The average power dissipated in AC circuit is 2W. If a current flowing throuh a circuit is 2A, impedance is `1Omega`, then what is the power factor of the circuit?

Options

A. `0.5`
B. 1
C. Zero
D. `1/sqrt(2)`

Explanation

To find the power factor of the AC circuit, we can follow these steps:

### Step 1: Understand the relationship between power, current, voltage, and power factor
The average power (P) in an AC circuit is given by the formula:
\[ P = V_{RMS} \times I_{RMS} \times \cos(\phi) \]
where:
- \( P \) is the average power,
- \( V_{RMS} \) is the root mean square voltage,
- \( I_{RMS} \) is the root mean square current,
- \( \cos(\phi) \) is the power factor.

### Step 2: Identify the known values
From the question, we have:
- Average power \( P = 2 \, \text{W} \)
- Current \( I_{RMS} = 2 \, \text{A} \)
- Impedance \( Z = 1 \, \Omega \)

### Step 3: Calculate the RMS voltage
Using Ohm's law, we know that:
\[ V_{RMS} = I_{RMS} \times Z \]
Substituting the known values:
\[ V_{RMS} = 2 \, \text{A} \times 1 \, \Omega = 2 \, \text{V} \]

### Step 4: Substitute the values into the power formula
Now we can substitute \( P \), \( V_{RMS} \), and \( I_{RMS} \) into the power formula:
\[ 2 \, \text{W} = 2 \, \text{V} \times 2 \, \text{A} \times \cos(\phi) \]

### Step 5: Simplify the equation
This simplifies to:
\[ 2 = 4 \cos(\phi) \]

### Step 6: Solve for the power factor
Now, we can solve for \( \cos(\phi) \):
\[ \cos(\phi) = \frac{2}{4} = \frac{1}{2} \]

### Step 7: Conclusion
Thus, the power factor of the circuit is:
\[ \cos(\phi) = 0.5 \]