|Jan27-05, 11:27 AM||#1|
KCL / KVL to determine Vo
Quick question that is throwing me off....
If you have a dependant current source in a parallel circuit, how do you represent that in your KCL equation? The one in this problem is Vo/2000.
I took that as I=V/R so R=2000. Since it is in a parallel circuit on its own 'branch' I used Vs/2000. But when I plug it all into the equations, the answer is wrong.
The circuit is this....
Parallel circuit with 4 branches. The following are the branches from left to right.
1. Dep. Current Source = Vo/2000
2. 6k ohm resister with + on top. (Marked also with Vs).
3. 1k ohm resister and a 2k ohm resister. Vo is indicated across the 2k ohm. The + sign is between the two resisters and the - is at the bottom of the 2k ohm.
4. indep current source at 2mA.
I took Vs to be the voltage across each branch since they are all in parallel and it is stated that Vs is across the 6k ohm resister. I tried to do the....
Vs/2000 + Vs/6k + Vs/3k + 2mA = 0
and then solved for Vs which ended up being 2V. Then used that to find Vo... except it is way off as the answer says Vo is 8V.
So, I am either screwing up the representation of the Vo/2000 in the equation or I am screwing up the entire KCL for the equation.
If someone could give a hint, Id appreciate it.
|Jan27-05, 02:25 PM||#2|
Your mistake was the V/R thing for branch 1.
Branch one is an dependant current source whose value is Vo/2000. No ohm's law trickery is necessary.
Therefore KCL at node Vs is:
Vo/2000-Vs/6000-Vs/3000+2/1000=0 (using your sign convention)
You can relate Vs to Vo by noting branch 3 is a voltage divider.
I am sure you have it from here...
|Jan27-05, 03:22 PM||#3|
Yes, that did the trick.
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