Find Leqv For Each Of The Given Circuits

The equivalent resistance (Req or Leqv for inductors) of a circuit is the total resistance/inductance seen from a specific pair of terminals in that circuit. Simplifying complex circuits to their equivalent resistance is a fundamental technique in circuit analysis, allowing for easier calculations of current, voltage, and power. Finding Leqv involves combining inductors in series and parallel, similar to how resistors are handled. The key difference is that inductors store energy in a magnetic field rather than dissipating it as heat like resistors.

Inductors in Series and Parallel: A Quick Recap

Before diving into complex circuits, let's quickly review the basics:

• Inductors in Series: When inductors are connected in series, the total inductance is simply the sum of the individual inductances.
  • Leqv = L1 + L2 + L3 + ... + Ln
• Inductors in Parallel: When inductors are connected in parallel, the reciprocal of the total inductance is equal to the sum of the reciprocals of the individual inductances.
  • 1/Leqv = 1/L1 + 1/L2 + 1/L3 + ... + 1/Ln
  • For two inductors in parallel, this simplifies to: Leqv = (L1 * L2) / (L1 + L2)

Now, let's move on to some examples.

Finding Leqv: Example Circuits

We will work through several circuits, each with increasing complexity, to demonstrate the process of finding the equivalent inductance. Remember to always identify series and parallel combinations first.

Example 1: Simple Series-Parallel Combination

Consider a circuit with two inductors, L1 = 10 mH and L2 = 20 mH, connected in series. This series combination is then connected in parallel with another inductor, L3 = 30 mH.

• Step 1: Simplify the Series Combination:
  • The series combination of L1 and L2 can be simplified into a single equivalent inductor, Lseries.
  • Lseries = L1 + L2 = 10 mH + 20 mH = 30 mH
• Step 2: Simplify the Parallel Combination:
  • Now we have Lseries = 30 mH in parallel with L3 = 30 mH.
  • Since the inductors are equal, we can use the formula Leqv = (L1 * L2) / (L1 + L2) or recognize that the equivalent inductance will be half of one of the inductors.
  • Leqv = (30 mH * 30 mH) / (30 mH + 30 mH) = 900 mH^2 / 60 mH = 15 mH
  • Therefore, the equivalent inductance of the circuit is 15 mH.

Example 2: More Complex Series-Parallel Network

Let's analyze a circuit with the following inductors: L1 = 5 mH, L2 = 10 mH, L3 = 15 mH, L4 = 20 mH. L1 and L2 are in series. This series combination is in parallel with L3. Finally, this entire parallel combination is in series with L4.

• Step 1: Simplify the First Series Combination:
  • L1 and L2 are in series, so we add their inductances.
  • Lseries1 = L1 + L2 = 5 mH + 10 mH = 15 mH
• Step 2: Simplify the Parallel Combination:
  • Now we have Lseries1 = 15 mH in parallel with L3 = 15 mH.
  • Lparallel = (Lseries1 * L3) / (Lseries1 + L3) = (15 mH * 15 mH) / (15 mH + 15 mH) = 225 mH^2 / 30 mH = 7.5 mH
• Step 3: Simplify the Second Series Combination:
  • Finally, we have Lparallel = 7.5 mH in series with L4 = 20 mH.
  • Leqv = Lparallel + L4 = 7.5 mH + 20 mH = 27.5 mH
  • Therefore, the equivalent inductance of the circuit is 27.5 mH.

Example 3: Delta-Wye Transformation (Star-Delta)

This example introduces a slightly more advanced technique when inductors aren't simply in series or parallel. Consider a delta (or pi) network of inductors. To simplify this, we can use a delta-wye (or pi-tee) transformation.

