Thevenin's Resistance Query

Factual Questions
1

So, I’m enrolled in an entry level Electronics/Electricity course and I’m doing well. Or I was, until tonight, when the professor demonstrated how to find the Thevenin Resistance for a circuit laid out similarly to the one shown here, about halfway down the page, just beneath the bolded Wheatstone bridge text. To make that circuit identical to the one in question, you need to-

  1. Connect points A and B with a jumper wire
  2. Add a load to the circuit outside of the “H”, on a wire contiguous with the centerline of the “H”, something like H—Load----Ground

His take was that Rth with respect to the Load was-

Doesn’t seem right to me. My take is that it would be-

Conceptually, my take is that if trace the ‘path’ of the resistance from the Negative side of the Load back to the Positive side, it will travel through all the resistors, hence Rth is calculated by placing all the resistors in parallel.

The professor’s take was that, tracing the ‘path’ of the resistance from the Positive side of RL, it will split once, at the ‘first’ node, then run an end around through the shorted V source, ignoring the resistors labeled R1 and R3 in the example, then something else happens, and voila, R2 ll R4.

It seems to me that the only way this approach would work is if we didn’t add the wire between points A and B in the linked picture.

BTW, the answer was left out of the book, so the professor was winging it and there was no way to check, and frankly, I’ve found enough errors in the book that I don’t really trust it anyway.

So, who’s right?

And any further insight, web resources, etc that you’d care to add insofar as the conceptual side of Thevenin’s Resistance is concerned would be greatly appreciated…

2

Perhaps someone will jump to answer this : but can you please create the exact circuit in paint so there’s no ambiguity about what the problem is?

3

I’ll try to work something up, scan something, etc…

4

The “something else” is that the load resistance is set to zero, and a simplification of the formula.

You are calculating the thevinin EQUIVALENT circuit, which has both a voltage source and resistance in SERIES.
To calculate the equivalent circuit, the resistance and voltage are needed, so two equations are required. These two equations are usually obtained by using the following steps, but any conditions placed on the terminals of the circuit should also work:

  1. Calculate the output voltage, when in open circuit condition (no load resistor—meaning infinite resistance). This is VTh.
  2. Calculate the output current when the output terminals are short circuited (load resistance is 0). RTh equals VTh divided by this current.

So what do we have ? Vth will be a voltage divider between (R1 || R2 ) and (R3 || R4 ). Ith will be V/(R1||R2). … Then Rth is Vth/Rth

5

Here’s a google doc of the circuit in question…

Isilder, I understand where you’re coming from, what the Thevenin Equivalent Circuit is, why it is useful, etc, but my question concerns Thevenin Resistance very specifically along these lines-

Thevenin Resistance can be calculated using only only the circuit diagram and the component resistors, without involving Ohm’s Law at all.

The book is very vague on this, and tries to demonstrate how to do so by using arrows to delineate a ‘path’ of this resistance, but without some rules (which either aren’t there or I’ve overlooked), it quickly becomes something of a guessing game as to which ‘path’ is taken, and hence what the calculation is.

The professor, frankly, is over his head on this matter. He may well be right about this problem, but he’s been wrong often enough for me to ask the 'Net.

But thanks for the insight that Rth = Vth / Ith, which will give me something to check results by and should have been patently obvious :smack:

6

I’d say you’re clearly correct. Once you short the supply, clearly all the resistors are connected on one side to positive load and the other side to negative load. So the equivalent must be all four in parallel, as you said.

This is maybe a tad easier to see if you pre-transform with Ra = R1 || R2 and Rb = R3 || R4. The final resistance is then Ra || Rb, or R1 || R2 || R3 || R4.

7

Combining Dr. Strangelove’s simplification strategy with Islider’s insight that Rth = Vth / In, and inserting the numbers from the problem gives me this proof that the prof. was off the mark.

The original numbers were 48 V, R1 = 47 ohms, R2 = 15 ohms, R3 = 10 ohms, R4 = 30 ohms, if anyone is interested.

Dr. Strangelove, you seem to have some practical insight into how to figure this; care to lay out a quick and dirty step by step guide?

Seems like your first thought was converting the parallel Resistors into R equivalents, which is probably my weakest point. Obviously I do it all the time, I just forget that after I do it I can leave it that way and consider the equivalent circuit instead of the original circuit.

8

I build lots of simple circuits in my spare time, but I’m by no means an expert!

I’m not sure I can give advice that’s useful for everyone. I’m a highly visual person, so I imagine circuits as being connected by stretchy wires, and connection points being a bit like beads on a string, so that they can slide around. So when I have two resistors in parallel, I imagine them getting closer and closer together until I can sorta pinch them into a single resistor. Likewise for series, except that I slide the resistors into each other. This approach works for me but probably isn’t for everyone.

