# How does the presence of a cube resistor affect current flow in a circuit?

• johnio09
In summary,The voltage across E and B is 2A (6/3 because there are three wires the current can go to), but this does not seem to be correct.Calculate the voltage across E and B. What do you think the voltage at F is then? (Think symmetry.)Probably because guessing is no substitution for writing loop or node equations.
johnio09
Homework Statement
I am given this cube of resistors, each with the same resistance of 4 Ω. The current that enters at E and leaves B is 6 A. I have found that the equivalent resistance is 3 Ω. However, I am unsure of how to find the current through the resistor between E and F.

Note that there are resistors between each of the two letters in the image.
Relevant Equations
Kirchoff's Laws

I would think that it is 2 A (6/3 because there are three wires the current can go to), but this does not seem to be correct.

Calculate the voltage across E and B. What do you think the voltage at F is then? (Think symmetry.)

johnio09 said:
I would think that it is 2 A (6/3 because there are three wires the current can go to), but this does not seem to be correct.
Probably because guessing is no substitution for writing loop or node equations.

johnio09 said:
because there are three wires the current can go to
But they are not equivalent. Having fixed B and E as the pair of interest, EA and EF are equivalent, but a path starting EH has further to go.
You do not say how you found the resistance to be 3Ω. Using the method that is obvious to me, it easy to determine what fraction of the current flows in each path.

Yes. Guessing is pointless.

OTOH, looking for symmetry to simplify things before you write the equations makes problems like this a lot simpler.

It's hard for me to think in 3D for this stuff. If the answer isn't obvious, the first thing I'll do is redraw the schematic as a planer (2D) network making it as symmetric as possible.

Frankly, I really dislike this sort of quiz question. There's no reason this needs to be a cube, it's an unnecessary trick IMO. In fact the planer diagram doesn't even have wires that cross. In my 30+ years as a practicing analog EE, I've drawn and solved lots of networks. I don't ever recall 3D geometry being a part of a circuit network problem. Yes, you can invent some, they do exist, but it's a rare thing that a good EE can solve somehow, even without practice.

This is a great example of the sort of electronics question you'll see in a physics class, but probably never in an EE class.

DaveE said:
Yes. Guessing is pointless.

OTOH, looking for symmetry to simplify things before you write the equations makes problems like this a lot simpler.
Agree ! Old Dutch expression: laziness makes inventive...

It's hard for me to think in 3D for this stuff. If the answer isn't obvious, the first thing I'll do is redraw the schematic as a planer (2D) network making it as symmetric as possible.
I think that's the core of this exercise. More brain gymnastics than anything else. And mentally flattening the cube is so much fun ! I can't resist to post this spoiler:

Frankly, I really dislike this sort of quiz question. There's no reason this needs to be a cube, it's an unnecessary trick IMO. In fact the planer diagram doesn't even have wires that cross. In my 30+ years as a practicing analog EE, I've drawn and solved lots of networks. I don't ever recall 3D geometry being a part of a circuit network problem. Yes, you can invent some, they do exist, but it's a rare thing that a good EE can solve somehow, even without practice.

This is a great example of the sort of electronics question you'll see in a physics class, but probably never in an EE class.
Again: agree. But the world is so much greater than just EE

##\ ##

phinds and DaveE
Also note that resistors that have no voltage drop across them also have no current through them and can effectively be removed from the network for analysis. It does make a different network, but it has the same solution. This is a problem that can be solved with no equations.

## What is a cube resistor, and how does it differ from a standard resistor?

A cube resistor is a resistive element shaped like a cube, where each face can potentially connect to a circuit. Unlike standard cylindrical or rectangular resistors, cube resistors can have multiple connections, which can affect how the current flows through them depending on the configuration of the circuit.

## How does the geometry of a cube resistor influence current distribution?

The geometry of a cube resistor can cause the current to distribute unevenly across its faces, depending on the points of connection. The three-dimensional shape means that current can flow through multiple paths, potentially altering the resistance experienced by the circuit compared to a one-dimensional resistor.

## How do you calculate the equivalent resistance of a cube resistor in a circuit?

Calculating the equivalent resistance of a cube resistor involves considering the resistances along the various paths between the points of connection. This often requires using principles from network theory, such as Kirchhoff's laws, to solve for the total resistance based on the specific connections and resistances of each path.

## Does the presence of a cube resistor affect the total current in a circuit?

Yes, the presence of a cube resistor can affect the total current in a circuit. The unique geometry and multiple connection points can change the overall resistance, which in turn affects the current according to Ohm's Law (I = V/R). The specific impact depends on how the cube resistor is integrated into the circuit.

## Can cube resistors be used to create more complex circuit designs?

Yes, cube resistors can be used to create more complex circuit designs due to their multiple connection points and three-dimensional nature. They allow for more intricate configurations and can be useful in applications requiring specific resistance values or current paths that are difficult to achieve with standard resistors.

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