For an AC circuit the potential difference and current are given by `V = 10 sqrt(2) sin omegat` (in V) and `i = 2 sqrt(2) cos omegat` (in A) respectively. The power dissipated in the instrument is

Options

A. `20 W`
B. `40 W`
C. `40sqrt(2)" W"`
D. Zero

Answer

To solve the problem of finding the power dissipated in the AC circuit given the potential difference and current, we can follow these steps:

### Step 1: Identify the given equations
The potential difference \( V \) and current \( i \) are given as:
- \( V = 10 \sqrt{2} \sin(\omega t) \) (in volts)
- \( i = 2 \sqrt{2} \cos(\omega t) \) (in amperes)

### Step 2: Convert the current equation to sine form
The current can be rewritten in terms of sine:
\[ i = 2 \sqrt{2} \cos(\omega t) = 2 \sqrt{2} \sin\left(\omega t + \frac{\pi}{2}\right) \]
This shows that the current leads the voltage by \( \frac{\pi}{2} \).

### Step 3: Identify the peak values
From the equations, we can identify:
- \( V_0 = 10 \sqrt{2} \) (peak voltage)
- \( I_0 = 2 \sqrt{2} \) (peak current)

### Step 4: Calculate RMS values
The RMS (Root Mean Square) values are calculated as follows:
\[ V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{10 \sqrt{2}}{\sqrt{2}} = 10 \, \text{V} \]
\[ I_{rms} = \frac{I_0}{\sqrt{2}} = \frac{2 \sqrt{2}}{\sqrt{2}} = 2 \, \text{A} \]

### Step 5: Determine the phase difference
From the current equation, we see that the phase difference \( \phi \) between the voltage and current is:
\[ \phi = \frac{\pi}{2} \]

### Step 6: Calculate the power dissipated
The power dissipated in the circuit can be calculated using the formula:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \]
Substituting the values we found:
\[ P = 10 \, \text{V} \cdot 2 \, \text{A} \cdot \cos\left(\frac{\pi}{2}\right) \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ P = 10 \cdot 2 \cdot 0 = 0 \, \text{W} \]

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

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