Let's say we have a delta network with:

• L_AB = 30 mH
• L_BC = 40 mH
• L_CA = 50 mH

We want to transform this into a wye (or tee) network with inductors L_A, L_B, and L_C connected to a common node. The transformation formulas are:

• L_A = (L_AB * L_CA) / (L_AB + L_BC + L_CA)
• L_B = (L_AB * L_BC) / (L_AB + L_BC + L_CA)
• L_C = (L_BC * L_CA) / (L_AB + L_BC + L_CA)

Plugging in the values:

• L_A = (30 mH * 50 mH) / (30 mH + 40 mH + 50 mH) = 1500 mH^2 / 120 mH = 12.5 mH
• L_B = (30 mH * 40 mH) / (30 mH + 40 mH + 50 mH) = 1200 mH^2 / 120 mH = 10 mH
• L_C = (40 mH * 50 mH) / (30 mH + 40 mH + 50 mH) = 2000 mH^2 / 120 mH = 16.67 mH (approximately)

After performing the delta-wye transformation, the circuit is now in a form that is potentially easier to analyze, as the wye configuration might create series or parallel combinations with other elements in the larger circuit. This transformation helps to break down circuits that are not immediately solvable through simple series/parallel reductions.

Example 4: A Ladder Network

Ladder networks often appear more intimidating than they actually are. The key is to start at the end of the ladder and work your way back towards the terminals. Consider a ladder network with the following inductors:

• L1 = 2 mH (closest to the input terminals)
• L2 = 4 mH
• L3 = 6 mH
• L4 = 8 mH (furthest from the input terminals)

The configuration is as follows: L4 is in series with a parallel combination of L3 and a series combination of L2 and L1.

• Step 1: Simplify the Innermost Series Combination:
  • L1 and L2 are in series.
  • Lseries1 = L1 + L2 = 2 mH + 4 mH = 6 mH
• Step 2: Simplify the Parallel Combination:
  • Lseries1 = 6 mH is in parallel with L3 = 6 mH.
  • Lparallel = (Lseries1 * L3) / (Lseries1 + L3) = (6 mH * 6 mH) / (6 mH + 6 mH) = 36 mH^2 / 12 mH = 3 mH
• Step 3: Simplify the Final Series Combination:
  • Lparallel = 3 mH is in series with L4 = 8 mH.
  • Leqv = Lparallel + L4 = 3 mH + 8 mH = 11 mH
  • Therefore, the equivalent inductance of the ladder network is 11 mH.

Example 5: Bridge Circuit

Bridge circuits are a classic circuit configuration. Analyzing them often requires careful consideration to determine if they can be simplified directly or if a transformation is needed. Let's analyze a bridge circuit with the following inductors:

• L1 = 10 mH (Upper Left)

• L2 = 20 mH (Upper Right)

• L3 = 15 mH (Lower Left)

• L4 = 30 mH (Lower Right)

• L5 = 25 mH (Center)

• Step 1: Check for a Balanced Bridge:

  • A bridge circuit is considered balanced if the ratio of the inductances in the two upper branches is equal to the ratio of the inductances in the two lower branches. In other words, if L1/L2 = L3/L4, then the bridge is balanced, and no current flows through L5. Therefore, L5 can be ignored when calculating the equivalent inductance from the input terminals.

  • Let's check:

    • L1/L2 = 10 mH / 20 mH = 0.5
    • L3/L4 = 15 mH / 30 mH = 0.5
    • Since L1/L2 = L3/L4, the bridge is balanced.

• Step 2: Simplify (Assuming Balanced Bridge):

  • Since the bridge is balanced, we can ignore L5. The circuit simplifies to L1 in series with L2, and L3 in series with L4, and these two series combinations are in parallel.

    • Lseries1 = L1 + L2 = 10 mH + 20 mH = 30 mH
    • Lseries2 = L3 + L4 = 15 mH + 30 mH = 45 mH
    • Leqv = (Lseries1 * Lseries2) / (Lseries1 + Lseries2) = (30 mH * 45 mH) / (30 mH + 45 mH) = 1350 mH^2 / 75 mH = 18 mH

  • Therefore, if the bridge is balanced, the equivalent inductance is 18 mH.

• Step 3: What if the bridge is NOT balanced?

  • If the bridge is NOT balanced (L1/L2 != L3/L4), then we can not ignore L5. There are several ways to solve the problem, but they involve more complex calculations and circuit transformations.

  • Delta-Wye Transformations (Multiple): The most common method involves performing delta-wye transformations to simplify the network. You can convert either the delta formed by L1, L5, and L3 or the delta formed by L2, L5, and L4 into a wye configuration. After the transformation, you will have series and parallel combinations that can be further simplified. This process is lengthy, requires careful algebraic manipulation, and is prone to errors.

