Ok I see it too but isn't the same as I posted earlier?

 

Ok I see it too but isn't the same as I posted earlier?

Isn't what the same as what that you posted earlier? You need to try to be more explicit so that people aren't always having to read between the lines of what you are saying.

If you are asking if the circuit that Tesla23 posted the same as yours, the answer is no. Your circuit ONLY allowed you to find the voltage difference between the two ends of R5, while his actually allows you to find the voltage at each end of R5 (i.e., relative to ground).

 

But it seems like a good approach when you can see a way to break the circuit up that way. Don't know how often that is possible, though.

Agree - but it is a useful trick - a bridge is a common circuit and you can always split it like this.

Ok I see it too but isn't the same as I posted earlier?

You should try to understand why it is different. Mine is an exact equivalent circuit relative to ground (you seem to have moved that), it will correctly (and easily) give you the voltages at the nodes at the ends of R5 with R5 present, removed and shorted. I'm not sure which are needed but all seem to have been discussed in this thread.

Just because someone says Thevenin doesn't mean that you should jump in and start 'Thevenining'. Once you start to get into trouble, sit back and see if there is an easier way - it often pays great dividends.

 

Let's clarify some theoretical aspects: Thevenin's theorem is used when the source signal is voltage. If the signal source is current Norton's theorem is used (rarely encountered in practice).

Now, how to apply Thevenin's theory? It simply, consists in the following two steps - the third step is optional (for circuit from the first post above):

1) Eth is determined for the case when the V1 source works in idle (this case consists in the removal of R5). I note with E3 voltage across the resistor R3 and with E4 voltage on the resistor R4:

Eth = E3 - E4 = [R3 / (R1 + R3)]*V1 - [R4 / (R2 + R4)]*V1;
Eth = V1 * [[R3 / (R1 + R3)]- [R4 / (R2 + R4)]]

2) Equivalent resistance Rth is determined passivated V1 source (V1 is replaced by a short circuit) and calculated resistance seen across the resistor R5:

Rth = [R1*R3 / (R1 + R3)] + [R2*R4 / (R2 + R4)]

3) Circuit output voltage (Uo) is the voltage across the resistor R5, that Eth, and if we are interested in the output current, it can be determined simply:

I0 = Eth / Rth

In general, this is all about.

Edit later: If you want to know the voltage on each resistor:

E1 = V1 - E3;
E2 = V1 - E4;
E3 = [R3 / (R1 + R3)]*V1;
E4 = [R4 / (R2 + R4)]*V1;
E5 = Eth = E3 - E4.

 

Let's clarify some theoretical aspects: Thevenin's theorem is used when the source signal is voltage. If the signal source is current Norton's theorem is used (rarely encountered in practice).

The nature of the signal source is irrelevant. If it weren't, then what would you do if you had a both a voltage source and a current source?

Regardless of what the circuit is in the black box that you are finding the equivalent for (as long as it is linear), you can model it using either a Thevenin equivalent or a Norton equivalent circuit.