# Electric Circuit - Equivalent Resistance

• leslielau
In summary, the discussion revolves around finding the equivalent resistance for a resistor network provided in two questions. The equations for resistors in series and parallel are provided. For question 1, the two paths shown are considered, one through the blue resistors and one through the orange resistors. Since these two paths are in parallel, the equivalent resistance is calculated to be 1 ohm. However, the provided answer is 2/3 ohm. For question 2, the upper and lower paths are identified as having 10 ohm resistance each, but the provided answer uses a different method to calculate the equivalent resistance. The concept of a wire having zero resistance is discussed and it is determined that the second path through the middle resistor is
leslielau

## 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 !

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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 !

## 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!

## What is the definition of equivalent resistance?

Equivalent resistance is the total resistance of a circuit component or combination of components, measured in ohms, that would produce the same effect as the original circuit.

## How is equivalent resistance calculated for series circuits?

The equivalent resistance for series circuits is calculated by adding the individual resistances together.

## How is equivalent resistance calculated for parallel circuits?

The equivalent resistance for parallel circuits is calculated by taking the reciprocal of each individual resistance and adding them together, then taking the reciprocal of the sum.

## What is the relationship between equivalent resistance and current in a series circuit?

In a series circuit, the equivalent resistance and current have a direct relationship. As the equivalent resistance increases, the current decreases and vice versa.

## What is the relationship between equivalent resistance and current in a parallel circuit?

In a parallel circuit, the equivalent resistance and current have an inverse relationship. As the equivalent resistance increases, the current also increases and vice versa.

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