Q1e) Find the equivalent inductance, Leq, at terminals a-b of the circuit given. Show all work. do 25 mH 10 mH Lea 60 mH 20 mH 30 mH b

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Chapter20: Capacitance In Ac Circuits
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**Question 1(c): Equivalent Inductance Calculation**

**Problem Statement:**

Find the equivalent inductance, \( L_{eq} \), at terminals \( a-b \) of the circuit given. Show all work.

**Circuit Diagram:**

The provided circuit diagram contains the following components:
- An inductor of 10 mH connected in series.
- An inductor of 60 mH, 25 mH, 20 mH, and 30 mH connected in a combination of series and parallel.

Here is the circuit layout in detail:
- Inductor 10 mH connected in series with a parallel combination
- The left branch of the parallel combination has an inductor of 25 mH.
- The right branch parallel to the 25 mH inductor consists of a series combination of 60 mH, 20 mH, and 30 mH inductors.

**Equivalent Inductance Calculation:**

1. **Identify Series and Parallel Combinations:**
   - In the right branch:
     - Inductors 60 mH, 20 mH, and 30 mH are in series.
     - The equivalent inductance of series inductors ( \( L_s \) ):
       \( L_s = 60 + 20 + 30 = 110 \) mH

2. **Combine Parallel Inductors:**
   - The inductors in the left and right branches are in parallel.
   - Inductors in parallel combine as:
     \[
     \frac{1}{L_p} = \frac{1}{L_1} + \frac{1}{L_2}
     \]
     Where \( L_1 \) is 25 mH, and \( L_2 \) is 110 mH.
     \[
     \frac{1}{L_p} = \frac{1}{25} + \frac{1}{110} = 0.04 + 0.0091 \approx 0.0491 \ (\text{in } \text{mH}^{-1})
     \]
     Therefore:
     \[
     L_p \approx \frac{1}{0.0491} \approx 20.37 \ \text{mH}
     \]

3. **Add the Series Inductance:**
   - Finally, add the 10 mH series inductor to the parallel combination:
Transcribed Image Text:**Question 1(c): Equivalent Inductance Calculation** **Problem Statement:** Find the equivalent inductance, \( L_{eq} \), at terminals \( a-b \) of the circuit given. Show all work. **Circuit Diagram:** The provided circuit diagram contains the following components: - An inductor of 10 mH connected in series. - An inductor of 60 mH, 25 mH, 20 mH, and 30 mH connected in a combination of series and parallel. Here is the circuit layout in detail: - Inductor 10 mH connected in series with a parallel combination - The left branch of the parallel combination has an inductor of 25 mH. - The right branch parallel to the 25 mH inductor consists of a series combination of 60 mH, 20 mH, and 30 mH inductors. **Equivalent Inductance Calculation:** 1. **Identify Series and Parallel Combinations:** - In the right branch: - Inductors 60 mH, 20 mH, and 30 mH are in series. - The equivalent inductance of series inductors ( \( L_s \) ): \( L_s = 60 + 20 + 30 = 110 \) mH 2. **Combine Parallel Inductors:** - The inductors in the left and right branches are in parallel. - Inductors in parallel combine as: \[ \frac{1}{L_p} = \frac{1}{L_1} + \frac{1}{L_2} \] Where \( L_1 \) is 25 mH, and \( L_2 \) is 110 mH. \[ \frac{1}{L_p} = \frac{1}{25} + \frac{1}{110} = 0.04 + 0.0091 \approx 0.0491 \ (\text{in } \text{mH}^{-1}) \] Therefore: \[ L_p \approx \frac{1}{0.0491} \approx 20.37 \ \text{mH} \] 3. **Add the Series Inductance:** - Finally, add the 10 mH series inductor to the parallel combination:
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