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 solve the problem, we need to find the average power dissipated in the instrument given the voltage and current equations. Let's go through the solution step by step.

### Step 1: Identify the given equations
The voltage \( E \) and current \( I \) are given by:
\[ E = 2 \cos(\omega t) \quad \text{(in volts)} \]
\[ I = 2 \sin(\omega t) \quad \text{(in amperes)} \]

### Step 2: Find the phase difference
The voltage is expressed as a cosine function, while the current is expressed as a sine function. We can express the sine function in terms of cosine to find the phase difference. Recall that:
\[ \sin(\theta) = \cos\left(\theta - \frac{\pi}{2}\right) \]
Thus, we can rewrite the current as:
\[ I = 2 \sin(\omega t) = 2 \cos\left(\omega t - \frac{\pi}{2}\right) \]
This indicates that the current lags behind the voltage by \( \frac{\pi}{2} \) radians.

### Step 3: Determine the phase difference \( \phi \)
From the equations, we see that the phase difference \( \phi \) between voltage and current is:
\[ \phi = \frac{\pi}{2} \]

### Step 4: Calculate RMS values
The RMS (Root Mean Square) values for voltage and current can be calculated as follows:
\[ V_{\text{rms}} = \frac{E_{\text{max}}}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \, \text{V} \]
\[ I_{\text{rms}} = \frac{I_{\text{max}}}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \, \text{A} \]

### Step 5: Use the average power formula
The average power \( P \) dissipated in an AC circuit is given by:
\[ P = V_{\text{rms}} \cdot I_{\text{rms}} \cdot \cos(\phi) \]
Substituting the values we have:
\[ P = \sqrt{2} \cdot \sqrt{2} \cdot \cos\left(\frac{\pi}{2}\right) \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ P = \sqrt{2} \cdot \sqrt{2} \cdot 0 = 0 \]

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