/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Resistances in a cube ** A cub... [FREE SOLUTION] | 91Ó°ÊÓ

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Resistances in a cube ** A cube has a resistor \(R\) along each edge. Find the equivalent resistance between two nodes that correspond to: (a) diagonally opposite corners of the cube; (b) diagonally opposite corners of a face; (c) adjacent corners. You do not need to solve a number of simultaneous equations; instead use symmetry arguments. Hint: If two vertices are at the same potential, they can be collapsed to one point without changing the equivalent resistance between the two given nodes.

Short Answer

Expert verified
(a) The equivalent resistance between diagonally opposite corners of the cube is \(R\). (b) The equivalent resistance between diagonally opposite corners of a face of the cube is \(\frac{2}{3}R\). (c) The equivalent resistance between adjacent corners of the cube is \(3R\).

Step by step solution

01

(a) Diagonally opposite corners of the cube

We can collapse the points at the same potential to simplify the circuit into a parallel circuit of two resistances \(2R\). So the equivalent resistance, \(R_{eq}\) is given by \(\frac{1}{R_{eq}} = \frac{1}{2R} + \frac{1}{2R}\), which upon solving gives \(R_{eq} = R\)
02

(b) Diagonally opposite corners of a face

To simplify the network, we can collapse the points at the same potential as before. The circuit now forms a bridge network. However, we can remove the resistance in the middle as the points it joins are at the same potential. Thus, we get a parallel circuit of two resistances \(R\) and \(2R\). So, \(R_{eq}\) is given by \(\frac{1}{R_{eq}} = \frac{1}{R} + \frac{1}{2R}\), which upon solving gives \(R_{eq} = \frac{2}{3}R\)
03

(c) Adjacent corners

Collapsing the points at the same potential, we see that the resulting network forms three resistances in series, each of resistance \(R\). So, the equivalent resistance is simply the sum of the three resistances, \(R_{eq} = 3R\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Network Simplification
When faced with a complex circuit like a cube with resistors on each edge, network simplification allows us to easily find the equivalent resistance. This technique involves reducing a complex circuit to simpler forms without altering the resistance between specified nodes.
The idea is to identify parts of the circuit that can be combined:
  • If resistors are in series, their resistances are added together.
  • If resistors are in parallel, their reciprocals are summed and the reciprocal of that sum gives the equivalent resistance.
Network simplification is particularly useful in circuits like the resistor cube where visualizing the interconnections can be challenging. By simplifying the circuit, we can focus on the relationship between the potential differences and the connections between nodes, making the problem more tractable.
Symmetry Arguments in Physics
Symmetry arguments in physics leverage the uniformity and balance in a circuit design to simplify solving for unknowns, such as the equivalent resistance. In the resistor cube problem, symmetry simplifies the circuit analysis by identifying nodes that are at the same potential.
When two nodes have the same potential, they can be treated as a single point, allowing parts of the network to be "collapsed." This technique reduces the complexity of the network by:
  • Identifying parts of the circuit that do not affect the resistance between the selected endpoints.
  • Enabling an accurate application of parallel and series circuit rules.
Symmetry is a powerful tool because it takes advantage of the inherent redundancy in the network. It reveals that even complicated resistor arrangements can often be reimagined as a much simpler system, providing insightful results with minimal calculation.
Resistor Cube Problem
The resistor cube problem uses a cube with a resistor along each edge. The challenge is to find the equivalent resistance between nodes without solving multiple equations. By applying symmetry arguments and network simplification, this problem becomes manageable.
For each query:
  • (a) Diagonally Opposite Corners of the Cube: Identifying nodes of equal potential allows reducing the circuit to a simple parallel arrangement, where each path contains a resistance of value \(2R\), leading to \(R_{eq} = R\).
  • (b) Diagonally Opposite Corners of a Face: After collapsing symmetric nodes and removing null resistors (where potential difference is zero), the circuit forms a mix of resistors in series and parallel, resulting in \(R_{eq} = \frac{2}{3}R\).
  • (c) Adjacent Corners: Simplification results in a straightforward series formation, adding the resistances along the path for a total of \(R_{eq} = 3R\).
This approach not only simplifies the process but also enhances understanding of potential and current distribution across the network. By identifying and using repetitive patterns, students can tackle similar complex structures in circuit theory effectively.

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