( -\frac{\pi}{2} \).
- Thus, the phase difference \( \phi \) between voltage and current is:
\[ \phi = -\frac{\pi}{2} \]

4. **Calculate the RMS values of voltage and current:**
- The maximum voltage \( V_0 = 5 \) V, so the RMS voltage \( V_{\text{rms}} \) is:
\[ V_{\text{rms}} = \frac{V_0}{\sqrt{2}} = \frac{5}{\sqrt{2}} = \frac{5\sqrt{2}}{2} \]
- The maximum current \( I_0 = 2 \) A, so the RMS current \( I_{\text{rms}} \) is:
\[ I_{\text{rms}} = \frac{I_0}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

5. **Calculate the power factor:**
- The power factor is given by \( \cos(\phi) \):
\[ \cos\left(-\frac{\pi}{2}\right) = 0 \]

6. **Calculate the average power (\( P \)):**
- The average power dissipated in the instrument is given by:
\[ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi) \]
- Substituting the values:
\[ P = \left(\frac{5\sqrt{2}}{2}\right) \cdot (\sqrt{2}) \cdot 0 = 0 \]

### Final Answer:
The power dissipated in the instrument is \( \boxed{0} \) watts.