The r.m.s current in an `AC` circuit is `2A`. If the wattless current be `sqrt(3)A`, what is the power factor?

Options

A. `1/2`
B. `1/3`
C. `1/sqrt(3)`
D. `1/sqrt(2)`

Explanation

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

### Step 1: Understand the relationship between RMS current and wattless current.
The wattless current (\(I_{WL}\)) is given by the formula:
\[ I_{WL} = I_{rms} \cdot \sin(\phi) \]
where:
- \(I_{WL}\) is the wattless current,
- \(I_{rms}\) is the root mean square current,
- \(\phi\) is the phase angle.

### Step 2: Substitute the known values into the formula.
From the problem, we know:
- \(I_{rms} = 2 \, A\)
- \(I_{WL} = \sqrt{3} \, A\)

Substituting these values into the equation, we get:
\[ \sqrt{3} = 2 \cdot \sin(\phi) \]

### Step 3: Solve for \(\sin(\phi)\).
Rearranging the equation to find \(\sin(\phi)\):
\[ \sin(\phi) = \frac{\sqrt{3}}{2} \]

### Step 4: Determine the angle \(\phi\).
From trigonometric values, we know that:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
Thus, \(\phi = 60^\circ\).

### Step 5: Calculate the power factor.
The power factor (\(PF\)) is given by:
\[ PF = \cos(\phi) \]
Substituting \(\phi = 60^\circ\):
\[ PF = \cos(60^\circ) = \frac{1}{2} \]

### Final Answer:
The power factor is \(\frac{1}{2}\).

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