Two inductors each of inductance L are connected in parallel. One more inductor of value 5 mH is connected in series of this configuration then the effective inductance is 15 mH . The value of L is ………..mH.

Options

A. 10
B. `5.0`
C. `2.5`
D. 20

Answer

To solve the problem step by step, we will follow the concepts of inductance in parallel and series configurations.

### Step 1: Identify the inductance in parallel
We have two inductors, each of inductance \( L \), connected in parallel. The formula for the equivalent inductance \( L' \) of two inductors in parallel is given by:

\[ \frac{1}{L'} = \frac{1}{L_1} + \frac{1}{L_2} \]

Since both inductors have the same inductance \( L \):

\[ \frac{1}{L'} = \frac{1}{L} + \frac{1}{L} = \frac{2}{L} \]

Thus, we can rearrange this to find \( L' \):

\[ L' = \frac{L}{2} \]

### Step 2: Add the series inductor
Next, we have another inductor of value \( 5 \, \text{mH} \) connected in series with the equivalent inductance \( L' \). The formula for the total inductance \( L_{\text{total}} \) in series is:

\[ L_{\text{total}} = L' + L_{\text{series}} \]

Substituting \( L' \) and the value of the series inductor:

\[ L_{\text{total}} = \frac{L}{2} + 5 \, \text{mH} \]

### Step 3: Set the total inductance equal to the given value
According to the problem, the effective inductance is \( 15 \, \text{mH} \). Therefore, we can set up the equation:

\[ \frac{L}{2} + 5 = 15 \]

### Step 4: Solve for \( L \)
Now, we will solve for \( L \):

1. Subtract \( 5 \) from both sides:

\[ \frac{L}{2} = 15 - 5 \]
\[ \frac{L}{2} = 10 \]

2. Multiply both sides by \( 2 \):

\[ L = 20 \, \text{mH} \]

### Final Answer
The value of \( L \) is \( 20 \, \text{mH} \).

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