- #1
The path with least resistance would be "R"!cepheid said:There are a couple of shorts in that diagram. So...what's the path of *least* resistance between A and B?
Delta said:Are you familar with transformations with resistor/impedance networks? This looks to be be one of those.
I recommend drawing the circuit another way.
Delta said:I was thinking star-delta transformations, but I've just moved them around again and the circuit is simpler than I though ... doh!
Where do you get 5R from?
ritwik06 said:i can't understand what you all mean to say. Will you please be so good to draw the simplified diagram in paint for me? Please.
cepheid said:Have you tried drawing it out yourself? It's definitely worth learning how to do this. The good part is that it's simple. Start at node A, and trace out all possible paths from node A to node B. For example, you can go through the leftmost resistor having 2R resistance, and that will take you from A to B. That's one path. What other paths are there?
ritwik06 said:yes, i tried it. from A to B:
1. through 2R and then through short circuit to B
2. through short circuit and the via R to B
ritwik06 said:3. via all three resistances without using any short circuits
cepheid said:Good! Can you see a third path that passes through ONLY the middle resistor?
I'm afraid this path is not valid. Here is a very important point in circuit analysis: if two points are connected directly by ideal wires (i.e. they are shorted together), then those two points are electrically connected and can be considered to be the same point. What I mean by this is that the two points and the entire wire connecting them are all at the same potential (there is no voltage between them). As a result, from the point of view of circuit analysis, they can be considered a single "node". I have modified your diagram to illustrate this:
http://img74.imageshack.us/my.php?image=resistorsif6.jpg
Everything circled in red is at the same potential as point B, and therefore is, in a sense, electrical point B. Likewise for everything circled in blue being A electrically. As a result, can you see that if current passes through the leftmost 2R resistor, it has already traveled from A to B in the sense of having lost potential equivalent to the voltage between A and B?
Now for the point all of this was building up to: if current goes through the left resistor, and then the middle one, then it will have gone from A, to B, and then back to A again! Clearly this won't happen in practice: either A or B will be at a higher potential if current is to flow. As a result, the path passing through all three resistors is not valid, for it goes from A to B to A to B. This is what I was trying to explain in my edit to post #8, but redrawing the circuit makes this point really clear.
cepheid said:Since you've clearly been racking your brains for some time now, and have put a fair amount of thought into the quesiton, I thought I'd help you out. This post will show you how you can redraw the circuit, step by step, just by considering all possible paths from A to B. So, current can either pass from A, through the left 2R resistor, and continue to B along the wire, or it can pass through the other wire, and then through the rightmost R resistor to B. That allows us to draw the following so far:
http://img76.imageshack.us/my.php?image=resistorsqt7.png
So, that takes care of the leftmost and rightmost resistors. What about the middle one? We can add it to our diagram by carefully noting where it starts and ends. From the argument in my previous post, it starts at B and ends at A.
Another way of looking at it: Once you've gone through the left resistor, there are two choices: 1. go straight to B through the wire. 2. Go back to A through the middle resistor (which could never actually happen). So clearly we need to add a junction on the far side of the left resistor. This junction is the starting point for the middle resistor, like so:
http://img64.imageshack.us/my.php?image=resistors2dl8.png
However, it's pretty clear that the diagram above is equivalent to the one below (and I figured out how to do thumbnails):
http://img64.imageshack.us/img64/6791/resistors3if9.th.png
Redrawing the circuit in this way makes two things very clear that were hard to see in your original picture:
1. There is a current path from A to B that passes ONLY through the middle resistor.
2. The current path through all three resistors is NOT VALID, because it would entail current going from A to B, then back to A, then back to B.
The effective resistance between points A and B is a measure of the overall resistance that an electric current encounters while traveling between the two points. It takes into account both the individual resistances of the components in the circuit as well as the arrangement and configuration of those components.
The effective resistance can be calculated using Ohm's Law, which states that the resistance (R) is equal to the voltage (V) divided by the current (I). This can be applied to both series and parallel circuits to determine the total resistance between points A and B.
The effective resistance is affected by the individual resistances of the components in the circuit, as well as the type of circuit (series or parallel) and the arrangement of the components within the circuit. Temperature, material conductivity, and wire thickness can also impact the effective resistance.
The higher the effective resistance between points A and B, the more difficult it is for current to flow between them. This means that a larger voltage is required to maintain the same amount of current, and the power dissipated in the circuit increases. Conversely, a lower effective resistance allows for a larger flow of current and less power dissipation.
No, the effective resistance can never be zero between points A and B. Even in a perfect conducting material, there will always be some resistance due to the physical dimensions and properties of the material. Additionally, in a circuit with multiple components, even if one has a resistance of zero, there will still be resistance from the other components in the circuit.