Electric Circuit - Equivalent Resistance

Click For Summary

Homework Help Overview

The discussion revolves around finding the equivalent resistance of a resistor network, specifically focusing on two questions related to the arrangement of resistors in series and parallel configurations.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss their interpretations of the resistor paths and equivalent resistance calculations, noting discrepancies between their results and provided answers. Questions arise regarding the flow of current in the circuit and the implications of a middle wire on resistance calculations.

Discussion Status

Participants are actively exploring different paths within the circuit and questioning the assumptions made about current flow. Some guidance has been offered regarding visualizing the circuit and considering the implications of wire resistance, but no consensus has been reached on the correct approach.

Contextual Notes

There is mention of a third path in the circuit that some participants have not considered, and discussions about the implications of current flow in the middle wire suggest a need for clearer definitions of circuit behavior.

leslielau
Messages
8
Reaction score
0

Homework Statement



For 1 and 2 : Find the equivalent resistance for the resistor network provided


Homework Equations



For resistor in series : Req = R_{1}+R_{2}+...R_{n}

For resistors in parallel : Req = (1/R_{1}+1/R_{2}+... 1/R_{n})^{-1}

The Attempt at a Solution



For question 1 :

I have considered the two paths shown :

1st path (blue) : total resistance = 2 ohm
2nd path (orange) : total resistance = 2 ohm

so since these two paths are in parallel , Req = 1 ohm

But answer provided = 2/3 ohm

For Question 2 :

Since no current passes through the middle wire, upper path = 10 ohm
lower path = 10 ohm

So, total resistance = 5 ohm

However , the answer used the method of (1/4+1/8)^-1 + (1/2+1/6)^-1 = 4.17 ohm

But I don't understand this part, if there is current flowing in the middle wire, there would be one current flowing down and one current flowing up. Won't they cancel out each other? Thanks !
 

Attachments

  • Resistor.jpg
    Resistor.jpg
    16.8 KB · Views: 440
Physics news on Phys.org
leslielau said:

Homework Statement



For 1 and 2 : Find the equivalent resistance for the resistor network provided


Homework Equations



For resistor in series : Req = R_{1}+R_{2}+...R_{n}

For resistors in parallel : Req = (1/R_{1}+1/R_{2}+... 1/R_{n})^{-1}

The Attempt at a Solution



For question 1 :

I have considered the two paths shown :

1st path (blue) : total resistance = 2 ohm
2nd path (orange) : total resistance = 2 ohm

so since these two paths are in parallel , Req = 1 ohm

But answer provided = 2/3 ohm

For Question 2 :

Since no current passes through the middle wire, upper path = 10 ohm
lower path = 10 ohm

So, total resistance = 5 ohm

However , the answer used the method of (1/4+1/8)^-1 + (1/2+1/6)^-1 = 4.17 ohm

But I don't understand this part, if there is current flowing in the middle wire, there would be one current flowing down and one current flowing up. Won't they cancel out each other? Thanks !

For question 1 there is a third path you have not considered.
 
I have thought of two : One through the middle through all 3 resistors, and another one , turning left between middle and rightmost resistor to pass through the middle resistor.

Which one is correct?
 
leslielau said:

Homework Statement



For 1 and 2 : Find the equivalent resistance for the resistor network provided

Homework Equations



For resistor in series : Req = R_{1}+R_{2}+...R_{n}

For resistors in parallel : Req = (1/R_{1}+1/R_{2}+... 1/R_{n})^{-1}

The Attempt at a Solution



For question 1 :

I have considered the two paths shown :

1st path (blue) : total resistance = 2 ohm
2nd path (orange) : total resistance = 2 ohm

so since these two paths are in parallel , Req = 1 ohm

But answer provided = 2/3 ohm

For Question 2 :

Since no current passes through the middle wire, upper path = 10 ohm
lower path = 10 ohm

So, total resistance = 5 ohm

However , the answer used the method of (1/4+1/8)^-1 + (1/2+1/6)^-1 = 4.17 ohm

But I don't understand this part, if there is current flowing in the middle wire, there would be one current flowing down and one current flowing up. Won't they cancel out each other? Thanks !

Homework Statement


Homework Equations


The Attempt at a Solution


For the second part, there may well be current flowing through the middle wire - but since we consider a wire to have zero resistance, the potential at each end will be the same.
re-draw the circuit with the centre wire much, much shorter [like zero length] and see what the circuit looks like. you should then see where their answer comes from.

EDIT: Sorry but i don't know how to load pictures
 
leslielau said:
I have thought of two : One through the middle through all 3 resistors, and another one , turning left between middle and rightmost resistor to pass through the middle resistor.

Which one is correct?

The second - as it offers another path back to the battery having passed through only one of the resistors - the same as the two paths you identified at first.
 
leslielau said:
I have thought of two : One through the middle through all 3 resistors, and another one , turning left between middle and rightmost resistor to pass through the middle resistor.

Which one is correct?

Try this: use a high-lite pen to high-lite the wire from the positive terminal of the battery. continue shading as far as you can along any wires without passing through a resistor, then have a look at what you have shaded.

Preferably with a different colour, start at the negative terminal and shade back along wires - but again don't go through any resistor.

What do you notice?

EDIT: It is 2:30 am here, I am off to bed!
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 42 ·
2
Replies
42
Views
6K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
10
Views
3K
Replies
4
Views
2K
Replies
3
Views
2K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 14 ·
Replies
14
Views
1K
Replies
9
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K