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Question:

Twelve wire each having a resistance of 3Ω are connected to form a cubical network. A battery of 10V and negligible internal resistance is connected across the diagonally opposite corners of this network. Determine its equivalent resistance and current along each edge of the cube.

Solution:

Let the vertices of the cube be A, B, C, D, E, F, G, H. Let the resistance of each edge be R = 3Ω. The potential difference across the diagonally opposite corners is 10V.

Due to symmetry, the current distribution is as follows:

  • The current from one corner splits equally into three branches.
  • The current flowing through each edge connected to the starting corner is I/3.
  • The current then combines at the other corners before reaching the opposite corner.

Consider the path from one corner (say A) to the opposite corner (say G). The equivalent resistance can be found using several methods, including superposition or symmetry arguments. One common approach is to use symmetry to simplify the circuit. Due to the symmetry of the cube and the equal resistances, the potential at points B, D, and F will be equal. Similarly, the potential at points C, E, and H will be equal.

We can simplify the circuit by considering the parallel combination of resistances. Let's analyze the current distribution from A to G:

  1. The current I divides equally into three branches AB, AD, and AF. The current in each branch is I/3.
  2. At B, D, and F the current further splits into two paths. Let's focus on the path from B. Half of the current (I/6) goes to C and the other half (I/6) to E. The same happens at D and F.
  3. The currents combine at G. The total current reaching G is the sum of the currents from each path.

Applying symmetry and Kirchhoff's laws becomes complex for direct calculation. Let's instead use the method of equivalent resistance.

Imagine injecting a current I at A and extracting it at G. The equivalent resistance between A and G is given by R_eq = V/I, where V = 10V.

By applying the principles of network reduction and symmetry, the equivalent resistance between opposite corners of a cube with edge resistance R is 5R/6.

Therefore, the equivalent resistance is:
R_eq = (5/6) * R = (5/6) * 3Ω = 2.5Ω

The total current flowing through the circuit is I = V/R_eq = 10V / 2.5Ω = 4A.

The current along each edge of the cube is I_edge = I/6 = 4A/6 = 2/3 A

Therefore:

  • Equivalent Resistance: 2.5Ω
  • Current along each edge: 2/3 A