  • Nodal Analysis/Mesh Analysis: An alternative method is to use circuit analysis techniques like nodal analysis or mesh analysis. This involves setting up a system of equations and solving for the currents and voltages in the circuit. While this method doesn't directly give you Leqv, you can determine the impedance of the circuit, and from that, infer the equivalent inductance at a given frequency.

Example 6: Circuit with Dependent Sources (Conceptual)

If the circuit contains dependent sources (voltage-controlled or current-controlled sources), the approach to finding Leqv is significantly different. You cannot simply combine inductors in series and parallel. Instead, you must use a test source method.

1. Deactivate Independent Sources: Set all independent voltage sources to zero (replace with a short circuit) and all independent current sources to zero (replace with an open circuit).

2. Apply a Test Source: Apply either a test voltage source (Vt) or a test current source (It) at the terminals where you want to find Leqv.

3. Calculate the Response: If you applied Vt, calculate the resulting current It flowing into the circuit from the voltage source. If you applied It, calculate the resulting voltage Vt across the terminals of the current source. This step will likely involve solving a circuit with dependent sources, requiring techniques like nodal or mesh analysis.

4. Determine the Input Impedance: Calculate the input impedance Zin = Vt / It.

5. Find Leqv: The equivalent inductance is then found using the relationship Leqv = Zin / (jω), where j is the imaginary unit and ω is the angular frequency of the test source. This formula highlights the fact that inductance is frequency-dependent in AC circuits. If the circuit contains only inductors and dependent sources, the resulting Zin will be of the form jωLeqv, allowing you to directly identify Leqv.

Important Considerations and Tips

• Frequency Dependence: Inductance is inherently frequency-dependent. The equivalent inductance Leqv is only truly equivalent at a specific frequency (or range of frequencies). When analyzing AC circuits, you need to consider the impedance of the inductors (Z = jωL), where j is the imaginary unit and ω is the angular frequency.

• Mutual Inductance: The examples above assume there is no mutual inductance between the inductors. If mutual inductance is present, the calculations become much more complex, and you need to consider the mutual inductance term (M) in the equivalent inductance formulas. For example, for two inductors L1 and L2 in series with mutual inductance M, the total inductance is Ltotal = L1 + L2 ± 2M (the sign depends on the winding direction).

• Simplification Strategy: Always look for the simplest combinations first. Start by identifying series and parallel combinations that can be easily reduced. If you encounter a delta or wye configuration, consider using a delta-wye transformation. If dependent sources are present, use the test source method.

• Redrawing the Circuit: Redrawing the circuit after each simplification step can make it easier to visualize the remaining circuit elements and identify further simplifications.

• Units: Always pay attention to units. Ensure that all inductances are in the same unit (e.g., mH, H) before performing calculations.

Common Mistakes to Avoid

• Incorrectly Identifying Series and Parallel: Carefully trace the current path to determine if components are truly in series or parallel. Components in series share the same current, while components in parallel have the same voltage across them.

• Forgetting the Reciprocal for Parallel Inductors: Remember that for inductors in parallel, you need to use the reciprocal formula (1/Leqv = 1/L1 + 1/L2 + ...). A common mistake is to simply add the inductances together.

• Ignoring Balanced Bridge Conditions: If you encounter a bridge circuit, always check if it is balanced before attempting to simplify it. Ignoring a balanced bridge can lead to significantly more complex and unnecessary calculations.

• Applying Series/Parallel Rules with Dependent Sources: You cannot use the standard series and parallel combination rules when dependent sources are present. The test source method is essential in these cases.

• Neglecting Mutual Inductance: If the problem specifies mutual inductance, you must include it in your calculations. Ignoring mutual inductance will result in an incorrect equivalent inductance.

Conclusion

Finding the equivalent inductance (Leqv) of a circuit is a crucial skill in circuit analysis. By systematically simplifying series and parallel combinations, applying delta-wye transformations when necessary, and using the test source method for circuits with dependent sources, you can reduce complex networks to their equivalent inductance. Remember to pay attention to units, double-check your calculations, and carefully consider the presence of mutual inductance or dependent sources. Mastering these techniques will allow you to analyze and design a wide range of inductive circuits with confidence. The key is practice; work through numerous examples to solidify your understanding and develop your problem-solving skills.