Find the resistance between 2 points

Thread Starter

UndoubtedCrow

Joined Jun 2, 2020
6
Hi,
I am working on the homework from "https://www.allaboutcircuits.com/worksheets/series-parallel-dc-circuits/" Question number 5. I thought I had a decent understanding of resistance, but since I can't figure these out, maybe not. The question asks me to find the resistance between 2 points. I assumed that the question wanted me to add the resistances of the shortest path from A to B, but that's not the answer provided. Neither is the equivalent resistance of the network. This problem has me stumped. Did I miss something?
 

Thread Starter

UndoubtedCrow

Joined Jun 2, 2020
6
Combining the two resistors in parallel first,
R = (1/2.2kohm +1/ 2.2kohm )^-1 = 1.1 k ohms
Then adding the resistors in series, Rt = 1.1 k ohm + 2(2.2 k ohms) = 5.5 k ohms
 

WBahn

Joined Mar 31, 2012
25,751
Combining the two resistors in parallel first,
R = (1/2.2kohm +1/ 2.2kohm )^-1 = 1.1 k ohms
Then adding the resistors in series, Rt = 1.1 k ohm + 2(2.2 k ohms) = 5.5 k ohms
In order for the 1.1 kΩ resistor (the parallel equivalent of the two 2.2 kΩ resistors) to be in series with the 4.4 kΩ resistor (the series equivalent of the two 2.2 kΩ resistors), whatever current that flows through the 1.1 kΩ resistor MUST also flow throw the 4.4 kΩ resistor? Is that the situation?

Two resistors are in series ONLY if whatever current flows through one must flow through the other.

The resistors are in parallel ONLY if whatever voltage appears across one must appear across the other.
 

WBahn

Joined Mar 31, 2012
25,751
I assumed that the question wanted me to add the resistances of the shortest path from A to B, but that's not the answer provided.
It's important that you realize why this is completely incorrect reasoning. If I have multiple paths between A and B, current will flow in ALL of them and the equivalent resistance is the ratio of the voltage across ALL of them to the sum of the currents flowing in ALL of them.

Think of the simple case -- two resistors in parallel that are slightly different. Is the total resistance just the value of the smaller of the two since it is the "shortest path", or is it the parallel combination of the two?

Think of a bucket of water with a bunch of different sized holes in the bottom. If you take another bucket the same size and put a hole in the bottom the same size as the biggest hole in the original one, would you expect both buckets to drain water at the same rate? No, because the original bucket also has all those other holes that are also draining water. To a first approximation, you need to put a hole in the second bucket whose area is equal to the sum of the areas of ALL the holes in the original bucket.

Neither is the equivalent resistance of the network. This problem has me stumped. Did I miss something?
How are you determining the equivalent resistance of the network? It needs to be the equivalent resistance of the network AS SEEN between points A and B.
 

dl324

Joined Mar 30, 2015
10,734
Then adding the resistors in series, Rt = 1.1 k ohm + 2(2.2 k ohms) = 5.5 k ohms
The 2 2.2k resistors in series are in parallel with the 2 parallel 2.2k resistors. It might be helpful if you labeled the resistors and wrote the expression in terms of resistor labels, then substituted values.

1591118873620.png
R1 || R2 || (R3+R4)
 
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