If a current I given by `l_(o) sin(omegat-pi//2)` flows in an AC circuit across which an AC potential of `E_(0)`.` sin (omegat)` has been applied, then the power consumption P in the circuit will be

Options

A. `E-(0)I_(0)// 2`
B. `E_(0)I-(0)//2`
C. `El//sqrt2`
D. zero

Answer

To solve the problem, we need to analyze the given information about the AC circuit.

### Step-by-Step Solution:

1. **Identify the Current and Voltage Functions**:
- The current \( I \) is given by:
\[ I = I_0 \sin(\omega t - \frac{\pi}{2}) \]
- The voltage \( E \) is given by:
\[ E = E_0 \sin(\omega t) \]

2. **Determine the Phase Difference**:
- The current lags the voltage by \( \frac{\pi}{2} \) radians (or 90 degrees). This is evident from the sine function of the current, which has a phase shift of \( -\frac{\pi}{2} \).
- Therefore, the phase difference \( \theta \) between the voltage and the current is:
\[ \theta = \frac{\pi}{2} \]

3. **Use the Power Formula**:
- The average power \( P \) consumed in an AC circuit can be calculated using the formula:
\[ P = V \cdot I \cdot \cos(\theta) \]
- Here, \( V \) is the peak voltage \( E_0 \), \( I \) is the peak current \( I_0 \), and \( \theta \) is the phase difference.

4. **Substitute the Values**:
- Substitute the known values into the power formula:
\[ P = E_0 \cdot I_0 \cdot \cos\left(\frac{\pi}{2}\right) \]

5. **Calculate the Cosine**:
- We know that:
\[ \cos\left(\frac{\pi}{2}\right) = 0 \]
- Therefore, substituting this value gives:
\[ P = E_0 \cdot I_0 \cdot 0 = 0 \]

6. **Conclusion**:
- The power consumption \( P \) in the circuit is:
\[ P = 0 \]

### Final Answer:
The power consumption in the circuit is \( P = 0 \).

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