Basic Engineering Circuit Analysis 10th Ed., Problem 2.87

Thread Starter

jbates

Joined Feb 2, 2011
3
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.

Rich (BB code):
            + 5V -
 |-------|----R----|-----|
 |       |         |     |
4kΩ   Is/|\       6kΩ   3kΩ
 |       |         |     |
 |-------|---------|-----|
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.
Rich (BB code):
            + 5V -
 |-------|----R----|
 |       |         |
4kΩ   Is/|\       2kΩ
 |       |         |
 |-------|---------|
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.
 
Last edited:

Thread Starter

jbates

Joined Feb 2, 2011
3
It's a resistor but the resistance isn't given, only the voltage across it.

So it really can't be solved with the given information?
 

Jony130

Joined Feb 17, 2009
5,155
So it really can't be solved with the given information?
If R is unknown, the only think you can do is this


Is = I1 + I2

I2 = 5V/R

I1 = ( 5V + ((5V/R)*2K) ) / 4K

Is = 5V/R + ( 5V + ((5V/R)*2K) ) / 4K = 5V/R1 + (5V + 10K/R1) / 4k =

= (6000 + R1) /(800 * R1)
 

steveb

Joined Jul 3, 2008
2,436
I agree. You need to find Is as a function of R.

It's obvious that the value of R is needed because if R=0, there is no solution (Is goes to infinity), and if R is very large then current through R is negligible and the current source has 5V across it resulting in 5V across the 4K resistor and Is is 1.25 mA in the limit as R goes to infinity.
 
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