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

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The discussion revolves around whether inductors L1 and L2 can be replaced by an equivalent inductor Leq in a specific circuit configuration. The formula for Leq is provided, which incorporates mutual inductance (M) and the individual inductances (L1 and L2). The user expresses difficulty in solving the problem and seeks validation of their approach. References to previous discussions on similar topics indicate that others have successfully navigated this problem. Overall, the thread emphasizes the complexity of inductance calculations and the importance of mutual inductance in circuit analysis.
james123

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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