In an `AC` circuit, the current is given by `i=5sin(100t-(pi)/2)` and the `AC` potential is `V=200sin(100t)`volt. Then the power consumption is

Options

A. 20 watts
B. 40 watts
C. 1000Watts
D. O watts

Answer

To solve the problem of power consumption in the given AC circuit, we can follow these steps:

### Step 1: Identify the given parameters
The current \( i \) and voltage \( V \) are given as:
- Current: \( i = 5 \sin(100t - \frac{\pi}{2}) \)
- Voltage: \( V = 200 \sin(100t) \)

### Step 2: Determine the phase difference
From the current equation, we can see that the term \( -\frac{\pi}{2} \) indicates a phase shift. Thus, the phase difference \( \theta \) between the current and voltage is:
\[ \theta = -\frac{\pi}{2} \]
This means that the current lags the voltage by \( \frac{\pi}{2} \) radians (or 90 degrees).

### Step 3: Calculate the average power consumption
The average power \( P \) consumed in an AC circuit can be calculated using the formula:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\theta) \]
Where:
- \( V_{rms} \) is the root mean square voltage
- \( I_{rms} \) is the root mean square current
- \( \cos(\theta) \) is the cosine of the phase difference

### Step 4: Calculate \( V_{rms} \) and \( I_{rms} \)
For sinusoidal signals:
- The \( V_{rms} \) for voltage \( V = 200 \sin(100t) \) is:
\[ V_{rms} = \frac{V_{max}}{\sqrt{2}} = \frac{200}{\sqrt{2}} = 100\sqrt{2} \]
- The \( I_{rms} \) for current \( i = 5 \sin(100t - \frac{\pi}{2}) \) is:
\[ I_{rms} = \frac{I_{max}}{\sqrt{2}} = \frac{5}{\sqrt{2}} \approx 3.54 \]

### Step 5: Determine \( \cos(\theta) \)
Since \( \theta = -\frac{\pi}{2} \), we calculate:
\[ \cos(-\frac{\pi}{2}) = 0 \]

### Step 6: Substitute values into the power formula
Now substituting the values into the power formula:
\[ P = V_{rms} \cdot I_{rms} \cdot \cos(\theta) = (100\sqrt{2}) \cdot (3.54) \cdot 0 \]
Since \( \cos(\theta) = 0 \), we find:
\[ P = 0 \]

### Final Answer
Thus, the power consumption in the circuit is:
\[ \boxed{0} \text{ watts} \]