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 find the value of L in the given problem, we can follow these steps:

### Step 1: Understand the Configuration
We have two inductors, each with inductance L, connected in parallel. The equivalent inductance (L_parallel) of two inductors in parallel is given by the formula:
\[ \frac{1}{L_{\text{parallel}}} = \frac{1}{L} + \frac{1}{L} \]
This simplifies to:
\[ L_{\text{parallel}} = \frac{L}{2} \]

### Step 2: Combine with the Series Inductor
Next, we have another inductor of 5 mH connected in series with the parallel combination. The total inductance (L_total) of inductors in series is simply the sum of their inductances:
\[ L_{\text{total}} = L_{\text{parallel}} + L_{\text{series}} = \frac{L}{2} + 5 \text{ mH} \]

### Step 3: Set Up the Equation
According to the problem, the effective inductance is given as 15 mH. Therefore, we can set up the equation:
\[ \frac{L}{2} + 5 = 15 \]

### Step 4: Solve for L
Now, we will solve the equation for L:
1. Subtract 5 from both sides:
\[ \frac{L}{2} = 15 - 5 \]
\[ \frac{L}{2} = 10 \]
2. Multiply both sides by 2 to isolate L:
\[ L = 20 \text{ mH} \]

### Conclusion
The value of L is 20 mH.

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