current sources not adding up

Not sure where I'm making an assumption or math error here. I pulled a simple problem from a book but I keep getting a wrong answer when I calc. Problem is find v and v1. Solution is v=15v and v1=4v. (I am not in school) 2k ___ +-----+-|___|+-------+-------------+ | | | + + + .-. .-. - .-. 1k | | | | 3k v1 | | 4k | | | | + | | '-' '-' '-' | 2/5 v1 | + + _ | | | / \\ | | | +--+(-->>++ | | | | \\_/ | | + | | | | .-. | | | | | | 2k | | | | | | | | _ | | '-' | | / \\ | | + +---+--+(-->>-++-----|+------------+ \\_/

- v + (created by AACircuit v1.28.4 beta 13/12/04

I think total supply current should be (2/5 * v1 + 1.4mA)... right? Using that and Requiv of 5k I intended to calculate v which is the voltage across the supplies (answer is 15v - given). But using 2/5 * v1 + 1.4mA and the equiv. resistance Re = 5k I end up w v=(2k * v1 + 1.4mA)(5k) however if v1 =

Correction:

....v=(2k * v1 + 1.4mA)(5k)...

should read:

....v=(2/5 * v1 + 1.4mA)(5k) and thus, v=(2k*v1 + 7)...

sorry for confusion.

school)

across

Try the solution with the upper current source equal to (2/5)*V1*ma.

Then the total supply current is ((2/5)*v1 + 1.4)*ma.

Thanks John, I was apparently having problems understanding the question.

Ok.

Then current is (1.4 + 2/5 * v1). Again, the current divides in a 2 to 1 ratio, so the current through the branch with v1 is

(1.4mA + 0.4 mA/V * v1)/3

Thus, v1 is

v1 = 4k * (1.4mA + 0.4mA/V * v1) / 3

Solve for v1 to get v1 = 4V

Thus the total current out of the sources is

(1.4m + 0.4*4) = 3mA

Since we already figured out the total resistance they see is 5k, the voltage across them is

v0 = 3m * 5k = 15V

calc.

school)

Using

across

v1 =

about

The top supply is a dependant source

Two current sources

Yes, but you don't have 15V to work with -- you must solve for it

Again -- you must solve for it .. so you either need v1 first or v first to get the other.

calc.

school)

Using

across

v1 =

about

Is that an assumption? Could not the top source be providing amps instead of mA? (It's obvious from the solutions that it is indeed mA, but if you didn't know the solutions how would you justify the *mA?

Yes. It is an assumption that the author thinks you can read his mind. ;-) Knowing the 15 volt result left no other choice.

Now you have two ways to define relationship between the voltage across the current sources (lets call it v0) and v1.

((1.4+(2/5)*v1)*Rtotal)=v0

And v1=voltage divided fraction of v0.

Then it is just a matter of rearranging one of those to solve for v1 in terms of v0 and then substituting that expression into the other for v1. Then v0 can be solved for.

OK. That did it for me. Making the assumption that it's mA/V as you and Mr. Popelish have suggested allows the math to work out. Thanks.