Mutual inductance / equivalent inductor

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SUMMARY

The discussion focuses on the calculation of the equivalent inductance, Leq, for two inductors, L1 and L2, with mutual inductance M. The formula provided for Leq is Leq = (L1L2 - M2) / (L1 + L2 - 2M). Participants emphasize the importance of understanding mutual inductance when simplifying the inductor network to a single equivalent inductor, similar to reducing a resistor network. The discussion encourages using an AC test voltage to analyze the current supplied by the circuit.

PREREQUISITES
  • Understanding of mutual inductance in electrical circuits
  • Familiarity with the concept of equivalent inductance
  • Basic knowledge of AC circuit analysis
  • Ability to manipulate algebraic expressions in electrical engineering
NEXT STEPS
  • Study the derivation of the equivalent inductance formula for coupled inductors
  • Learn about AC circuit analysis techniques, particularly using test voltages
  • Explore the concept of mutual inductance and its effects on circuit behavior
  • Practice problems involving the reduction of inductor networks to equivalent inductors
USEFUL FOR

Electrical engineering students, circuit designers, and anyone involved in analyzing or designing circuits with inductors and mutual inductance.

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Homework Statement


q4_tma_diagram.jpg


Homework Equations


See if you can show that L##_1## and L##_2## as in (a) can be replaced by the equivalent inductor L##_{eq}## as in (b):

$$L_{eq} = \frac {L_1 L_2 - M^2 } {L_1 + L_2 - 2M }
$$

The Attempt at a Solution


Um, I don't really know where to start because I don't really understand what the question is asking. Am I supposed to provide some 'proof' in algebraic form? Thus I don't know the form that the answer should take, as in what equals what.

A gentle nudge would get me moving :)
 
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Just like reducing a resistor network to a single equivalent resistance, they want you to reduce the inductor network to a single equivalent inductor. The tricky bit is handling the mutual inductance. How you go about it is up to you, but consider driving the circuit with ac test voltage V and finding the resulting current it supplies. Then compare to the expression for the current if the load was just a single inductor Leq.
 
gneill said:
Just like reducing a resistor network to a single equivalent resistance, they want you to reduce the inductor network to a single equivalent inductor. The tricky bit is handling the mutual inductance. How you go about it is up to you, but consider driving the circuit with ac test voltage V and finding the resulting current it supplies. Then compare to the expression for the current if the load was just a single inductor Leq.
Thank you. I shall post my attempt for scrutiny!
 

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