Engineering Series-Parallel Combinations of Inductors

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The discussion focuses on correcting a mistake in combining inductors in a circuit. The error involved treating two resistors as if they were in series while neglecting a parallel resistor. The correct approach involves combining the top-right resistor with the second resistor from the right in parallel. After addressing the mistake, the correct calculations led to the final answer of 15 H for the inductance. The clarification helped resolve the confusion and achieve the right result.
johnsmith7565
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Homework Statement
Assume that the initial energy stored in the inductors of Fig. P6.21 is zero. Find the equivalent inductance with respect to the terminals a, b.
Relevant Equations
L(series)= L1 + L2 + ... + Ln
1/(L(parallel)) = 1/L1 + 1/L2 + ... + 1/Ln
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I am not sure where I made the mistake. If someone could point that out that would be much appreciated!
 
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That next-to-last step on the first page looks wrong. You are combining two resisters as though they are in series and forgetting the one in parallel on the far top-right. Instead, that step should be to combine the far top-right with the parallel one second from the right.
 
FactChecker said:
That next-to-last step on the first page looks wrong. You are combining two resisters as though they are in series and forgetting the one in parallel on the far top-right. Instead, that step should be to combine the far top-right with the parallel one second from the right.
Thanks a million! I got the correct answer.
 
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20 // 30 = 12
+ 8 = 20
// 80 = 16
+14 = 30
// 60 = 20
+10 = 30
// 15 = 10
+ 5 = 15 H
 
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