Unfortunately, I don’t have a step-by-step on how to see the areas of simplification; it’s mainly intuition for me. Obviously practice helps quite a bit! Fortunately, it sounds like you have a good start on the intuition if the professor’s answer didn’t sit right with you.

Something I just thought of, though–have you tried color coding? Sometimes it’s hard to follow where all the wires go and work out what’s really connected. Color might make it easier. Just give each connected network of wires a unique color. If you do this on your original diagram, with shorted supply and open load, you’ll find that you only need two colors for the whole thing. That’s a big hint that everything is in parallel and can be simplified! And in fact, no matter where the components are, if their two terminals are the same pair of colors, then they must be in parallel.

BTW, good job with your solution. It’s hard to overstate the utility of arriving at the same answer with two different methods. It’s not always possible but it’s a great form of error-checking.

9

I couldn’t help but smile at this:)

10

Yep, this is it.

Just for reference, here’s how I would do it, step-by-step:

Before proceeding with the problem, I would lump R[sub]1[/sub] and R[sub]2[/sub] together (which we’ll call R[sub]A[/sub]), and then lump R[sub]3[/sub] and R[sub]4[/sub] together (which we’ll call R[sub]B[/sub]). These can stay lumped for the duration of the problem:

R[sub]A[/sub] = R[sub]1[/sub]||R[sub]2[/sub] = R[sub]1[/sub]R[sub]2[/sub] / (R[sub]1[/sub] + R[sub]2[/sub])

R[sub]B[/sub] = R[sub]3[/sub]||R[sub]4[/sub] = R[sub]3[/sub]R[sub]4[/sub] / (R[sub]3[/sub] + R[sub]4[/sub])

So now we just have two resistors in series (a.k.a. voltage divider), and we want to find the Thévenin equivalent when the “output” of the circuit is across R[sub]B[/sub].

I always start with finding V[sub]Th[/sub]. This is simply the open-circuit (R[sub]L[/sub] = infinite ohms) voltage across R[sub]B[/sub]:

V[sub]Th[/sub] = V[sub]S[/sub]R[sub]B[/sub]/(R[sub]A[/sub] + R[sub]B[/sub]) = V[sub]S[/sub](R[sub]1[/sub] + R[sub]2[/sub])R[sub]3[/sub]R[sub]4[/sub] / ( R[sub]1[/sub]R[sub]2[/sub](R[sub]3[/sub] + R[sub]4[/sub]) + R[sub]3[/sub]R[sub]4[/sub](R[sub]1[/sub] + R[sub]2[/sub]) )

Next I find R[sub]Th[/sub]. To do this I make V[sub]S[/sub] = 0 (i.e. replace it with a wire), and then calculate the total resistance across R[sub]B[/sub]. When you do this, it simply puts R[sub]A[/sub] and R[sub]B[/sub] in parallel, hence

R[sub]Th[/sub] = R[sub]A[/sub]||R[sub]B[/sub] = R[sub]1[/sub]R[sub]2[/sub]R[sub]3[/sub]R[sub]4[/sub] / ( R[sub]1[/sub]R[sub]2[/sub](R[sub]3[/sub] + R[sub]4[/sub]) + R[sub]3[/sub]R[sub]4[/sub](R[sub]1[/sub] + R[sub]2[/sub]) )

11

I do that too; for instance, in my original circuit, since the Vs is shorted, R1 and R2 effectively have their tops on the ground wire and their bottoms on the wire leading directly toward RL. I’d redraw them as a big parallel network of 5 ‘squares’.

Or, as, you mention with the connection points moving like beads on a string, I’d slide all the connections of R1, R2, R3, R4 together until they fanned out into a, well, a fan shape, again obviously in parallel visually.

Boxes and a leaf rake.

For the record, I’m really trying to come up with some hard and fast ‘rule’, so that I know if I violate it, I’m on the wrong track. But it seems like the closest I can come is this-

Rth equals the lowest R value possible given that the resistance ‘travels’ through every point on the circuit in a sensible manner.

The sensible manner clause would keep you from taking the resistance through a bunch of parallel networks ad nauseum until Rth was effectively 0.

12

Not to disrespect the need to learn basic theory, but as a practising electronics engineer, if I encounter a circuit problem I can’t solve in 1 minute, I hand it over to a simulator such as SPICE. It won’t make a mistake, (subject of course to inputting the circuit right) but I might well if I try to solve it by hand.

There are several good versions available for free down load on t’net.

13

Nice, thanks for posting Bert Nobbins…I’ve been looking for something along those lines to help me learn the theoretical side.

As for the rest, if I run into another Rth problem, I guess I’ll just go the Rth = Vth / In route.