Combine Resistors: Parallel Method to 1 Resistor

[DiamondV]

Homework Statement

Combine the resistors into one single resistor

Homework Equations

Resistors in parallel = 1/1/R1 + 1/R2 +1/R3...
Resistors in series = add

The Attempt at a Solution

I have two ways of getting the resistance.
One is by saying that the 80ohm is parallel 20ohm and then adding the 16ohm which gives me 32ohm(which is the correct given answer) (1/1/80 + 1/20)+16 = 32ohm (correct)Another way I thought of is to say that the 16ohm and 80ohm are in parallel and then adding the 20ohm which gives me 33.3ohms which isn't correct. (1/1/80 +1/16)+20 = 33.3ohm (not correct)

What I don't understand is how is the second method not giving me the same answer?

 

[phinds]

If you treat the voltage source as a short circuit, which apparently you are doing, and you have a short, as drawn, across the output, then the three resistors are in parallel. Your second method treats one of them as being in series, which is not correct.

 

[DiamondV]

If you treat the voltage source as a short circuit, which apparently you are doing, and you have a short, as drawn, across the output, then the three resistors are in parallel. Your second method treats one of them as being in series, which is not correct.

I'm not treating the voltage as a short circuit.

 

[phinds]

I'm not treating the voltage as a short circuit.

In that case, how do you ever get the 20 ohm resistor to be in parallel with anything?

 

[DiamondV]

In that case, how do you ever get the 20 ohm resistor to be in parallel with anything?

Good point. The way I've been told is that the resistors connected across the same two nodes are in parallel. Thats how I am getting the parallel sum of 16 and 80 ohm resistors and then adding that with the 20ohm resistor which is giving me 33.3 which is incorrect.

 

[phinds]

Good point. The way I've been told is that the resistors connected across the same two nodes are in parallel.

correct

Thats how I am getting the parallel sum of 16 and 80 ohm resistors ...

But the 16 and 80 ohm resistors don't share both nodes.

OOPS ... sorry. I got myself confused. Yes, that's how you got that far, but how did you then (in the first part) get the 80 and 20 to be in parallel, since THEY don't share both nodes

 

[CWatters]

The expression "Combine the resistors into one single resistor" does not compute without further information. Why are you trying to combine the resistors?

For example if you were trying to calculate the load seen by the voltage source (with the output short circuited as drawn) THEN method 2 would be correct eg (16//80)+20=33.3. On the other hand if you were trying to calculate the output impedance then method 1 would be correct.

They give different answers because they are the resistance at different places in the circuit.

 

[DiamondV]

correct
But the 16 and 80 ohm resistors don't share both nodes.

OOPS ... sorry. I got myself confused. Yes, that's how you got that far, but how did you then (in the first part) get the 80 and 20 to be in parallel, since THEY don't share both nodes

Yeah. I looked through my notes again and realized that was wrong. after I posted it. I think the answer of 32ohms given may be wrong.

This is MY final solution(idk if correct):

The 16ohm and 80ohm resistor both share the two red nodes thus they are in parallel. So I apply the resistors in parallel formula.

(1/1/16 +1/80) + 20 = 33.3ohms

 

[DiamondV]

The question is actually from here http://people.clarkson.edu/~jsvoboda/eta/dcWorkout/thevenin.pdf
However I don't need to find the voltage as I know how to do that. My issue is in the understanding of the parallel parts. So I adapated that example 1 from that link to show you guys what I am having trouble with exactly with no extra stuff. They get the thevenin resistance to be 32ohms and I am getting it as 33.3. The method they use of changing voltage sources into current etc. is something that we haven't done so I don't understand their solution fully.

 

[CWatters]

They are correct the answer is 32R. All will be clear once you have covered Thévenin and Norton Equivalent Circuits.

Note that their final circuit has a 16V voltage source rather than the original 20V so they haven't just combined the resistors. There is more to it.

 

[DiamondV]

They are correct the answer is 32R. All will be clear once you have covered Thévenin and Norton Equivalent Circuits.

We've done thevenin but not norton. I am just trying to find more examples since I haven't really been given many examples for circuits

 

[DiamondV]

If I short

They are correct the answer is 32R. All will be clear once you have covered Thévenin and Norton Equivalent Circuits.

Note that their final circuit has a 16V voltage source rather than the original 20V so they haven't just combined the resistors. There is more to it.

Can you tell how you would do this? getting the equivalent resistance here.

 

[phinds]

What can you tell about the shared nodes of each of the resistors?

 

[gneill]

So, it seems that what you're really looking for is the Thevenin Equivalent for your circuit (without the short circuit on the output, which I presume was added as part of the implied circuit analysis)?

Otherwise, what would be the point since the short circuit would impose a 0 Ω result no matter what, right?

The usual approach is to suppress the sources and determine the resistance presented to the output terminals by the remainder of the network. That would mean the the 20 V source is replaced with a short, thus making parallel the 20 and 80 Ohm resistors as you first surmised.

 

[donpacino]

Homework Statement

Combine the resistors into one single resistor

Homework Equations

Resistors in parallel = 1/1/R1 + 1/R2 +1/R3...
Resistors in series = add

The Attempt at a Solution

I have two ways of getting the resistance.
One is by saying that the 80ohm is parallel 20ohm and then adding the 16ohm which gives me 32ohm(which is the correct given answer) (1/1/80 + 1/20)+16 = 32ohm (correct)Another way I thought of is to say that the 16ohm and 80ohm are in parallel and then adding the 20ohm which gives me 33.3ohms which isn't correct. (1/1/80 +1/16)+20 = 33.3ohm (not correct)

What I don't understand is how is the second method not giving me the same answer?

That black line on the far right of the circuit. why is it a different thickness and style?
should it be there

 

[CWatters]

If I short

Can you tell how you would do this? getting the equivalent resistance here.

See the reply by phinds. Mark up the nodes with a red dot.

 

[CWatters]

Can I suggest you practice simplifying some other series/parallel resistor circuits and return to the problem in that PDF another time. That particular problem is somewhat different. It's not really about simplifying a resistor network on it's own and using it as the basis for that has ended up confusing you.