The problem is to determine the total resistance for the network. I came up with roughly 25 ohms, and I will detail each step. I am asking someone to verify this please. I am not sure if I needed to use any delta-wye transforms. The numbers are not "nice," unlike the other problems, so it is possible there is an error in my method, or not.

series resistors on far right (working towards nodes a & b)
(30 +60 +18) ohm = 108 ohm
108 ohm || 16 ohm = (1/108 +1/16)^-1 = 432/31 ohm
432/31 ohm in series with 28 ohm = (432/31 +28) ohm = 1300/31 ohm
1300/31 ohm || 40 ohm = (31/1300 +1/40)^-1 = 2600/127 ohm
2600/127 ohm series with 20 ohm = (2600/127 +20) ohm = 5140/127 ohm
5140/127 ohm || 24 ohm = (127/5140 +1/24)^-1 = 30,840/2047 ohm
30,840/2047 series with 25 ohm, 10 ohm = 30,840/2047 +25 ohm +10 ohm = 102,485/2047 ohm
102,485/2047 ohm || 50 ohm = (2047/102,485 +1/50)^-1 = 241,409/9650 ohm ~= 25 ohm

I thought it would be too cluttered to include a re-drawn circuit at each step. I worked from right to left in the picture only making series and parallel simplifications. I was not sure if a delta to wye transformation was needed, since I missed class due to health reasons that day. Any help is welcome. Thanks for reading. See you soon.

 

Hello,

Is there a wire parallel to the 60 Ohms on the end?
If so, the 60 Ohms has no meaning and only the wire counts as the 60 Ohms is shorted.

Bertus

 

Hello,

Is there a wire parallel to the 60 Ohms on the end?
If so, the 60 Ohms has no meaning and only the wire counts as the 60 Ohms is shorted.

Bertus

Yes, I take it that wires have negligible or zero resistance, so then the current would flow through the wire instead of the 60 ohm resistor, shorting it as you said. Then would total resistance be 100 ohm (a nice number)?
Thanks Bertus!

 

No, total resistance is exactly equal to 25Ω

 

No, total resistance is exactly equal to 25Ω

Just making sure everyone is paying attention...

 

The hundred ohms refers to the stack paralleling the 24ohm. The result for the whole thing is slightly greater than 26ohm.

 

The hundred ohms refers to the stack paralleling the 24ohm. The result for the whole thing is slightly greater than 26ohm.

Wrong check your math. As I said early the total resistance is exactly equal to 25Ω

 

Wrong check your math. As I said early the total resistance is exactly equal to 25Ω

Without giving direct answers to the post: I get 100-ohms for the stack as noted, paralleling 24-ohms. This is not going to give you an even number....

 

Are we not allowed to post solutions? (I'm kinda new here... ) Because I worked it out and got exactly 25 ohm total equivalent resistance for the network. Hint is you end up with some equivalent resistance in series with the 25ohm and 10 ohm resistor, and then that parallels the 50 ohm one. Big trick was to recognize the 60 ohm resistor is completely bypassed by current because of that extra wire, so that can be removed from the simplifications. So, if I took an ammeter through the 60 ohm resistor and the 30 ohm resistor, I would get 0A current? Would I read any voltage on a voltmeter on the ends of that resistor?

 

Hello,

As you are the Thread Starter, you may post the answer, so we can see of you have created the correct solution.
Others will comment if there are still things wrong.

The members that are helping you are not allowed to give the full answer, but give you hints on getting the correct solution.

Bertus

 

Alright. Since you said the 60 ohm resistor is effectively meaningless due to the wire in parallel to it, I assume this takes all of the current since it has negligible resistance compared to the 60 ohm resistor, there is a series combo. of (30 +18)ohm = 48 ohm. Then this is in parallel with the 16 ohm resistor, giving (1/16 +1/48)^-1 ohms = 12 ohm. Next there is a series connection with the 28 ohm resistor, so (12 +28) ohm = 40 ohm. This is connected in parallel to the 'diagonal' 40 ohm resistor, so you get half the identical resistors, or (1/40 +1/40)^-1 ohm = 20 ohm. This is in series through the bottom resistor of 20 ohm also, for (20 +20) ohm = 40 ohms. This is in parallel to the 24 ohm resistor, so (1/24 +1/40)^-1 ohm = 15 ohm. Then the series simplified circuit is with a 25 ohm and 10 ohm resistor, so (10 +25 +15) ohm = 50 ohm. This parallels the final 50 ohm resistor, for half the identical resistance, (1/50 +1/50)^-1 ohm = 25 ohm.

I assume the reason the 60 ohm resistor has no current is that there is no potential difference between its two endpoints.

