3D Resistor Cube Network Analysis
Visualizing current distribution in a cubic resistor network
Simulation Parameters
Battery Voltage
10V (A to C′)
Resistor Value
1Ω each
Total Current
12A
Equivalent Resistance
5/6 Ω
Step-by-Step Calculation
1
Problem Setup
A 10V battery is connected across opposite corners A and C′ of a cube with 12 resistors, each of 1Ω.
2
Symmetry Analysis
Due to the cube's symmetry, currents in equivalent branches are equal:
Current in edges AA′, AD, and AB = I
Current in edges A′B′, B′C′, etc. = I/2
Current in edges A′B′, B′C′, etc. = I/2
3
Kirchhoff's Loop Law
Applying Kirchhoff's second law to loop ABCC′EA:
−IR − (1/2)IR − IR + ε = 0
⇒ ε = (5/2)IR
⇒ ε = (5/2)IR
4
Equivalent Resistance
Calculating the equivalent resistance of the network:
Req = ε / (3I) = (5/2)IR / (3I) = 5R / 6
5
Numerical Solution
With R = 1Ω and ε = 10V:
Req = 5/6 Ω
Total current: 3I = 10 / (5/6) = 12A ⇒ I = 4A
Total current: 3I = 10 / (5/6) = 12A ⇒ I = 4A
Concept Explanation
This is a classic problem in circuit analysis where resistors form the edges of a cube. The solution exploits the cube's high degree of symmetry to simplify the analysis.
Key insights:
- At node A, total current (12A) splits equally into 3 branches (I = 4A each)
- At each middle node (like A′, B, D), the incoming current I splits into two I/2 branches
- The symmetry allows us to determine currents without solving 12 simultaneous equations
- The equivalent resistance (5/6Ω) is less than any single resistor due to parallel paths
This method demonstrates how symmetry considerations can dramatically simplify complex network analysis problems that would otherwise require extensive matrix calculations.
