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### Network # Something wrong

W

#### Wong

Jan 1, 1970
0
Hi,
Based on this circuit, I have some strange results from simulations.

5V
___
|
|
\
/ 2K
\
|
|--------------------------o C
| |
| |
\ |
/ 2K |
\ |
| |
| |
| |
o o
A B

J

#### John Larkin

Jan 1, 1970
0
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

Aren't B and C the same node?

John

W

#### Wong

Jan 1, 1970
0
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 ?

J

#### John Miller

Jan 1, 1970
0
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 ?

If the ASCII drawing is correct, B and C will always be the same; they are
directly connected.

--
John Miller

A diplomat is a man who can convince his wife she'd look stout in a fur
coat.

M

#### Michael A. Covington

Jan 1, 1970
0
Do you have a connection to ground in your circuit? Electronics Workbench
(Multisim) and similar simulators need one, in order to calculate voltages.

R

#### Randy Day

Jan 1, 1970
0
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

This is correct, as long as *no voltage* is being
applied at point B. The resistance between B and C
is 0 ohms, so Vc will always equal Vb.
I am not so sure with the 'mix' of 2.5v and 5v (point B).
Is point C still 5v under this circumstances ?

The voltage chart is correct for the circuit as drawn.

If you connect B to 5v, you're putting a 0 (that's zero)
ohm resistance in parallel with the upper 2k resistor.

If you connect B to A (0v), you're putting a 0 (that's zero)
ohm resistance in parallel with the *lower* 2k resistor.

J

#### John Larkin

Jan 1, 1970
0
But with point A equal to 0v. The node between 2 resistors should have
2.5v according to the voltage divider rule.

It would if nothing else is connected to it. But when you apply a hard
voltage source to the junction, the source wins.

Please don't top post.

John

J

#### Jamie

Jan 1, 1970
0
eh?
Point B and C are connected! why shouldn't they be the same?
just think about that.
if Point B is applied voltage from a source its ovbious that
Point C will look the same.

W

#### Wong

Jan 1, 1970
0
Jamie said:
eh?
Point B and C are connected! why shouldn't they be the same?
just think about that.
if Point B is applied voltage from a source its ovbious that
Point C will look the same.

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.

M

#### Mortel

Jan 1, 1970
0
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

W

#### Wong

Jan 1, 1970
0
Mortel said:
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. Putting the point to ground.

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.
Mind to explain what is measured voltage and applied voltage? What's
the different ?

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