# [SOLVED] Total Resistance of Circuit

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#### Deleted member 983787

Joined Dec 31, 1969
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Hello, I am solving some problems to gain a decent understanding of the basics of circuits.

I am stuck in that question. I know that I have to perform a star to triangle transform, but I do not seem to be able to solve it correctly.

Can someone explain in detail the solution?

#### SamR

Joined Mar 19, 2019
5,078
Welcome to AAC! We will not do your homework for you but if you show us what you have done to solve the problem we will comment on your work. I will suggest that you start by simplifying the circuit.

#### MrChips

Joined Oct 2, 2009
31,076
Hello, I am solving some problems to gain a decent understanding of the basics of circuits.

I am stuck in that question. I know that I have to perform a star to triangle transform, but I do not seem to be able to solve it correctly.

Can someone explain in detail the solution?
There is no "star to a triangle transform".
Just redraw the diagram and solve using series and parallel reduction.

#### Deleted member 983787

Joined Dec 31, 1969
0
There is no "star to a triangle transform".
Just redraw the diagram and solve using series and parallel reduction.
[/QU
There is no "star to a triangle transform".
Just redraw the diagram and solve using series and parallel reduction.
Don't 400, 25, 10 Ohm resistors form a star?

I will do what you said, it is far easier even if the transformation I proposed is possible.

#### Deleted member 983787

Joined Dec 31, 1969
0
Welcome to AAC! We will not do your homework for you but if you show us what you have done to solve the problem we will comment on your work. I will suggest that you start by simplifying the circuit.
Yes of course, thank you!

#### crutschow

Joined Mar 14, 2008
34,837
Hint:
Start by identifying all the resistors in parallel, and combine them into their equivalent resistance.

#### MrChips

Joined Oct 2, 2009
31,076
Don't 400, 25, 10 Ohm resistors form a star?

I will do what you said, it is far easier even if the transformation I proposed is possible.
You mean 400, 20 and 100? It looks like a star but it isn't.
400 and 100 are in parallel.

#### WBahn

Joined Mar 31, 2012
30,293
There are several techniques you can use to solve a circuit like this -- some easier, some harder. Leaving this particular circuit aside, it's hard to make recommendations without knowing what your background is, particular which analysis techniques you've been exposed to.

One thing to keep in mind is that problems are often intentionally drawn so as to look unconventional and to either tempt you into using a very suboptimal analysis choice, or to throw up your hands in resignation. Either result indicates that you don't yet have a firm enough grasp on the fundamentals and how to apply them.

Whenever you see circuits with lines crossing in problems, it's worth taking a few minutes to at least try to redraw the circuit by rearranging components so that the crossing wires are removed, and that components all lie along vertical and horizontal grid lines. This tends to make the relationships clear.

Also, it can be very helpful to color the various nets. This will help you identify components that are in series (a given colored wire is connected to exactly two component terminals -- assuming we are talking about two-terminal devices) or in parallel (two components are each connected to the same two colored wires).

Another thing that should become routine habit is to see if you can quickly come up with some bounds on what the final result has to be. In some cases, this alone gets you close enough to the answer to call it a day (particularly on multiple-choice questions, but even in the real world it happens).

Let's see that in action in this circuit.

The first thing you see is that the 10 Ω and 30 Ω resistors are in series with everything else. So even if everything else somehow gets reduced to 0 Ω, the absolute lowest that Rin can be is 40 Ω. Any answered you are offered, or any answer you come up with, that is less than that and you know it is wrong.

Now consider the following:

Call everything in the red region R1, as seen at the two nets that cross the boundary.

Even if we know absolutely nothing about R1, we see that it is in parallel with the 100 Ω resistor, and we know that two resistors in parallel have an equivalent resistance that is no greater than the smallest resistor, and no larger than half the value of the larger. So that means that the parallel resistance of R1 and 100 Ω can be no more than 100 Ω, and that that is in series with our 40 Ω from earlier.

So now we know that

40 Ω ≤ Rin ≤ 140 Ω

But we can do better.

By pulling the 80 Ω resistor out of the red region, and noting that what is left is in series with it, we establish that the smallest R1 can be is 80 Ω. Given to resistors in parallel, the equivalent resistance has to be at least half of whatever the smallest one is. so that means that now we have a lower bound of Rin that is 80 Ω, so

80 Ω ≤ Rin ≤ 140 Ω

We can actually push this up by doing the math to determine that 100 Ω || 80 Ω = 44.4 Ω, making the lower bound 84.4, but we generally like to keep the estimations simple so that we can do them easily, preferably by inspection while doing the math on the fly in our heads.

By pulling another one of the resistors out (I won't say which one, because that would be giving too much of a hint on the problem you are actually working on, which in and of itself is a bit of a hint), we can narrow our range all the way down to

80 Ω ≤ Rin ≤ 93 Ω

Carrying this the rest of the way, we get that

86 Ω ≤ Rin ≤ 93 Ω

That's a pretty tight set of bounds (less than ±5% of the geometric mean, which is about 89.4 Ω, but using the arithmetic mean of 89.5 Ω is just as good) and it is unlikely that a mistake in your detailed calculations would lead to a value that is wrong, but within those bounds. It's also likely that, if that did turn out to be the case, you would get significant partial credit and, in the real world, the result, while wrong, would probably be close enough to be completely acceptable.

It's been my experience when doing these kinds of estimations that the actual answer tends to be within, very roughly, 10% of the spread. So here our spread is ±5%, so the actual answer is likely within about ±0.5%. This is because the behavior of resistances in parallel has a very strong crowding out effect and it requires rather extreme values to move the the actual answer very far away from the mean.

So I would expect the actual value to be between about 89 Ω and 90 Ω and would be a bit surprised if it were less than about 88 Ω or more than about 91 Ω, which is certainly not to say that this can't happen.

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#### Deleted member 983787

Joined Dec 31, 1969
0
I think I have solved the problem thanks to your help dear users!
400//100=>R=80
80 in series with 20=>R=100
100//25=>R=20
100//(20+80)=>R=50
So the final result is Rtotal=50+30+10=90 Ohm

#### crutschow

Joined Mar 14, 2008
34,837
So the final result is Rtotal=50+30+10=90 Ohm
You got it grasshopper.

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