Resistance total in combination circuit

Thread Starter

Eddey

Joined Sep 9, 2020
10
I am currently new to circuits and i am struggling to calculate this circuit, i dont know what to consider in paralell and what is series, to put in the Rt=1/(1/r1+1/r2+1/r3) formula. A correct answer and calculations would be much appreciated so that i may understand how to caluclate combination circuits in the future.
 

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jpanhalt

Joined Jan 18, 2008
11,087
Break it into smaller parts. For example, solve the obviously series resistors first. Then the resulting parallel pair, then series again, then parallel.
 

Thread Starter

Eddey

Joined Sep 9, 2020
10
Break it into smaller parts. For example, solve the obviously series resistors first. Then the resulting parallel pair, then series again, then parallel.
I tried this and still think i have the wrong answer i got 5.7k
 

sagor

Joined Mar 10, 2019
786
I get 5.747k as an answer, so your 5.7k seems correct to one decimal point....
Do 22k and 10k in series first, then parallel calculate that total with the 15k. Result is "X". Add "X" to 3.3k and end up with total resistance "Y" for the entire leg on the right side.
Now. calculate parallel resistance with 10k and "Y"
 

WBahn

Joined Mar 31, 2012
27,875
I am currently new to circuits and i am struggling to calculate this circuit, i dont know what to consider in paralell and what is series, to put in the Rt=1/(1/r1+1/r2+1/r3) formula. A correct answer and calculations would be much appreciated so that i may understand how to caluclate combination circuits in the future.
It really helps us help you if you show your work as fully as possible. That way we can see what approach you are taking, what you are doing right, and where you are going wrong.

Since it appears that you managed to get the right answer, but seem unsure that it IS the right answer, let's walf through it.

As for what is in parallel and what is in series, both have a very simple definition you can rely on. Two components are in series if and only if the current flowing in one IS the current flowing in the other. Two components are in parallel if and only if the voltage across one IS the voltage across the other. It's important to keep in mind that two components may be in neither series nor parallel.

With that in mind, let's see what we can do with this circuit. First, trying to "calculate this circuit" is highly ambiguous. You want to start being far more specific in describing what you are trying to do. The inference in this case is that you are trying to find the total (or effective) resistance as seen by the voltage source. Later you will discover that we often talk about the resistance seen by difference components.

It is often helpful to redraw the circuit with the component that is doing the "seeing" removed (in this case the voltage source) and color coding the various nodes.

Req.png
Notice that the blue node is only connected to two components. This means that whatever current enters this node from one component must flow out the node into the other component. Hence they are in series. So these can be combined into a 32 kΩ resistor that is connected to the yellow node and the red node. But the 15 kΩ resistor is also connected to the yellow and red nodes, so these two resistors have the same voltage across them and are therefore in parallel and can be combined into a single 10.21 kΩ resistor. But now the yellow node only connects this resistor and the 3.3 kΩ resistor, placing them in series and having an equivalent resistance of 13.51 kΩ. At this point we are left with two resistors, each connected to the green and red nodes, putting them in parallel and yielding the final answer of 5.75 kΩ.

If you need to, redraw the circuit after each combination is reduced.

Another very helpful tool is to estimate what the answer should be, particularly if you can put high and low bounds on it. Then if your answer isn't within those bounds, you know you have done something wrong. Also, sometimes the bounds end up being good enough to provide the answer without going any further.
 

Thread Starter

Eddey

Joined Sep 9, 2020
10
I get 5.747k as an answer, so your 5.7k seems correct to one decimal point....
Do 22k and 10k in series first, then parallel calculate that total with the 15k. Result is "X". Add "X" to 3.3k and end up with total resistance "Y" for the entire leg on the right side.
Now. calculate parallel resistance with 10k and "Y"
This is initally how i worked it out with the and got 10.2k for x, then adding the 3.3k resistor in series to x making it 13.5k then using that to calculate the resultant of 10k and effective 13.5k paralell circuit if that makes sense.
 

Thread Starter

Eddey

Joined Sep 9, 2020
10
It really helps us help you if you show your work as fully as possible. That way we can see what approach you are taking, what you are doing right, and where you are going wrong.

Since it appears that you managed to get the right answer, but seem unsure that it IS the right answer, let's walf through it.

As for what is in parallel and what is in series, both have a very simple definition you can rely on. Two components are in series if and only if the current flowing in one IS the current flowing in the other. Two components are in parallel if and only if the voltage across one IS the voltage across the other. It's important to keep in mind that two components may be in neither series nor parallel.

With that in mind, let's see what we can do with this circuit. First, trying to "calculate this circuit" is highly ambiguous. You want to start being far more specific in describing what you are trying to do. The inference in this case is that you are trying to find the total (or effective) resistance as seen by the voltage source. Later you will discover that we often talk about the resistance seen by difference components.

It is often helpful to redraw the circuit with the component that is doing the "seeing" removed (in this case the voltage source) and color coding the various nodes.

