# Voltage across an open circuit

1. Jun 19, 2011

### Ry122

The potential difference between the two grounds in this circuit can be determined
with the equation

AB = X + Z

where z is the voltage source and x is the node at the top of the circuit, and AB
is the potential difference between the two grounds.

Could someone please explain to me why this is the case?

http://img31.imageshack.us/img31/1401/circuity.jpg [Broken]

Last edited by a moderator: May 5, 2017
2. Jun 19, 2011

### dlgoff

Ground is ground. They are at the same potential (assuming they are connected which is what the ground symbol means). Hence the difference is zero.

Why would it be otherwise?

3. Jun 19, 2011

### dlgoff

If the ground symbols you are showing are just points where a potential is taken (test points that are not connected), the difference would be the potential across R3 (R2 in the picture below).

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/voldiv.html" [Broken]

Last edited by a moderator: May 5, 2017
4. Jun 19, 2011

### vk6kro

In your circuit, the voltage across R3 is

V * {R3 / (R3 + R2)}

This is normal voltage divider action.

There is no voltage across R1 because there is no current flowing in it. So, the voltage at each end of this resistor must be the same.

So, the output voltage is the same as the voltage across R3.

The equation in your post doesn't seem to be true.

5. Jun 19, 2011

### Ry122

The two grounds are just supposed to be terminal outputs actually. Sorry about that.

6. Jun 19, 2011

### davenn

exactly :) You have 1 "ground" the negative rail fron the supply labelled V

you have the positive rail labelled V1. DONT confuse the 2 arrows pointing away to the right as "ground connections". They are just arrows indicating output/ onward connection to further circuitry

Have a look at these that I quickly drew up :)

cheers
Dave

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