The voltage E and the current I in an instrument are represented by the equations:
`E=2cos omegatV`
`I=2sin omegat A`
The average power dissipated in the instrument will be

Options

A. zero
B. 1.0 W
C. 4 W
D. 2.0 W

Answer

To find the average power dissipated in the instrument given the voltage and current equations, we can follow these steps:

### Step-by-Step Solution:

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

2. **Determine the peak values**:
- The peak voltage \( V_0 \) is 2 volts.
- The peak current \( I_0 \) is 2 amperes.

3. **Calculate RMS values**:
- The RMS (Root Mean Square) voltage \( V_{RMS} \) is given by:
\[ V_{RMS} = \frac{V_0}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \text{ volts} \]
- The RMS current \( I_{RMS} \) is given by:
\[ I_{RMS} = \frac{I_0}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \text{ amperes} \]

4. **Determine the phase difference**:
- The voltage is represented by a cosine function and the current by a sine function. The phase difference \( \phi \) between them is:
\[ \phi = \frac{\pi}{2} \text{ radians} \quad (\text{or } 90^\circ) \]

5. **Use the formula for average power**:
- The average power \( P \) is calculated using the formula:
\[ P = V_{RMS} \cdot I_{RMS} \cdot \cos(\phi) \]

6. **Substitute the values**:
- Substituting the RMS values and the cosine of the phase difference:
\[ P = (\sqrt{2}) \cdot (\sqrt{2}) \cdot \cos\left(\frac{\pi}{2}\right) \]
- Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ P = (\sqrt{2}) \cdot (\sqrt{2}) \cdot 0 = 0 \text{ watts} \]

### Conclusion:
The average power dissipated in the instrument is \( 0 \) watts.