# Help on Combination circuit

#### expertz

Joined Oct 20, 2022
9

But as a check for validity, current in R2 and R3 must be the same (except for rounding errors).
current in R4 + current in R5 = current in R6
current in R3 + current in R6 = current in R1
I described what I did and attached a file. R2 and R3 are pretty much the same, and R4 + R5 is almost the same as R6. I am still confused why R3 + R6 doesn't equal to 1. Can you please give some steps of how to do this correctly?

#### MrChips

Joined Oct 2, 2009
28,151
If the voltage across R1 = 12.4V
What is the voltage across R2 + R3?
What is the current through R2 and R3?

#### WBahn

Joined Mar 31, 2012
28,185
I don't know how to get there, because what I tried was combine 49 + 51 which is 100. Then I used I = V/R (7.6/100) to get 0.076 which is the current at the top part of the "box". I couldn't really do V= IR because that didn't work for me.

Instead I did 0.155 (total current) - 0.076 to get 0.079 A on the other branch, then I applied 0.079 A on the lone R6 then found the voltage (5.925 V) from there. I then did 7.6 - 5.925 which was 1.675 V, which I used to solve that side. I noticed the bottom part used 5.925 so then I did 7.6 - 5.925 and I got 1.675 V AGAIN, then I used it to solve the top side.

I don't know if it's correct since R2 and R3's current did not add to 0.076 A - if not can you explain what I could do?

View attachment 278973
You're pretty much on the right path.

Do you know what significant figures are? If not, do a quick Google search on it.

To prevent excessive roundoff errors, always carry one or two extra significant figures in your calculations. Your final answers should generally be given to three sig figs unless there's something specific that requires something different. When counting sig figs, many people (myself included) do not count a leading 1 (but other people do). Since this is a rule of thumb, there is no rigid, correct answer.

Look at your table and you can see some of the problems. You know that the voltage across the top branch in the box must total to (20 V - 12.4 V) or 7.6 V. Yet you only have (1.675 V + 1.675 V) or 3.35 V. So you know something is wrong.

Let's walk through your work and see if we can spot it (and what might have caused it).

I don't know how to get there, because what I tried was combine 49 + 51 which is 100. Then I used I = V/R (7.6/100) to get 0.076 which is the current at the top part of the "box". I couldn't really do V= IR because that didn't work for me.
This is fine, but you really need to start tracking your units properly. You have 49 Ω + 51 Ω = 100 Ω. Then you have I = 7.6 V / 100 Ω = 0.076 A.

Instead I did 0.155 (total current) - 0.076 to get 0.079 A on the other branch, then I applied 0.079 A on the lone R6 then found the voltage (5.925 V) from there.
Other than the sloppiness with the units, this is fine.

I then did 7.6 - 5.925 which was 1.675 V, which I used to solve that side.
I'm having to read between the lines a bit here, but I think you did fine as far as what you intended to do and your results for R4 and R5 look find.

I noticed the bottom part used 5.925 so then I did 7.6 - 5.925 and I got 1.675 V AGAIN, then I used it to solve the top side.

I don't know if it's correct since R2 and R3's current did not add to 0.076 A - if not can you explain what I could do?
I don't know if by "bottom part" you are referring to R5 alone and by "top side" you are talking about R4, or whether your "bottom part" is R4, R5, and R6 and your "top side" is R2 and R3.

I'm suspicious that it might be the latter, in which case here is where you are going off the rails. The currents in R2 and R3 don't add to 76 mA. They are in series, so they both have the same current, namely the 76 mA you found previously.

Also, the 5.925 V is the voltage across R6. It has no bearing or utility at all when it comes to calculating anything involving R2 and R3.

Since you know the current through R2 (and R3), you can use V=IR to calculate the voltages across each. Those voltages need to add to the 7.6 V that appears across your "box".

You were quite close.

#### WBahn

Joined Mar 31, 2012
28,185
I am still confused why R3 + R6 doesn't equal to 1. Can you please give some steps of how to do this correctly?
Why would they equal 1 (I assume you mean 1 Ω??) PLEASE treat units properly.

