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, we need to find the average power dissipated in the AC circuit given the instantaneous voltage and current.

### Step-by-Step Solution:

1. **Identify the given equations**:
- The voltage \( V(t) = 10\sqrt{2} \sin(\omega t) \) (in volts)
- The current \( i(t) = 2\sqrt{2} \cos(\omega t) \) (in amperes)

2. **Calculate the instantaneous power**:
The instantaneous power \( P(t) \) is given by the product of voltage and current:
\[ P(t) = V(t) \cdot i(t) = (10\sqrt{2} \sin(\omega t)) \cdot (2\sqrt{2} \cos(\omega t)) \]
Simplifying this:
\[ P(t) = 20 \cdot 2 \cdot \sin(\omega t) \cdot \cos(\omega t) = 40 \sin(\omega t) \cos(\omega t) \]

3. **Use the trigonometric identity**:
We can use the identity \( \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \) to simplify the power:
\[ P(t) = 20 \sin(2\omega t) \]

4. **Find the average power over one cycle**:
The average power \( P_{avg} \) over one complete cycle is calculated using:
\[ P_{avg} = \frac{1}{T} \int_0^T P(t) \, dt \]
where \( T = \frac{2\pi}{\omega} \) is the time period.

5. **Set up the integral**:
Substitute \( P(t) \) into the integral:
\[ P_{avg} = \frac{1}{T} \int_0^T 20 \sin(2\omega t) \, dt \]
\[ = \frac{20}{T} \int_0^T \sin(2\omega t) \, dt \]

6. **Evaluate the integral**:
The integral of \( \sin(2\omega t) \) over one complete cycle (from \( 0 \) to \( T \)) is zero because the positive and negative halves of the sine wave cancel each other out:
\[ \int_0^T \sin(2\omega t) \, dt = 0 \]

7. **Calculate the average power**:
Therefore, substituting back into the average power equation:
\[ P_{avg} = \frac{20}{T} \cdot 0 = 0 \]

### Final Answer:
The average power dissipated in the circuit is \( P_{avg} = 0 \) watts.