 

yes

 

∞≤

View attachment 80091
The problem is to determine the total resistance for the network. I came up with roughly 25 ohms, and I will detail each step. I am asking someone to verify this please. I am not sure if I needed to use any delta-wye transforms. The numbers are not "nice," unlike the other problems, so it is possible there is an error in my method, or not.

One thing that can help you sanity check your work is to place bounds on the answer. When you start out, you only know that

0 ≤ R ≤ ∞

Any answer you get is going to look okay (as long as it isn't negative). But what if we could place upper and lower bounds on that real quick? Then, if you get an answer that is outside those bounds you will KNOW that you did something wrong.

In this case you have a couple of obvious bounds. You have a 50Ω resistor in parallel with something, so you know the answer MUST be less than 50Ω.

0 ≤ R ≤ 50Ω

Notice that this simple step right here rules out what you though might be the answer in Post #3!

If you go one section further out, what is in parallel with the 50Ω has to be greater than 35Ω (since it has 25Ω and 10Ω in series with something), so the smallest the answer can be is 50Ω||35Ω which is 20.5Ω. (Note that in order to maintain the inequality, you should round away from the center).

20.5 ≤ R ≤ 50Ω

If we then take the 24Ω resistor into account, we know that the MOST resistance we can have in the parallel branch is 35Ω+24Ω or 59Ω. That means that the most the overall resistance can be is 50Ω||59Ω which is 27.1Ω.

20.5 ≤ R ≤ 27.1Ω

So with almost no effort you know that the answer must be between 20.5Ω and 27.1Ω. Many mistakes you make in working out the circuit will probably throw the answer outside of these bounds and let you catch them. Also, note that if we just pick the midpoint (geometric mean) of the limits we get 23.6Ω and the extremes of our bounds differ from this less than 15%. In many, many real world applications this would be "good enough".

You can walk the bounds in tighter pretty easily, too. The 24Ω resistor is in parallel with the path through the 40Ω and 20Ω resistor, so you have a max resistance for that path of 24Ω||60Ω which is 17.1Ω. That places the 50Ω resistor in parallel with a resistance of, at most, 52.2Ω giving an overall resistance of, at most, 25.6Ω.

20.5 ≤ R ≤ 25.6Ω

This is a tight enough bound to rule out the conclusion offered by Denesius in Post #6.

Now everything that is left is in parallel with the 40Ω resistance and the smallest that this can be is 28Ω, so our new lower limit is 24.8Ω.

24.8Ω ≤ R ≤ 25.6Ω

Notice how we now have a very tight bound. The midpoint of 25.2Ω places us well within 2% of either bound.

 

Nice way of working from the source out into the circuit! I wouldn't say, "almost no effort," though. It does take some thought to arrive at those bounds.

 

Nice way of working from the source out into the circuit! I wouldn't say, "almost no effort," though. It does take some thought to arrive at those bounds.

Some, but not a lot. You can even do them in such a way that everything is in your head (at least the math) by using a couple of simple rules about parallel branches. If you have two resistors in parallel, Ra and Rb, then the combined resistance has to lie between half of the smaller of the two and the smaller of the two. You can even bound this tighter because the most it can be is also half of the larger of the two.

Using this approach you get right off the bat that the resistance has to be between 25Ω and 50Ω. Adding the 24Ω resistor to the mix you know the most it can be is half of 59Ω, so call that 30Ω. So by quick inspection you have the bounds to between 25Ω and 30Ω. That's almost certainly good enough bounds to provide a good check on the answer since, in the real world, if we are wrong but within those bounds it is probably "good enough".

But let's walk the bounds from the right to the left and see what we get.

We note immediately that the 60Ω is shorted out, so we have 16Ω||48Ω. That has to lie between 8Ω and 16Ω. That is in series with 28Ω so we have something between 36Ω and 44Ω. That is in parallel with 40Ω, so we have something that is between 20Ω and 22Ω. That is in series with 20Ω so we have something between 40Ω and 42Ω. That is in parallel with 24Ω, so now we have something that is between 12Ω and 21Ω. That is in series with 35Ω giving us something between 47Ω and 56Ω. That is in parallel with 50Ω giving us a final bound that is between 25Ω and 28Ω. Notice how no calculator was needed and all of the math was super-simple. Yes, we didn't get nearly as tight a bound, but we got these bounds within, literally, under a minute (took me 50 seconds) and we know that we can guess an answer, 26.5Ω, that is within 1.5Ω of the correct answer, which is roughly 5% max error.

 

Ah, great. Using this method of bounds, I think you can learn more about the circuit than just relying on the formulas given. And it's a lot quicker!

Also, is there any way of finding bounds for a delta or wye connection? Those have some known transform equations, but is there any way of checking your answer without using the equations for them to get the new resistance values?

 

....... = 25 ohm....

My bad!