EE - Finding Effective Resistance in Parallel Circuits

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Discussion Overview

The discussion revolves around calculating the effective resistance in a circuit containing both series and parallel resistors. Participants are trying to understand the correct approach to find the equivalent resistance, particularly in the context of a homework problem involving an 18 Ohm resistor in series with three parallel resistors of 100 Ohm, 25 Ohm, and 30 Ohm.

Discussion Character

  • Homework-related
  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion over their calculations for the effective resistance, stating they calculated the equivalent resistance of the parallel resistors to be 12 Ohms, leading to a total of 30 Ohms when combined with the series resistor.
  • Another participant suggests that the textbook answer of 6 Ohms might be derived from the current calculation based on a 5V source and the total resistance of 30 Ohms.
  • There is a discussion about the interpretation of the problem statement, with one participant questioning if they are missing something since a classmate achieved the textbook answer.
  • Another participant emphasizes that the equivalent resistance seen by the source cannot be less than 18 Ohms due to the series resistor, suggesting a misunderstanding of the circuit configuration.
  • Participants discuss the importance of accurately representing the circuit schematic to clarify the calculations.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct equivalent resistance value, with multiple interpretations of the problem and calculations presented. There is disagreement regarding the implications of the series resistor on the total resistance.

Contextual Notes

Some participants mention potential mistakes in textbooks and the need for clarity in circuit schematics. There is also uncertainty about whether the problem was interpreted correctly, particularly regarding the equivalent resistance seen by the source.

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EE -- Effective Resistance

Homework Statement


I don't know how to post schematics so I hope this makes sense. There are four resistors in the circuit. An 18 Ohm resistor is the first resistor out of the 5V voltage source followed by three resistors hooked up in parallel. They have a resistance of 100ohms 25ohms and 30ohms. Find the equalization resistance.

Homework Equations


Hey guys this is my first post to bear with me while I figure things out. And thanks for any help in advance. So I'm having trouble calculating the effective resistance across circuits and I can't figure out why! The math seems to easy but I seem to be getting them wrong. Could someone explain to me what I am doing wrong.

for 2 resistors in parallel... Req= R1*R2/(R1+R2) (will yield a resistance less than the lower resistor)
for more than 2 resistors in parallel... 1/Req = 1/R1 + 1/R2 + 1/R3... (will yield a resistance less than the lower resistor)
for resistors in series... Req = R1+R2 (will yield a resistance greater than the largest resistors)

The Attempt at a Solution


So I started by find the Req for the parallel resistors. 1/Req = 1/r1 + 1/r2 + 1/r3 = (1/100) + (1/25) + 1/(30) which I got meant Req = 12ohms. I then noticed the first resistor and the combined resistors are in series. So I added R1 + R2 = Req = 30ohms. The back of the book says the answer is 6ohms =/ is there something I need to be doing with the voltage? I'm just so confused, simple math and I'm coming up short every time. If you guys need a diagram please let me know how and I will post one. Thanks again
 
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Take a picture of the schematic diagram and post it here. Use the "manage attachments" below the reply window.
 


Hope this works
 

Attachments



Guess I should explain how I simplified the whole circuit now.

I combined the 10ohm and 40ohm (in parallel) ... (10)(40)/50 = 8ohm
I combined the 8ohm and the 22ohm (in series)... 8+22 = 30ohm
I combined the 100ohm the 25ohm and the 30ohm (in parallel)... (1/100)+(1/25)+(1/30)=12ohms
I combined the 12ohm and the 18ohm(in series)... 12+18 = 30ohm
 


You'll notice the image says "pending approval". It can't be viewed until one of the moderators notices it and approves it.

If you upload an image to one of the free image hosting sites, and post the link here, you can get a quicker response.
 


http://img30.imageshack.us/img30/3494/circuir.jpg there that should work I think, thanks for the advice
 
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Shepherd7 said:
Guess I should explain how I simplified the whole circuit now.

I combined the 10ohm and 40ohm (in parallel) ... (10)(40)/50 = 8ohm
I combined the 8ohm and the 22ohm (in series)... 8+22 = 30ohm
I combined the 100ohm the 25ohm and the 30ohm (in parallel)... (1/100)+(1/25)+(1/30)=12ohms
I combined the 12ohm and the 18ohm(in series)... 12+18 = 30ohm

Your calculations are correct. Textbooks are known to make mistakes.

You will note that if the source is 5 volts, loaded by 30 ohms, the current will be 1/6 amps.

That's the only way I see to come with a "6" in this problem.
 


The problem stated excatly is "Find the equivalent resistance seen by the source in the circuit." Does everything still appear to be the same. A kid in my class said that he was able to match the answers out of the book so I still think I'm missing something =/
 


Shepherd7 said:
Hope this works

Please post your attachment in PDF format next time. Fewer worries about macros and such. You can get a PDF writer for free from PrimoPDF if you don't have one already.
 
  • #10


Shepherd7 said:
The problem stated excatly is "Find the equivalent resistance seen by the source in the circuit." Does everything still appear to be the same. A kid in my class said that he was able to match the answers out of the book so I still think I'm missing something =/

Look at the circuit. There's an 18 ohm resistor in series with the source, first thing. There's no way you can do anything after that point (right end of the 18 ohm resistor) to create less than 18 ohms as seen by the source, if you can only add various combinations of real resistors.

You could do so if you allowed negative resistors, but I don't see that happening here.

Could you be looking at the wrong schematic?

You better ask the other guy for some details of what he did.

Please come back here and tell us what the problem was when you find out.
 

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