Thanks, but I ment can you show the maths....

 

You said you have already shown it using nodal and mesh analysis--isn't that enough?

It shouldn't be too hard to convert what I described in English to a mathematical description.

 

there was an answer in there somewhere

 

it gave 7 A. This exercise doesn't need even any calculator.


Edit: you have the Rth determined... For Vth we need to know the voltage of 10 V with the voltage across the R of 4 ohms. It is easy since the current drawn in the R4ohms only can be.... A (remember that the load - the resistor of 12 ohms MUST BE discarded always to calculate the Vth). To know the voltage in the resistor Ohm's "law" tell us that V=R.I so V= 4*... V
Well, finally you have Vth and Rth calculated... and once again we use Ohm's Law by using I = Vth / (12+4) = the answer.
Now.. it must be a piece of cake.

 

Start with the basic equation for I:

\(I\,=\, \frac{V_{ab}}{12}\)

Next, by inspection, form the Millman's Theorem equation for the voltage Vab:

\(V_{ab}\,=\,\frac{\frac{80}{4}\,+\,8}{\frac{1}{4} \,+ \, \frac{1}{12}}\)

Simplify equation:

\(V_{ab}\,=\,\frac{20\,+\,8}{\frac{1}{4} \,+ \, \frac{1}{12}}\)

Simplify equation:

\(V_{ab}\,=\,\frac{28}{\frac{1}{4} \,+ \, \frac{1}{12}}\)

Substitute above equation into the expression for I at the top of this post:

\(I\,=\, \frac{1}{12} *\, \frac{28}{\frac{1}{4} \,+ \, \frac{1}{12}}\)

Simplify equation:

\(I\,=\, \frac{28}{\frac{12}{4} \,+ \, \frac{12}{12}}\)

Simplify equation:

\(I\,=\, \frac{28}{3 \,+ \, 1}\)

Simplify equation:

\(I\,=\, \frac{28}{4}\)

Simplify equation:

\(I\,=\,7 \, \)Amps

hgmjr

 

For this question the load is given, but not all thevinin questions have the load given. If no load was given the question would just state find Eth and Rth for this circuit and hence find 'I' if a certain R(load) is connected. Is that possible with this circuit? that was what I was trying to ask.. as i dont think you can use nodal analysis and mesh current analysis if there is no load. Also would the current source cause I problem, because it is that current due to the load is it not???

 

For this question the load is given, but not all thevinin questions have the load given. If no load was given the question would just state find Eth and Rth for this circuit and hence find 'I' if a certain R(load) is connected. Is that possible with this circuit?

In post #9, I said "In your circuit if you remove the 12Ω resistor from the a-b terminals, and then replace the 80 volt source with a short circuit, and the 8 amp source with an open circuit, the resistance looking into the a-b terminals will be 4Ω; this is Rth (or Req).".

So, Rth is found without the load resistor being connected.

Then I said "To calculate the open circuit voltage, notice that the 2Ω resistor has no effect. Then calculate the voltage drop across the 4Ω resistor. If the 12Ω resistor is absent, how much current will there be in the 4Ω resistor? If you know the current in the 4Ω resistor, you know the voltage drop across it. That voltage drop is in series with the 80 volt source, so what will be the voltage seen at the a-b terminals? This will be Vth (or Veq).".

In post #20, I explained "...a resistor in series with a current source causes some extra voltage drop in the branch containing the current source and the resistor in series with it, but it does not change the current coming from the current source. It has no effect on the rest of the circuit".

Thus Eth is found with the load resistor disconnected.

This circuit is so simple that Rth and Eth can be found by inspection. But, there is no reason why it can't be done with nodal or mesh analysis. Just because a circuit has some particular resistor missing doesn't prevent the calculation of the voltage across the terminals where that resistor had been connected, by nodal or mesh analysis.

If you didn't know that this was a Thevenin problem, and there was no 12Ω resistor, why wouldn't you be able to solve the circuit for all the node voltages and mesh currents?

 

Then I said "To calculate the open circuit voltage, notice that the 2Ω resistor has no effect. Then calculate the voltage drop across the 4Ω resistor. If the 12Ω resistor is absent, how much current will there be in the 4Ω resistor?



If you know the current in the 4Ω resistor, you know the voltage drop across it.



That voltage drop is in series with the 80 volt source, so what will be the voltage seen at the a-b terminals? This will be Vth (or Veq).".



In post #20, I explained "...a resistor in series with a current source causes some extra voltage drop in the branch containing the current source and the resistor in series with it, but it does not change the current coming from the current source. It has no effect on the rest of the circuit".

Thus Eth is found with the load resistor disconnected.

This circuit is so simple that Rth and Eth can be found by inspection. But, there is no reason why it can't be done with nodal or mesh analysis. Just because a circuit has some particular resistor missing doesn't prevent the calculation of the voltage across the terminals where that resistor had been connected, by nodal or mesh analysis.

If you didn't know that this was a Thevenin problem, and there was no 12Ω resistor, why wouldn't you be able to solve the circuit for all the node voltages and mesh currents?

OK bare with me..

'I' in 4ohm is 8A??

vd in 4ohms is 8 x 4 = 32v??

if in series and is added to 80v = 112 which is wrong...

 

Wrong?
Eth=112V and Rth=4Ω
So know we connect 12Ω resistor and we end up with this simple schematics:

So easily can find current that is flow in this circuit:

I = 112V/(4Ω+12Ω) = 7A
And voltages on A - B terminal is equal
V_ab = 7A*12Ω = 84V

 

nice one bud.. forgot to add rth

 

nice one bud.. forgot to add rth

If you had produced your work, this discussion wouldn't have lasted 30 posts.

 

.............................

 

eh? i did...

OK bare with me..

'I' in 4ohm is 8A??

vd in 4ohms is 8 x 4 = 32v??

if in series and is added to 80v = 112 which is wrong...