I need some help with calculating Thevenin voltage

[Herr Masel]

In this schematic from wikipedia Vth is calculated using the voltage divider formula:

What happens if you add another voltage source underneath R2 and R3, working in the opposite direction of V1 (in the schematic it would be "looking" up)? Is the new formula  (R3+R2*V1)/(R2+R3+R4) -V2 ? Or is V2 subtracted from the top part of the equation only? Or is it as mr. Cleese put it, something completely different? My study books have shown some examples that I can't understand, they seem wrong. In a similar example they wrote that Uth=V2+IR3+IR2, which completely ignores the part of the schematic on the left! I reckon the logic behind this is that according to Kierkehoff's voltage law Uth (or AB) is equal to the voltage of the section parallel to it, but what happens when you add that second voltage source? I just realized that changing the value of E1 will change the current in the resistors therefore changing Vth. There is another example of theirs I don't get but I'll wait a little before posting it, I'm sure that after reading your answers it will become clearer.

[scaesic]

i'm not sure if this will help or not, but calculating the thev voltage is done from the otuside of a system, you don't care whats happening inside the system, just consider the whole circuit a black box, then put a voltmeter over the output of the circuit, thats Vth

Once you've done that, put a variable resitor over the output of the circuit and measure the voltage across that, once V is half Vth, then the value of the variable resistor is Rth.

[davebungo]

Is the new formula  (R3+R2*V1)/(R2+R3+R4) -V2

This doesn't make much sense since R2*V1 isn't a resistance...

Assuming as I think you said, V2 is in the same direction i.e. +ve pointing up then:

Vth = V2 + (V1-V2)*(R2+R3)/(R2+R3+R4)

and Rth = R1 + R4*(R2+R3)/(R2+R3+R4)

[Herr Masel]

Rth I understood myself, but

This doesn't make much sense since R2*V1 isn't a resistance...

....
Vth = V2 + (V1-V2)*(R2+R3)/(R2+R3+R4)

The equation I meant is for the voltage divider ((R3+R2)*V1)/(R2+R3+R4) I simply placed V1 inside the brackets by mistake. Could you explain why you subtracted V2 from V1 inside the brackets and added V2 to the whole equation? I'm not so clear on that one.

i'm not sure if this will help or not, but calculating the thev voltage is done from the otuside of a system, you don't care whats happening inside the system, just consider the whole circuit a black box, then put a voltmeter over the output of the circuit, thats Vth

Once you've done that, put a variable resitor over the output of the circuit and measure the voltage across that, once V is half Vth, then the value of the variable resistor is Rth.

Ooh, that makes some sense of things. You mean Vth only "sees" the outside/the row of components in parallel to it that is the closest. Like the wall of a box. Now I understand why it is constantly referred to as physically seeing. The inside of the circuit is used to calculate current and Rth but for the Vth only the external part is required. Ignoring half the circuit to calculate voltage kind of makes sense and doesn't make sense at the same time... I will need to digest this - and your way of measuring Rth which I haven't heard of yet - overnight. Sweet dreams.

[scaesic]

aaaaaaaaaaaaaaayye

[davebungo]

OK, to explain:

The definition of the Thevenin voltage is the open circuit voltage of the network.
The definition of the Thevenin resistance is that resistance (or impedance) seen looking into the network with all voltage sources shorted out and all current sources open circuited.  (As Scaesic pointed out, this can be measured if you are examining a real live circuit where you don't know the inards or they are too complicated to work out using algebra).

In your circuit:
R1 can play no part since no current can flow through R1.
The current flowing between V1 and V2 = (V1-V2)/(R2+R3+R4)
The voltage developed across R2 and R3 = (R2+R3) * (V1-V2)/(R2+R3+R4)
The voltage seen on the output consists of V2 and the voltage across R2 and R3 i.e. V2 + (R2+R3)*(V1-V2)/(R2+R3+R4)

I hope this explains.

