Hi, Based on this circuit, I have some strange results from simulations.
5V ___ | | \ / 2K \ | |--------------------------o C | | | | \ | / 2K | \ | | | | | | | o o A B
---------------------------- A B C 0v 0v 0v 5v 0v 0v x 5v 5v (something wrong here)
---------------------------- I think that when point B is 5V when point A either 5 or 0v, point C should not get 5v as well. But I just cant figure out why the simulator behave like this. I am using Electronics Workbench 5.12. Thanks in advance
Think that I am confused by the word 'input' and 'output'. FYI, A & B label as input points and C point is output. And I have been given the waveforms of C in the middle of 0v to 5v(half Vcc) for A=0v and B=5v. Yup, u guys are correct since output C is followed input of point B. Thanks.
When you say "0v" do you mean putting the point to ground or letting it floating. There's a big difference between the two options.
If point A is floating (not connected to anything...) you have 0v connected to this point BUT there is no current flowing on the branch... so you can remove the entire line from point A to the next node (with the 2k resistor) since this part of the circuit is doing nothing.
If A is connected to Ground (zero volt), current can flow between point A and the rest of the circuit. You can then apply voltage dividing and current dividing rules.
Also, when you say " [...] pint A equal 0v [...]" is it a measured voltage or an applied voltage ? You have to do the distinction between these notions for us (and you) to understand what you are doing.
I think that you are on a good way to succeed your simulation, you're probably not doing exactly what you want to do. Be shure that, in all time, all your points are connected (either to ground, source or instruments).
If point B = 5v then point C = 5v (same node), the upper resistor is shortcutted (no voltage difference on it). You can then ignore this branch.
If point A = 5v and point B = 5v then point C = 5v and no current is flowing on the circuit. If point A = gnd and point B = 5v then C = 5V and a current is inducted in the 2K resistor (I let you do the math... I = V/R).
For clarity, be shure to define the conditions on all nodes an any time.