Even the notion of adding R3 to R6 makes no sense -- they are NOT in series.

#### MrChips

Joined Oct 2, 2009
28,151
R2 and R3 are close in values but for the purpose of this exercise, they don't have to be.
Given R2 = 49, R3 = 51
You can make
R2 = 50, R3 =50
R2 = 10, R3 = 90
R2 = 0, R3 = 100
It makes no difference to the rest of the circuit.

#### MrChips

Joined Oct 2, 2009
28,151
I do not see any file which you claim shows your steps.
This is an essential part of the exercise. Show your steps.

Step #1: R2 + R3 =
Step #2: R4//R5 =
Step #3: R4//R5 + R6 =

etc.
(Show the remaing two steps).

#### WBahn

Joined Mar 31, 2012
28,185
He has correctly calculated the total resistance and, indeed, done a pretty decent job of then walking through the voltages and currents of most of the components.

I suspect his mistake in his final step was more of a momentary conceptual goof than a fundamental misunderstanding; he seems to know the concepts, just not firmly enough yet to consistently identify and apply them correctly.

#### MrChips

Joined Oct 2, 2009
28,151
Yes, TS has arrived at the correct total effective resistance.
My point is, you need to show the intermediate results in order to work backwards to find individual voltages and currents.

#### WBahn

Joined Mar 31, 2012
28,185
Yes, TS has arrived at the correct total effective resistance.
My point is, you need to show the intermediate results in order to work backwards to find individual voltages and currents.
And also (which I thought might be your main point) is that the work really needs to be shown so that we can provide much more effective guidance.

#### MrChips

Joined Oct 2, 2009
28,151
And also (which I thought might be your main point) is that the work really needs to be shown so that we can provide much more effective guidance.
That too, as requested at the introduction of the exercise. But since TS arrived at the correct effective resistance we can assume TS did it correctly. We still need the intermediate results in order to continue.

#### Jerry-Hat-Trick

Joined Aug 31, 2022
232
In this circuit, a quick visual glance should tell you that the total resistance can be no more than 180 Ω.
In fact, I'd suggest it's imediately possible to see the overall resistance is less than 130 Ω
For a quick result, different resistors in parallel are always less than half the average, more so as the two values differ

#### expertz

Joined Oct 20, 2022
9
You're pretty much on the right path.

Do you know what significant figures are? If not, do a quick Google search on it.

To prevent excessive roundoff errors, always carry one or two extra significant figures in your calculations. Your final answers should generally be given to three sig figs unless there's something specific that requires something different. When counting sig figs, many people (myself included) do not count a leading 1 (but other people do). Since this is a rule of thumb, there is no rigid, correct answer.

Look at your table and you can see some of the problems. You know that the voltage across the top branch in the box must total to (20 V - 12.4 V) or 7.6 V. Yet you only have (1.675 V + 1.675 V) or 3.35 V. So you know something is wrong.

Let's walk through your work and see if we can spot it (and what might have caused it).

This is fine, but you really need to start tracking your units properly. You have 49 Ω + 51 Ω = 100 Ω. Then you have I = 7.6 V / 100 Ω = 0.076 A.

Other than the sloppiness with the units, this is fine.

I'm having to read between the lines a bit here, but I think you did fine as far as what you intended to do and your results for R4 and R5 look find.

I don't know if by "bottom part" you are referring to R5 alone and by "top side" you are talking about R4, or whether your "bottom part" is R4, R5, and R6 and your "top side" is R2 and R3.

I'm suspicious that it might be the latter, in which case here is where you are going off the rails. The currents in R2 and R3 don't add to 76 mA. They are in series, so they both have the same current, namely the 76 mA you found previously.

Also, the 5.925 V is the voltage across R6. It has no bearing or utility at all when it comes to calculating anything involving R2 and R3.

Since you know the current through R2 (and R3), you can use V=IR to calculate the voltages across each. Those voltages need to add to the 7.6 V that appears across your "box".

You were quite close.
Wait, so current is the same through R2 and R3? If so then I may have arrived at the correct answer?

#### MrChips

Joined Oct 2, 2009
28,151
Current in R4 + Current in R5 = Current in R6?