Two inductances connected in parallel are equivalent to a single inductance of `1.5H` and when connected in series are equivalent to a single inductance of 8 H. The difference in their inductance is

Options

A. 3 H
B. 7.5 H
C. 2 H
D. 4 H

Explanation

To solve the problem of finding the difference in inductance between two inductors connected in parallel and series, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Inductances**:
Let the two inductances be \( L_1 \) and \( L_2 \).

2. **Use the Given Information**:
- When connected in parallel, the equivalent inductance is given by:
\[ \frac{1}{L_{eq, parallel}} = \frac{1}{L_1} + \frac{1}{L_2} \]
Given \( L_{eq, parallel} = 1.5 \, H \), we can write:
\[ \frac{1}{1.5} = \frac{1}{L_1} + \frac{1}{L_2} \]
This simplifies to:
\[ \frac{2}{3} = \frac{1}{L_1} + \frac{1}{L_2} \]

- When connected in series, the equivalent inductance is:
\[ L_{eq, series} = L_1 + L_2 \]
Given \( L_{eq, series} = 8 \, H \), we have:
\[ L_1 + L_2 = 8 \]

3. **Set Up the Equations**:
From the parallel combination, we can rearrange the equation:
\[ \frac{1}{L_1} + \frac{1}{L_2} = \frac{2}{3} \]
This can be rewritten as:
\[ \frac{L_1 + L_2}{L_1 L_2} = \frac{2}{3} \]
Substituting \( L_1 + L_2 = 8 \):
\[ \frac{8}{L_1 L_2} = \frac{2}{3} \]
Cross-multiplying gives:
\[ 8 \cdot 3 = 2 \cdot L_1 L_2 \implies 24 = 2 L_1 L_2 \implies L_1 L_2 = 12 \]

4. **Form a Quadratic Equation**:
Now we have two equations:
- \( L_1 + L_2 = 8 \)
- \( L_1 L_2 = 12 \)

We can use these to form a quadratic equation:
\[ x^2 - (L_1 + L_2)x + L_1 L_2 = 0 \]
Substituting the known values:
\[ x^2 - 8x + 12 = 0 \]

5. **Solve the Quadratic Equation**:
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here \( a = 1, b = -8, c = 12 \):
\[ x = \frac{8 \pm \sqrt{(-8)^2 - 4 \cdot 1 \cdot 12}}{2 \cdot 1} \]
\[ x = \frac{8 \pm \sqrt{64 - 48}}{2} \]
\[ x = \frac{8 \pm \sqrt{16}}{2} \]
\[ x = \frac{8 \pm 4}{2} \]
This gives us two solutions:
\[ x = \frac{12}{2} = 6 \quad \text{and} \quad x = \frac{4}{2} = 2 \]
Thus, \( L_1 = 6 \, H \) and \( L_2 = 2 \, H \) (or vice versa).

6. **Find the Difference**:
The difference in their inductances is:
\[ |L_1 - L_2| = |6 - 2| = 4 \, H \]

### Final Answer:
The difference in their inductance is \( 4 \, H \).

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