# Circuit Analysis: Parallel Series component failure.

#### VinceClortho

Joined Mar 30, 2016
20
Hello all. I'm a first time poster here with a problem from my Circuits 1 class that has stumped me all day. Please forgive me for any improper protocol or such. Part b is what I am stuck on. My first instinct is telling me that R3 has shorted (indicated by the increase in current). I have considered R1 being shorted/open and R2 being open (and honestly I think I'm missing something) but I don't think that's the right answer. Please disregard my redrawn circuit on the right. Thanks for any help in advance.

#### VinceClortho

Joined Mar 30, 2016
20
Maybe an updated picture of the solved circuit would help?

#### shteii01

Joined Feb 19, 2010
4,644
There are something like 7 scenarios. Are you supposed to outline and explain each scenario or do a numeric solution?

#### VinceClortho

Joined Mar 30, 2016
20
The question says to show all work. But I think the work that shows the faulty element will suffice.

#### shteii01

Joined Feb 19, 2010
4,644
The question says to show all work. But I think the work that shows the faulty element will suffice.
Ok. Like you said, assume that R3 changed, but all other components are the same. Replace R3=4.7 kOhm with an X.
You know that you have 64.2 V across R4+X. You know that you have 12.77 mA through R4+X.
Apply Ohm's Law: V=IR
64.2 V = 12.77 mA (5 kOhm +X)
Solve for X. That will tell you the new value of R3. Then take old R3, subtract X, and that will illustrate how much the R3 decreased.

#### VinceClortho

Joined Mar 30, 2016
20
solving for X gave me 27.4 (ohms correct?). Subtracting that from the original value of 4700 ohms resulted in 4.6726 kohms. Rounding up would give 4.7k, the original value. Going by what you said about that being how much R3 decreased does that indicate a short in R3?

#### WBahn

Joined Mar 31, 2012
29,167
Solve for the current in R3 symbolically.

Now ask what type of change in each component would result in that current increasing. For each component, find out what the current in R3 would be if the change in that component was extreme. In other words, if you determine that decreasing R1 would increase the current in R3, then set R1 equal to zero. But if you determine that increasing R1 would increase the current in R3, then set R1 equal to infinity. This lets you quickly evaluate the low hanging fruit -- it is unlikely that the "fault" will be anything other than a single resistor either failing as a short or failing as an open.

#### WBahn

Joined Mar 31, 2012
29,167
solving for X gave me 27.4 (ohms correct?). Subtracting that from the original value of 4700 ohms resulted in 4.6726 kohms. Rounding up would give 4.7k, the original value. Going by what you said about that being how much R3 decreased does that indicate a short in R3?
Well, check it. Replace R3 by a short and calculate how much current would flow in it.

#### VinceClortho

Joined Mar 30, 2016
20
Ok, following your advice the current came out to be 12.09mA. So that is wrong. But I did the same thing with R4 and the current came out to be 12.76mA. Which isn't exactly what I'm looking for but is deadly close. Also, using the method mentioned by shteii01 and making substitutions I was able to come up with 4.7 kohms. I'm hoping this suggests that it is R4 that has shorted and that one digit difference is just the result of some rounding.

#### JoeJester

Joined Apr 26, 2005
4,390

On the second question, one change increased the current through R3 to 12.76 e-2 amperes, close to the problem statement of 12.77 mA.

#### VinceClortho

Joined Mar 30, 2016
20
My answer to the first question was "Yes, the circuit is operating properly because V2=64.2V."

I did the calculations for R3 another way and got the correct answer of 12.77mA. So the answer is R4 has shorted. Thank you all so much for your help.

#### WBahn

Joined Mar 31, 2012
29,167
Ok, following your advice the current came out to be 12.09mA. So that is wrong. But I did the same thing with R4 and the current came out to be 12.76mA. Which isn't exactly what I'm looking for but is deadly close. Also, using the method mentioned by shteii01 and making substitutions I was able to come up with 4.7 kohms. I'm hoping this suggests that it is R4 that has shorted and that one digit difference is just the result of some rounding.
Yes, that is correct -- and you are correct that the tiny difference is merely an issue of rounding differences between the way you did it and the way the person that determined the "correct" answer did it.

#### The Electrician

Joined Oct 9, 2007
2,936
Vince, you say in another thread that your instructor likes to throw you curveballs. Throw one back.

Point out that this problem has an infinite number of solutions. For example, if R3 decreased in value to 2200 ohms, and R4 decreased to 2500 ohms, you get a current of 12.76 mA in R3--the same current as if R4 is shorted. Or any other changes in R3 and R4 such that their sum is 4700 ohms.

#### WBahn

Joined Mar 31, 2012
29,167
While true, the instructor could simply throw back at him Occam's Razor and ask which is the more likely fault -- that a single component would fail as a short or that two or more components would fail in exactly just the right partial way as to reproduce the same result? Yes, multiple component failures can and do happen and certainly partial component failures are far from impossible, but the simplest explanation that fits the observed data is generally the most likely solution and almost always the one that should at least be suggested/explored first. And that is without considering the likely context of the question which probably only deals with short/open failure of single components.

