Engineering Equivalent Resistance in series-parallel circuit

AI Thread Summary
The discussion focuses on solving for the equivalent resistance in a series-parallel circuit, specifically finding the unknown resistor R2 given R1 and R3 values. Participants express frustration over the lack of guidance in their notes and share their calculations, which yield results close to the expected answer of 2250 ohms but not exact. A suggested method involves using the formula for equivalent resistance, emphasizing the importance of proper mathematical representation to avoid errors. The conversation highlights the challenges of online homework systems that do not allow for partial credit. Overall, participants appreciate the collaborative effort in tackling the problem despite the difficulties faced.
JustMerc
Messages
3
Reaction score
0

Homework Statement


This is a sample problem. I'm trying to figure it out before I do the actual homework. The issue is, the way to figure it out is not included in our notes. I'm absolutely lost (math isn't my strong point) I've also included my attempt at how I think it should be done below, which got my really close. We're rounding to the nearest whole number, but I'm still off by 2. If it's hard to see: R1 = 2900 ohms, R3 = 800 ohms, Req = 3490 ohms. Find R2 in ohms.
Math126_Sample2.jpg


Homework Equations


The only equations given to us for equivalent resistance in series-parallel is GT = G1 + G2 and Rx = 1/GT which is really not helping to figure this out.

The Attempt at a Solution


Here's my attempt. I subtracted R1 from RT (590 or .001694915) and divided by 1. Then subtracted R3 (.00125) from that (.000444915). Divided by 1 and got within 2 of the answer after rounding (2248). I'm so lost on this.
attempt.jpg
 
Physics news on Phys.org
I got the same thing as you. If you're given that the answer is 2250, it's likely that it simply has to do with the significant figures. Either way, I'd say you're correct.
 
  • Like
Likes berkeman
So then I'm actually doing it correctly? The problem with that is, the homework is online. There's only one answer which works. If you're close...it's still wrong and there's no partial credit, because we don't show our work. Regardless to that fact, I appreciate you looking at it and leading me to believe I'm on the right path.
 
Agreed. I got 2247.6, which rounds to your answer.
 
BTW, a better, more visual way to do this problem is like this:

\frac{1}{\frac{1}{x}+\frac{1}{800}} = 590

Then put the two terms in the denominator over a common denominator, and solve for x. :-)
 
berkeman said:
BTW, a better, more visual way to do this problem is like this:

\frac{1}{\frac{1}{x}+\frac{1}{800}} = 590

Then put the two terms in the denominator over a common denominator, and solve for x. :)

Nice, thank you berkeman! Even though I couldn't get the answer (2250) on the sample question, I attempted the homework and thus far it is working. I appreciate the help from you all. It's not fun going into this stuff blindly.
 
When you turn in written work, do yourself a favor and don't write stuff like R3 = 800 = 0.00125 because because 800 doesn't equal 0.00125 and R3 isn't 0.00125 ohms. Chances are whoever is grading your work pays attention to mathematical detail and may find what you wrote irritating, and it's best not to irritate the grader.
 
  • Like
Likes berkeman

Similar threads

Replies
2
Views
2K
Replies
4
Views
1K
Replies
14
Views
5K
Replies
11
Views
7K
Replies
3
Views
1K
Replies
8
Views
2K
Replies
3
Views
2K
Replies
4
Views
2K
Replies
5
Views
2K
Back
Top