If the power factor in an AC circuit changes from `(1)/(3)` to `(1)/(9)` then by what percent reactance will change (approximately), If resistance remains constant?

Options

A. Increase by 200%
B. Decrease by 200%
C. Increased by 100%
D. Decrease by 100%

Answer

To solve the problem, we need to find the percentage change in reactance when the power factor changes from \( \frac{1}{3} \) to \( \frac{1}{9} \), while keeping the resistance constant.

### Step-by-Step Solution:

1. **Understanding Power Factor**:
The power factor (PF) is defined as:
\[ \text{PF} = \cos(\phi) \]
where \( \phi \) is the phase angle between the voltage and current.

2. **Calculate the Angles**:
For the initial power factor of \( \frac{1}{3} \):
\[ \cos(\phi_1) = \frac{1}{3} \implies \phi_1 = \cos^{-1}\left(\frac{1}{3}\right) \]
For the new power factor of \( \frac{1}{9} \):
\[ \cos(\phi_2) = \frac{1}{9} \implies \phi_2 = \cos^{-1}\left(\frac{1}{9}\right) \]

3. **Relate Reactance to Resistance and Angle**:
The reactance \( X \) can be expressed as:
\[ X = R \tan(\phi) \]
Therefore, we can express the reactance for both cases:
\[ X_1 = R \tan(\phi_1) \]
\[ X_2 = R \tan(\phi_2) \]

4. **Calculate Tangent Values**:
Using the identity \( \tan(\phi) = \frac{\sin(\phi)}{\cos(\phi)} \) and knowing \( \sin^2(\phi) + \cos^2(\phi) = 1 \):
- For \( \phi_1 \):
\[ \sin(\phi_1) = \sqrt{1 - \left(\frac{1}{3}\right)^2} = \sqrt{1 - \frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \]
\[ \tan(\phi_1) = \frac{\sin(\phi_1)}{\cos(\phi_1)} = \frac{\frac{2\sqrt{2}}{3}}{\frac{1}{3}} = 2\sqrt{2} \]
- For \( \phi_2 \):
\[ \sin(\phi_2) = \sqrt{1 - \left(\frac{1}{9}\right)^2} = \sqrt{1 - \frac{1}{81}} = \sqrt{\frac{80}{81}} = \frac{4\sqrt{5}}{9} \]
\[ \tan(\phi_2) = \frac{\sin(\phi_2)}{\cos(\phi_2)} = \frac{\frac{4\sqrt{5}}{9}}{\frac{1}{9}} = 4\sqrt{5} \]

5. **Calculate Reactance Values**:
Now we can find the reactance values:
\[ X_1 = R \cdot 2\sqrt{2} \]
\[ X_2 = R \cdot 4\sqrt{5} \]

6. **Calculate Percentage Change**:
The percentage change in reactance is given by:
\[ \text{Percentage Change} = \frac{X_2 - X_1}{X_1} \times 100 \]
Substituting the values:
\[ \text{Percentage Change} = \frac{R \cdot 4\sqrt{5} - R \cdot 2\sqrt{2}}{R \cdot 2\sqrt{2}} \times 100 \]
Simplifying this:
\[ = \frac{4\sqrt{5} - 2\sqrt{2}}{2\sqrt{2}} \times 100 = \left(2\frac{4\sqrt{5} - 2\sqrt{2}}{2\sqrt{2}}\right) \times 100 \]
\[ = \left(2\frac{4\sqrt{5}}{2\sqrt{2}} - 1\right) \times 100 = \left(2\sqrt{10} - 1\right) \times 100 \]

Approximating \( \sqrt{10} \approx 3.16 \):
\[ = (6.32 - 1) \times 100 \approx 532\% \]

### Final Answer:
The percentage change in reactance is approximately **200%**.