Circuits - Calculating Effectice Resistance

Click For Summary

Homework Help Overview

The discussion revolves around calculating the effective resistance in circuits that involve both series and parallel configurations. The original poster presents two parts of a problem, seeking guidance on how to approach the calculations, particularly when both configurations are present.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the effective resistance of circuits with mixed series and parallel components. The original poster attempts to apply known formulas for series and parallel resistances but expresses uncertainty about combining them. Some participants suggest focusing on specific branches of the circuit to simplify the calculations.

Discussion Status

The discussion has progressed with participants providing hints and confirming understanding. One participant successfully calculates the effective resistance for part B by breaking down the circuit into individual branches. There is a collaborative atmosphere with participants offering support and clarification.

Contextual Notes

There is mention of a lack of prior instruction on calculating effective resistance in mixed configurations, which may influence the original poster's approach. The complexity of the circuit layout is noted as a challenge in understanding how to proceed.

FaraDazed
Messages
347
Reaction score
2
Sorry for my crude drawings but I thought it was easiest and quickest this way to post the question.

Homework Statement



Calculate the effective resistance for the circuits below.

A.
circuit_A.jpg


B:
circuit_B.jpg



Homework Equations


[tex]R_{series}=\sum\limits_{i=1}^n R_i \\<br /> \frac{1}{R_{parallel}}=\sum\limits_{i=1}^n \frac{1}{R_i}[/tex]


The Attempt at a Solution



We have never been told how to calculate the effective resistance where both series and parallel occur, only how to do it if it is series and how to do it if it is parallel. But using intuition for part A i just did the bit in series and then the bit in parallel and added them like below.

[tex] \sum\limits_{i=1}^n R_i = 2+2=4 \\<br /> \sum\limits_{i=1}^n \frac{1}{R_i} = \frac{1}{2}+\frac{1}{2} = 1 ∴ R=1 \\<br /> ∴R_{total}=4+1=5Ω[/tex]

For part B I am scratching my head as to where to start because the whole thing is in parallel and then there are parrellel bits inside of that iyswim. Any hints on how to approach part B would be much appreciated, thanks.
 
Last edited:
Physics news on Phys.org
Forget about the rest of the circuit .First consider only 12,4,16 Ohm resistors in the top branch .What do you think ? Are they in series or parallel ? Can you calculate their effective resistance ?
 
Last edited:
  • Like
Likes   Reactions: 1 person
Tanya Sharma said:
Forget about the rest of the circuit .First consider only 12,4,16 Ohm resistors in the top branch .What do you think ? Are they in series or parallel ? Can you calculate their effective resistance ?

Right ok, I think I see what your getting at. So as the 12,4 and 16 are in parrallel I can calculate the effective resistance of just that, ignoring the rest (for now), and replace the 12,4,16 branches with the effective resistance. Ok I will have a go now.
 
Yes...that is correct .
 
  • Like
Likes   Reactions: 1 person
OK so for part B then I broke it down to the individual branches to make it easier on the eye.

Top Branch:
[tex] 3+9+(\frac{1}{12}+\frac{1}{4}+\frac{1}{16})^{-1}=14.53Ω[/tex]

Middle Branch[tex] 20+10=30Ω[/tex]

Bottom Branch[tex] 5+(\frac{1}{8}+\frac{1}{33})^{-1}=6.44Ω[/tex]

Putting it all together[tex] (\frac{1}{14.53}+\frac{1}{30}+\frac{1}{6.44})^{-1}=3.88Ω[/tex]
 
Nice work :thumbs:
 
  • Like
Likes   Reactions: 1 person
Thanks for your help :)
 
You are welcome :smile:
 
  • Like
Likes   Reactions: 1 person

Similar threads

Replies
10
Views
3K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 6 ·
Replies
6
Views
1K
Replies
9
Views
1K
  • · Replies 31 ·
2
Replies
31
Views
5K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K