# Basic Engineering Circuit Analysis 10th Ed., Problem 2.87

Discussion in 'Homework Help' started by jbates, Feb 2, 2011.

1. ### jbates Thread Starter New Member

Feb 2, 2011
3
0
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.

Code ( (Unknown Language)):
1.             + 5V -
2.  |-------|----R----|-----|
3.  |       |         |     |
4. 4kΩ   Is/|\       6kΩ   3kΩ
5.  |       |         |     |
6.  |-------|---------|-----|
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.
Code ( (Unknown Language)):
1.             + 5V -
2.  |-------|----R----|
3.  |       |         |
4. 4kΩ   Is/|\       2kΩ
5.  |       |         |
6.  |-------|---------|
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: Feb 2, 2011
2. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
Is = I1 + I2

I2 = 5V/R

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

3. ### blah2222 Well-Known Member

May 3, 2010
554
33
A numerical solution will be determined by R.
JP

4. ### jbates Thread Starter New Member

Feb 2, 2011
3
0
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?

5. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
I think so too. Is or R should be given. You can always find a solution as a function of one of them though.

6. ### Jony130 AAC Fanatic!

Feb 17, 2009
3,957
1,097
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)

7. ### jbates Thread Starter New Member

Feb 2, 2011
3
0
Okay, I thought so.

Well, there goes two hours of my life.

Thanks everyone.

8. ### steveb Senior Member

Jul 3, 2008
2,433
469
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.