Voltage applied to an `AC` circuit and current flowing in it is given by
`V=200sqrt2sin(omegat+pi/4)` and `i=-sqrt2cos(omegat+pi/4)`
Then, power consumed in the circuited will be

Options

A. `200 W`
B. `400W`
C. `200sqrt2W`
D. none of these

Explanation

To find the power consumed in the given AC circuit, we will follow these steps:

### Step 1: Identify the given voltage and current equations
The voltage and current in the circuit are given as:
- Voltage: \( V = 200\sqrt{2} \sin(\omega t + \frac{\pi}{4}) \)
- Current: \( i = -\sqrt{2} \cos(\omega t + \frac{\pi}{4}) \)

### Step 2: Rewrite the current equation
We can rewrite the current equation using the identity \( \cos(\theta) = \sin(\theta + \frac{\pi}{2}) \):
\[ i = -\sqrt{2} \cos(\omega t + \frac{\pi}{4}) = -\sqrt{2} \sin\left(\omega t + \frac{\pi}{4} + \frac{\pi}{2}\right) \]
This simplifies to:
\[ i = \sqrt{2} \sin\left(\omega t + \frac{3\pi}{4}\right) \]

### Step 3: Determine the phase difference
Now we can compare the phase of the voltage and the current:
- Voltage phase: \( \frac{\pi}{4} \)
- Current phase: \( \frac{3\pi}{4} \)

The phase difference \( \phi \) is:
\[ \phi = \frac{3\pi}{4} - \frac{\pi}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \]

### Step 4: Calculate the power consumed
The average power \( P \) consumed in an AC circuit can be calculated using the formula:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\phi) \]
Where:
- \( V_{rms} = \frac{V_{peak}}{\sqrt{2}} = \frac{200\sqrt{2}}{\sqrt{2}} = 200 \)
- \( I_{rms} = \frac{I_{peak}}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}} = 1 \)

Now substituting the values:
\[ P = 200 \cdot 1 \cdot \cos\left(\frac{\pi}{2}\right) \]

Since \( \cos\left(\frac{\pi}{2}\right) = 0 \):
\[ P = 200 \cdot 1 \cdot 0 = 0 \]

### Conclusion
The power consumed in the circuit is \( P = 0 \).