From my perspective, if a student answered the question by finding multiple components that changed in just the right ways I would likely give them full marks, or nearly full marks, provided their solution approach was valid. If I got the feeling that they were just being a smart ass and almost certainly knew that the "right" answer should have involved finding a single component total failure, they would get full marks (the penalty they paid was wasting time that could have been used elsewhere), but if I got the feeling that they didn't grasp the notion of looking at complete failures of a single-component to begin with and thus were grasping at straws, especially if that had been the focus of that block of instruction, then they would lose at least some points for failing to demonstrate a grasp of the current material.

#### The Electrician

Joined Oct 9, 2007
2,936
While true, the instructor could simply throw back at him Occam's Razor and ask which is the more likely fault -- that a single component would fail as a short or that two or more components would fail in exactly just the right partial way as to reproduce the same result? Yes, multiple component failures can and do happen and certainly partial component failures are far from impossible, but the simplest explanation that fits the observed data is generally the most likely solution and almost always the one that should at least be suggested/explored first. And that is without considering the likely context of the question which probably only deals with short/open failure of single components.

From my perspective, if a student answered the question by finding multiple components that changed in just the right ways I would likely give them full marks, or nearly full marks, provided their solution approach was valid. If I got the feeling that they were just being a smart ass and almost certainly knew that the "right" answer should have involved finding a single component total failure, they would get full marks (the penalty they paid was wasting time that could have been used elsewhere), but if I got the feeling that they didn't grasp the notion of looking at complete failures of a single-component to begin with and thus were grasping at straws, especially if that had been the focus of that block of instruction, then they would lose at least some points for failing to demonstrate a grasp of the current material.
I didn't say that the student should assert that it was more likely that two resistors had changed in just the right way to give a solution, nor that the student should only submit a two resistor change solution as the answer. I merely suggested that the student could mention that there are other solutions.

Had I been the student, I would have submitted the "expected" solution, and pointed out that besides that, there are other solutions. That way full credit should be obtained for the "expected" solution, and the instructor will see that the student's "grasp" is excellent.

I think the student who realized that there are an infinite number of solutions would not likely be "grasping at straws", since that infinite panoply of solutions includes the "expected" single component failure, which would be obvious to this gifted student. If this were my student, I would think he had demonstrated an excellent grasp of the current material, going beyond the "expected" solution.

As far as the "smart ass" problem, it's up to the student as to what attitude he adopts. I certainly don't recommend a smart ass attitude.

#### SLK001

Joined Nov 29, 2011
1,548
My answer to the first question was "Yes, the circuit is operating properly because V2=64.2V."

I did the calculations for R3 another way and got the correct answer of 12.77mA. So the answer is R4 has shorted. Thank you all so much for your help.
Well, that is NOT the correct answer to the question. I see R2=2.2K, R3=4.7K and R4=5K. That would make the equivalent resistance to be ~1.15k. 1K in series with 1.15 no way results in 62V.

#### JoeJester

Joined Apr 26, 2005
4,390
Well, that is NOT the correct answer to the question. I see R2=2.2K, R3=4.7K and R4=5K. That would make the equivalent resistance to be ~1.15k. 1K in series with 1.15 no way results in 62V.
You might want to revisit your calculations or post your work here for others to review.

#### WBahn

Joined Mar 31, 2012
29,167
I didn't say that the student should assert that it was more likely that two resistors had changed in just the right way to give a solution, nor that the student should only submit a two resistor change solution as the answer. I merely suggested that the student could mention that there are other solutions.

Had I been the student, I would have submitted the "expected" solution, and pointed out that besides that, there are other solutions. That way full credit should be obtained for the "expected" solution, and the instructor will see that the student's "grasp" is excellent.

I think the student who realized that there are an infinite number of solutions would not likely be "grasping at straws", since that infinite panoply of solutions includes the "expected" single component failure, which would be obvious to this gifted student. If this were my student, I would think he had demonstrated an excellent grasp of the current material, going beyond the "expected" solution.

As far as the "smart ass" problem, it's up to the student as to what attitude he adopts. I certainly don't recommend a smart ass attitude.
Oh, I realize what you were suggesting and have no problem with it.

I have, sadly, seen more than a few students that have "grasped at straws" and stumbled upon a correct solution while having no comprehension of how they got there or why it is a correct solution. Usually, but not always, of course, you can get a feel for whether they fall in that category by hints throughout their work. But, on the bright side, that work often gives you a lot of inside into what they were thinking and which straws they were grasping at so that you can make good, meaningful comments that, if they read them and think about them (some do, some don't) can help clear up a lot of misconceptions.

Also, the "smart ass" categorization was made tongue-in-cheek. I love it when a student goes beyond what was asked in the assignment and points out subtleties or even outright weaknesses and errors in an assignment.

#### WBahn

Joined Mar 31, 2012
29,167
Well, that is NOT the correct answer to the question. I see R2=2.2K, R3=4.7K and R4=5K. That would make the equivalent resistance to be ~1.15k. 1K in series with 1.15 no way results in 62V.
Where's the 62 V figure coming from?

#### SLK001

Joined Nov 29, 2011
1,548
You might want to revisit your calculations or post your work here for others to review.
Oops... I was solving a different problem. I saw one thing and solved another. I'll blame "reading and scrolling"! 1.79k is the equivalent resistance, so 64.2V is correct.