View attachment 216773
Notice that the blue node is only connected to two components. This means that whatever current enters this node from one component must flow out the node into the other component. Hence they are in series. So these can be combined into a 32 kΩ resistor that is connected to the yellow node and the red node. But the 15 kΩ resistor is also connected to the yellow and red nodes, so these two resistors have the same voltage across them and are therefore in parallel and can be combined into a single 10.21 kΩ resistor. But now the yellow node only connects this resistor and the 3.3 kΩ resistor, placing them in series and having an equivalent resistance of 13.51 kΩ. At this point we are left with two resistors, each connected to the green and red nodes, putting them in parallel and yielding the final answer of 5.75 kΩ.

If you need to, redraw the circuit after each combination is reduced.

Another very helpful tool is to estimate what the answer should be, particularly if you can put high and low bounds on it. Then if your answer isn't within those bounds, you know you have done something wrong. Also, sometimes the bounds end up being good enough to provide the answer without going any further.
Thankyou for this explanation it has been incredibly helpful, i see now that my intial calculations were correct and gave me this result as i calculated in this manner but this is a good visual for future reference to help me understand how to determine what to calculate.
 

Thread Starter

Eddey

Joined Sep 9, 2020
10
I asked an engineering friend to help with this issue and was given this diagram and explanation, can someone explain why this is wrong because my friend got an answer that appeared to be wrong compared to the answers all of you have so kindly given me.
this seems like a way that may work to draw the diagram more simply but his calculations give an answer along the lines of 2k.
 

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WBahn

Joined Mar 31, 2012
27,875
You can keep the friend if you want, but find someone else for help on engineering topics.

His assertion that anytime there's nodes in between them that connect to other resistors means that they're in parallel is just simply wrong.

Look at his diagram -- he is claiming that the positive terminal of the battery is connected directly to every resistor except the 22 kΩ resistor. Look at the original diagram. Is that the case? He's claiming that both the 3.3 kΩ resistor and the 15 kΩ resistor are connected directly across the voltage source? Look at the original diagram. Is that the case?
 

Thread Starter

Eddey

Joined Sep 9, 2020
10
Ok thankyou all so much im glad i posted about this issue there was some real confusion about how to do it correctly and i think i understand fairly well now, i appreciate everyones time in responding to me.
 

WBahn

Joined Mar 31, 2012
27,875
I asked an engineering friend to help with this issue and was given this diagram and explanation, can someone explain why this is wrong because my friend got an answer that appeared to be wrong compared to the answers all of you have so kindly given me.
this seems like a way that may work to draw the diagram more simply but his calculations give an answer along the lines of 2k.
I started to mention this previously, but let it go. However, now is a good time to bring it up.

One thing you want to get in the habit of doing is estimating answers based on quick simplifications to the problem. This will give you a ballpark answer. Often, you can choose your simplifications so that the estimated answer is guaranteed to be too high or too low, giving you a one-sided bound that you know the correct answer has to be on the right side. The best situation is to do two estimates, one that is guaranteed to be too high and one that is guaranteed to be too low so that you know the correct answer has to be between them.

If we look at the original circuit, we can take the mess of three resistors in the bottom right and say that the equivalent value will be somewhere between 0 Ω (a short circuit) and ∞ Ω, an open circuit. If it's a short circuit, then we are left with 10 kΩ in parallel with 3.3 kΩ and that means that the absolutely lowest that the resistance can be is 2.48 kΩ. If it's an open circuit, that means the most the resistance can be is 10 kΩ. So the answer MUST be between those to limits -- and since 2 kΩ is well outside that, we know that that answer is wrong (or that our estimate has an error, which is also a possibility).

When we have two resistors in parallel, we know immediately that the highest resistance the combination can have is the smallest value of the two and that the smallest it can have is half the smallest value of the two.

We can narrow things down further pretty easily by inspection. We see that we have 10 kΩ in series with 22 kΩ giving us 32 kΩ in series with 15 kΩ. The lowest the result could be is half of 15 kΩ, or 7.5 kΩ, and the most is 15 kΩ. So when we add the 3.3 kΩ we get something between 10.3 kΩ and 18.3 kΩ. These are in parallel with 10 kΩ, so the smallest the combination can be is 5 kΩ and the biggest is 10 kΩ.

We can even go better than this.

Given two resistors in parallel, the largest that the combined resistance can be is the smaller of {the smaller resistance, half the larger resistance}. Think about this until you understand why this is the case.

So with these rules in mind, our 15 kΩ in parallel with 22 kΩ has to be between a minimum of 7.5 kΩ and 11 kΩ. When we add the 3.3 kΩ to these we get 10.3 kΩ and 14.3 kΩ. When we then place these in parallel with the remaining 10 kΩ resistor we get something in the range of 5 kΩ to 7.2 kΩ. That's getting pretty tight; in fact, in many situation finding the upper and lower bounds narrows the answer well enough that there's no need to go any further. On an exam, it might be because only one of the available answers is within that range. In the real world, it is often the case that any answer within those bounds is acceptable.
 
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