# Resistance Maze

#### Kyrol

I am trying to understand the difference between two paths of traveling current and why one is "right" and why the other is not.

So the situation is this: All resistance are 4ohm and all batteries are ideal and have an emf of 10V. I am trying to solve for the current, and I have drawn two paths.
The first picture displays a path with the starting point indicated by the blue dot. Traveling in this direction would give me:
-R -R/2 -R -R +V -V +V -V -V -R = 0. From here, I would have V = 9R/2. We know that V = IR, so V = (9R/2)I --> I = 2V/9R = 2(10)/9(4) = 20/36 = 0.556Amps

The second picture displays the "correct" path, where we travel and avoid all as much resistance as possible. Now I understand that we do this to make calculating things easier, but even if I took the first path, shouldn't the current be the same since they are both following a loop?
With the second path, we would get
-R -R/2 -R -V +V +V -V +V -V -V -V -V -V = 0 --> 4V = 5R/2 I --> I = 8V/5R = 8(10)/5(4) = 4Amps.

So what's the difference in each path?

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#### ChipB

PHF Helper
Why do you think the currents through the two paths should be the same? Consider a much simpler example of a circuit with two paths for current: two resistors connected in parallel to a battery, where the resistors have different values. Is the current through each resistor the same? Suppose the battery is 10V and the resistors are 2 ohms and 5 ohm, respectively. The current through the 2 ohm resistor will be 10/2 = 5 amps, and the current through the 5 ohm resistor is 10/5 = 2 amps. So, the calculation shows two different values of current for the two different paths. To get the actual total current flow you need to sum the currents flowing through the two different paths.

2 people

#### Woody

Path of Least Resistance

As a rule of approximation, the current will take the path of least resistance.
However, as ChipBs example illustrates, more current will flow down the path of least resistance (but not all).

It is sometimes instructive to re-imagine these circuits as pumps and pipes in a fluid flow circuit.
Thus the batteries become pumps and the resistors become narrow pipes,
then the electrical current becomes the fluid flow rate through the circuit.

Perhaps using this more mechanical analogy you can see that certain parts of the circuit will become "backwaters" with hardly any fluid flowing through them.

2 people

#### Kyrol

I would understand that it would be different amps if the resistant was different values, but they are all the same values. However, as Woody explained, current takes the path of least resistance which helps me makes a little bit more sense of this problem, but then I also have another problem with how one path doesn't influence the other? For example, looking at the second photo to the right, at where the blue mark is, how come the resister to the left of the blue dot doesn't affect the current flow of the red path? Wouldn't it also input some current to the path as well?

#### Woody

Another way to break down the problem is to look at the voltage difference across each resistor.

It is difficult to pick out in such a complex network,
but I suspect that if you calculate the voltage difference across the resistor to the left of the dot you will find it will be zero (or pretty close to it).

If I might ask, where did this problem come from?

1 person

#### Kyrol

Hey Woody, this was a problem within the Fundamental of Physics 10th edition. I believe it was problem #74 of chapter 27: Circuits.

I think the explanation you provided with figuring out the potential difference across each resistor makes the most sense. That would be why we would take the path where there is as little resistors as possible, so we dont have to calculate what the actual value of the resistor is through looking at all the potential differences that could influence the resistor.

Thank you Woody