4
\$\begingroup\$

I am reading up on Thevenin's Theorem and this has prompted a few questions that I hope somebody will be able to answer.

This is the tutorial I am reading.

The circuit below is then given as an example. (I can't post images yet so the image urls will have to do, I'm also limited in the number of hyperlinks, so I'll have to remove )

enter image description here

The first step involves removing the 40Ω resistor and shorting all the emfs to produce this.

enter image description here

It then says the 10Ω resistor is in parallel with the 20Ω resistor.

If there is no connection between A and B why are the resistors not in series, to me it looks like there is only one path?

Secondly does shorting out the emfs mean removing them? Presumably if they where still connected the resistors would stop a short circuit.

Later the voltages are added back in to produce this circuit:

enter image description here

Is the reason it is shown to go counterclockwise because the 20V battery is more powerful than the 10V battery causing current to flow in that direction?

I'd appreciate help with any of these questions. Thanks very much in advance!

\$\endgroup\$
  • \$\begingroup\$ DNN, please change "Thevenins" to "Thevenin's". The page doesn't let me make such a small change. The tutorial is wrong. See en.wikipedia.org/wiki/Th%C3%A9venin%27s_theorem \$\endgroup\$ – Telaclavo Apr 24 '12 at 10:27
  • \$\begingroup\$ @Telaclavo where exactly the tutorial is wrong? \$\endgroup\$ – clabacchio Apr 24 '12 at 12:44
  • \$\begingroup\$ @clabacchio In its title. It is not "Thevenins Theorem", but "Thevenin's Theorem" (or, even better, "Thévenin's Theorem", but I'm not asking that much). A teaching page should care about those details, too. \$\endgroup\$ – Telaclavo Apr 24 '12 at 13:26
  • 2
    \$\begingroup\$ @Telaclavo: Your comment "The tutorial is wrong" suggests that it contains major mistakes. Qualifying it as such for a minor spelling error is highly misleading! (BTW, the site OP refers to is very good!) \$\endgroup\$ – Federico Russo Apr 24 '12 at 14:39
  • 1
    \$\begingroup\$ @FedericoRusso Then I hope you've never used the expression "that book/paper/reference/guy is wrong" (or even "is right"), because it's the same. Come on. \$\endgroup\$ – Telaclavo Apr 24 '12 at 15:48
2
\$\begingroup\$

If there is no connection between A and B why are the resistors not in series, to me it looks like there is only one path?

They are in parallel because if you look at the circuit from AB, you have two paths: one passing through the 10 Ohms resistor, and the other passing through the 20 Ohms resistor.

As vicatcu pointed, they are in parallel because they share both terminals at the nodes A and B.

Secondly does shorting out the emfs mean removing them? Presumably if they where still connected the resistors would stop a short circuit.

Thévenin's analysis consists in synthesizing a linear circuit into a voltage supply and a resistor: being linear, the circuit will be characterized by a straight line, which will have a certain slope and offset. The slope is given by the equivalent resistance, and the offset is given by the equivalent voltage supply.

Mathemathically this coud be expressed as:

$$ V_{AB} = ( I_{AB} \cdot R_{th} ) + V_{th} $$

(considering Iab going into the Thévenin circuit)

Removing the voltage suppies means that you are analizing only the part of the circuit which depends by the current (and in this case it's represented by the resistors). Shorting them means "I remove all the offsets and consider just the slope"

It's like if you have a straight line passing for the point (0,2) in a graph: you can say that it's the sum of the same line passing in the origin (0,0) and the constant '2'.

The figure represents graphically the problem (click to enlarge):

enter image description here

on the x axis you have the current, and on the y you have the voltage. The Thévenin equivalent voltage source is a horizontal line, while the resistance determines the steepness of the total slope. The load resistor is a line passing for the origin, and they meet at the operating point.

Note that, arbitrarily fixed the direction of the current as going into the Thévenin equivalent, the resistor characteristic is inverted. You can also do the inverse.

Is the reason it is shown to go anti-clockwise because the 20V battery is more powerful than the 10V battery causing current to flow in that direction?

When you have a circuit to analyze, you first define a conventional sense for the current, and then you find the numerical value (positive or negative) with the corresponding sign. In this case, being a voltage source dominant, they choose that sense because it's more likely to reflect the real current flow.

\$\endgroup\$
  • \$\begingroup\$ Parallel Resistors Defined: Define Resistor "A" with Terminals "X" and "Y", and Resistor "B" with Terminals "J" and "K". The resistors are "in parallel" if either ((node A.X == node B.J) AND (node A.Y == node B.K)) OR ((node A.Y == node B.J) AND (node A.X == node B.K)). I'm using "==" here to mean that the two operands are aliases of the same electrical node. \$\endgroup\$ – vicatcu Apr 24 '12 at 14:24
  • \$\begingroup\$ @vicatcu in this case they share the nodes A and B \$\endgroup\$ – clabacchio Apr 24 '12 at 14:32
  • \$\begingroup\$ right on - I wasn't correcting you, just expanding upon your answer \$\endgroup\$ – vicatcu Apr 24 '12 at 15:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.