Wong said:
But with point A equal to 0v. The node between 2 resistors should have
2.5v according to the voltage divider rule. So this is something that
I am not so sure with the 'mix' of 2.5v and 5v (point B).
Is point C still 5v under this circumstances ?
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).
John Larkin <
[email protected]> wrote in
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.
Regards.
Jean-François Martel