# Equivalent Inductance Problem

1. Jun 30, 2013

### RoKr93

1. The problem statement, all variables and given/known data

2. Relevant equations
Leq = L1 + L2 + ... + Ln (series)

Leq = 1/((1/L1) + (1/L2) + ... + (1/Ln)) (parallel)

3. The attempt at a solution

For part A, I redrew the circuit (above), taking into account that the switch was open, but I'm not entirely certain I did it right. If that is correct, then it would appear that for part B, a redrawn circuit would be the same except for one more parallel branch with nothing on it (representing the closed switch), which would cause a short circuit. This doesn't seem right. Did I redraw the circuit incorrectly?

2. Jun 30, 2013

### Staff: Mentor

You did fine for part A.

Note that the two sets of 12H and 24H inductors remain parallel pairs regardless of the switch position, so you might as well replace them with their equivalents right away.

Now, for part B, when the switch closes it ties the top rail to the bottom rail, making them all one node. So you can "fold" the circuit about the horizontal mid-line bringing the bottom rail up to the top and merging the connections along the resulting single rail. That should put some more inductors in parallel...

3. Jun 30, 2013

### Staff: Mentor

I make it 7L for part (a), also. For (b), why don't you redraw it showing 4 inductors, two of them being 8L, and go from there? Simplify it in small, easily-manageable, absolutely-irrefutable steps and you can't go wrong!

4. Jun 30, 2013

### RoKr93

It's...the same thing? Combining the 12L and 24L inductors definitely helped clear up my mental picture. But am I correct in assuming that since those 3 branches are parallel, I can swing the empty one into the middle like I did?

5. Jun 30, 2013

### Staff: Mentor

Right.

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