Inductance: Can L1 and L2 be replaced by an equivalent inductor in Figure 4(b)?

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SUMMARY

The discussion focuses on the equivalence of inductors L1 and L2 with an equivalent inductor Leq in the context of circuit analysis. The formula for Leq is established as Leq = (L1L2 - M^2) / (L1 + L2 - 2M). Participants confirm the correctness of the calculations and provide insights into the relationship between the currents I, I1, and I2. The thread references a previous discussion on parallel mutual inductances for additional clarity.

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Homework Statement



(c) See if you can show that L1 and L2 can be replaced by the

equivalent inductor, Leq, of FIGURE 4(b) where Leq = L1L2-M^2/L1+L2-2M

Homework Equations

The Attempt at a Solution


[/B]
I+I1+I2

Leq = V(L2-M)/jω(L1*L2-M^2) + V(L1-M)/jω(L1*L2-M^2)

= V*(L1+L2)-2M/jω(L1*L2-M^2)

V/jω*Leq = L1*L2-M^2/L1+L2-2M

Can anybody tell me if I'm close with this one or not? Struggling for days with it!

Any advice is appreciated!
 
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Ahh great! So after reading through I presume 'his_tonyness' got part 'C' correct since he wasn't corrected by yourself apart from I=I1+I2?
 
james123 said:
Ahh great! So after reading through I presume 'his_tonyness' got part 'C' correct since he wasn't corrected by yourself apart from I=I1+I2?
I believe that is the case. It's been a while since I took part in that thread (2015).
 
Many thanks gneill!
 

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