[Sir H C]

If I am understanding you correctly, isn't the answer zero?  Assume point b is ground.  Point A is midway between V+ over 2k and a V- over 2k, so 0 volts.  Or do you have the supply going with the plus up?  In that case you could merge the supplys, have the series resistor and do it as 15V with 1k series.

[guitar_199]

The way I learned this ( and this was back in the seventies... ) ......"the sum of the voltages in a closed loop must be equal to zero."

So, starting from the lower left ground and going clockwise:

V1 - IR4 - IR3 -IR2 = 0
it follows that....
V1 = IR4 + IR3 + IR2
V1 = I (R2 + R3 + R4)
V1/(R2 + R3 + R4) = I

Now...... the voltage drop across any R in the circuit can be described as  I * R, so the drop across any resistor can be calculated that way.

THe voltage between A and B is the sum of the voltage drops across R1, R2, and R3.  Since A -> B is open circuit the current through R1 = 0 therefore the drop across R1 = 0 * 1K = 0v therefore R1 can be omitted and the voltage A-B is the sum of the drops across R2 and R3 with no regard for the V1 source or the drop across R4.

I the circuit is modified to add a V2 under R2, polarized the other direction the total becomes:

V1 - IR4 - I R3 - IR2 - V2 = 0v
again, it follows that
V1 = IR4 + IR3 + IR2 + V2

following the logic above......... the voltage from A -> B is composed of    the drops across R3, R4 PLUS the voltage of V2.

stealing that formula back a second.....

V1 = IR4 + IR3 + IR2 + V2

and pulling the R4 drop back to the left....

V1 - IR4 = IR3 + IR2 + V2

does support that the parallel branches have the same voltage potential.

I apologize if I have headed off in the wrong direction, but it just seems like this info might give some help.

Bob

[scaesic]

The way I learned this ( and this was back in the seventies... ) ......"the sum of the voltages in a closed loop must be equal to zero."

So, starting from the lower left ground and going clockwise:

V1 - IR4 - IR3 -IR2 = 0
it follows that....
V1 = IR4 + IR3 + IR2
V1 = I (R2 + R3 + R4)
V1/(R2 + R3 + R4) = I

Now...... the voltage drop across any R in the circuit can be described as  I * R, so the drop across any resistor can be calculated that way.

THe voltage between A and B is the sum of the voltage drops across R1, R2, and R3.  Since A -> B is open circuit the current through R1 = 0 therefore the drop across R1 = 0 * 1K = 0v therefore R1 can be omitted and the voltage A-B is the sum of the drops across R2 and R3 with no regard for the V1 source or the drop across R4.

I the circuit is modified to add a V2 under R2, polarized the other direction the total becomes:

V1 - IR4 - I R3 - IR2 - V2 = 0v
again, it follows that
V1 = IR4 + IR3 + IR2 + V2

following the logic above......... the voltage from A -> B is composed of    the drops across R3, R4 PLUS the voltage of V2.

stealing that formula back a second.....

V1 = IR4 + IR3 + IR2 + V2

and pulling the R4 drop back to the left....

V1 - IR4 = IR3 + IR2 + V2

does support that the parallel branches have the same voltage potential.

I apologize if I have headed off in the wrong direction, but it just seems like this info might give some help.

Bob

ahhhhh, loop analysis.

they tought us that at uni breifly before saying it has been superceded by nodal analysis.

[guitar_199]

Yes indeed.   This was at DeVry Dallas.  We  learned nodal analysis as well but, for some reason, this really stuck in my head better.  In ways, when you are addressing voltage drop issues.... I think this makes it a little easier to see.  But that's just me.....

[R.G.]

Both loop and nodal analysis are needed to really examine a circuit. They're both simple:

- the sum of voltages around any loop must be zero
- the sum of currents into any node must be zero

Using both together makes it easier in many cases to do solutions.

[Herr Masel]

And things magically become clearer once again! Thank you all, and thank you guitar_199!