In an AC circuit , `underset(o)(V)` , `underset(o)(I)` and cos`theta` are voltage amplitude , current amplitude and power factor respectively, the power consumption is

Options

A. 1/2`underset(o)(V)``underset(o)(I)` cos`theta`
B. `(1/sqrt(2)) V_(0)I_(0)costheta`
C. `V_(0)I_(0)costheta`
D. `(1/sqrt(2))V_(0)I_(0)sintheta`

Answer

To solve the problem of power consumption in an AC circuit given the voltage amplitude \( V_0 \), current amplitude \( I_0 \), and power factor \( \cos \theta \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Definitions**:
- Voltage amplitude \( V_0 \) is the maximum voltage in the AC circuit.
- Current amplitude \( I_0 \) is the maximum current in the AC circuit.
- Power factor \( \cos \theta \) is the cosine of the phase difference between the voltage and current.

2. **Expression for Instantaneous Power**:
- The instantaneous power \( P(t) \) in an AC circuit can be expressed as:
\[ P(t) = V(t) \times I(t) \]
- Where \( V(t) = V_0 \sin(\omega t + \phi) \) and \( I(t) = I_0 \sin(\omega t) \).

3. **Substituting the Expressions**:
- Substitute the expressions for \( V(t) \) and \( I(t) \):
\[ P(t) = V_0 \sin(\omega t + \phi) \times I_0 \sin(\omega t) \]

4. **Using the Trigonometric Identity**:
- Use the trigonometric identity for the product of sine functions:
\[ \sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)] \]
- Apply this identity to the expression for \( P(t) \):
\[ P(t) = V_0 I_0 \left(\frac{1}{2} [\cos(\phi) - \cos(2\omega t + \phi)]\right) \]

5. **Finding Average Power**:
- To find the average power \( P \) over one complete cycle, we integrate \( P(t) \) over one period \( T \):
\[ P = \frac{1}{T} \int_0^T P(t) dt \]
- The average value of \( \cos(2\omega t + \phi) \) over one complete cycle is zero, so we only consider the constant term:
\[ P = \frac{1}{2} V_0 I_0 \cos(\phi) \]

6. **Final Result**:
- Therefore, the average power consumed in the AC circuit is:
\[ P = \frac{1}{2} V_0 I_0 \cos \theta \]

### Conclusion:
The power consumption in the AC circuit is given by:
\[ P = \frac{1}{2} V_0 I_0 \cos \theta \]