I've been playing around with this problem for a couple hours. I think it might be impossible, so asking here is my last resort.
Find Is. Is is a current source, btw, in-case my text-art isn't clear enough.
First, I did the obvious and combined the two rightmost resistors into a 2kΩ equivalent resistor.
From the perspective of the node above Is, that leaves two paths to the node below Is where the voltage drops should be the same. Calling the current counter-clockwise through the left loop I1 and clockwise through the right loop I2, Is = I1+I2 and
5V + I2*2kΩ = I1*4kΩ
I've tried a million different things from this point, but I just can't solve for Is. I don't think it's possible. This is a new edition of the book and I think this problem must have a typo somewhere.
Rich (BB code):
+ 5V -
|-------|----R----|-----|
| | | |
4kΩ Is/|\ 6kΩ 3kΩ
| | | |
|-------|---------|-----|
First, I did the obvious and combined the two rightmost resistors into a 2kΩ equivalent resistor.
Rich (BB code):
+ 5V -
|-------|----R----|
| | |
4kΩ Is/|\ 2kΩ
| | |
|-------|---------|
5V + I2*2kΩ = I1*4kΩ
I've tried a million different things from this point, but I just can't solve for Is. I don't think it's possible. This is a new edition of the book and I think this problem must have a typo somewhere.
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