Hi

Please, I need help to understand this issue. I made some exercises on Thévenin equivalent but I'm stuck in this one.

Here I attached two files taken off from O'Malley's book ("Basic circuit analysis" edited by schaum...) (pages 103-104)


I have two doubts:

First - why we consider the 30 V from 30 voltage if we DON'T consider the 8-resistor? "The 8-22 resistor has no effect on this voltage because there is zero current flow through it as a result of the open circuit." It is ok, but if there is no current why we must consider the 30 V to calculate V th?
I= V / R if there is no current therefore 0 A = 30V / R internal (of the voltage source). Confused. :|

Second - Why you cannot consider the "5-R and 4-R resistors" to calculate the Vth? O'Malley said that "Notice that the 5-R and 4-R resistors have no effect on VTh." but no explanation given.

I thought that the correct would be:

(V - 100)/10 + V / 40 + 20 + V / 5 = 0 (the 4-resistor was not used, because we have the current known... I think.)

Why not put Vth = -20 * 4 = -80 V? Directly... I confused about the adding of 30 V if there is no current in that voltage source...


Concerning Rth, no problem.


Thanks for your help,
Jorge Almeida

 

Ohm's law (V = IR) only applies if there is a circuit. since the 8-ohm resistor is hanging (it is only connected at one end), there is no circuit, and it has the same voltage at both nodes.

*edit* I stated this wrong. Ohm's law always applies, but current can only flow if there is a circuit, therefore with no circuit there can be no voltage drop. V = 0A * R = 0v

The 4-ohm resistor has no effect on Vth because it has constant current flowing through it. The voltage across it will be a constant 80V (20A * 4-ohm), regardless of what the output voltage is.

The 5-ohm resistor has no effect on Vth because it is connected in parallel with the voltage source, which outputs constant voltage regardless of current draw. The voltage across the 5-ohm resistor is 100 V, even if you were to reduce its resistance to 1-ohm or increase it to 100k-ohm.

 

After all, thanks for clarification. Please see below just to finish at once the smaller doubts.

Ohm's law (V = IR) only applies if there is a circuit. since the 8-ohm resistor is hanging (it is only connected at one end), there is no circuit, and it has the same voltage at both nodes."

*edit* I stated this wrong. Ohm's law always applies, but current can only flow if there is a circuit, therefore with no circuit there can be no voltage drop. V = 0A * R = 0v"

That is why I am asking why 30 V is considering for the calculation of Vth...
and O'Malley considered the -80 V + 30 V for Vth...

"The 4-ohm resistor has no effect on Vth because it has constant current flowing through it. The voltage across it will be a constant 80V (20A * 4-ohm), regardless of what the output voltage is."

and what if there was not there the source current? The voltage drop would be 100 V??

"The 5-ohm resistor has no effect on Vth because it is connected in parallel with the voltage source, which outputs constant voltage regardless of current draw. The voltage across the 5-ohm resistor is 100 V, even if you were to reduce its resistance to 1-ohm or increase it to 100k-ohm."

I didn't fully understand this paragraph. Can you clarify, please? Do you mean that ANY resistor in paralell with the VOLTAGE source WILL HAVE *always* the same voltage as the VOLTAGE SOURCE? If so.. why?

"with the voltage source, which outputs constant voltage regardless of current draw." this is the same as we are considering the VOLTAGE SOURCE as an ideal? Is it right? I am not sure but in the real sources the voltage can change in SOURCE VOLTAGE with different current?...


Thanks a lot!

 

The 30V source is considered because it is in series with the output nodes, so it adds 30V to whatever the internal Vth is.

I don't understand what you mean by your second response. The voltage drop across the 4-ohm resistor is irrelevant because the voltage across the current source could be anything.

And to your third: Yes, I'm saying the voltage across a resistor in parallel with a voltage source will be contant exactly because we assume the voltage source is ideal.

 

The 30V source is considered because it is in series with the output nodes, so it adds 30V to whatever the internal Vth is.

I don't understand what you mean by your second response. The voltage drop across the 4-ohm resistor is irrelevant because the voltage across the current source could be anything.

And to your third: Yes, I'm saying the voltage across a resistor in parallel with a voltage source will be contant exactly because we assume the voltage source is ideal.

thanks.

"and what if there was not there the source current? The voltage drop would be 100 V??"

hmm... it cannot be 100 V. If you took off the source current, the voltage for 4 ohms resistor would be the same as Vth (as we see indicated in the image).


To confirm: ANY resistor in paralell with the VOLTAGE source WILL HAVE *always* the same voltage as the VOLTAGE SOURCE due the fact that voltage (from voltage source) is CONSTANT and there are no other resistors between the 5 ohms resistor.

ok, I think that's all for now concerning this exercise.

Thank you.