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Kirchhoff's Law
f-alpha.net » Physics » Electricity » Kirchhoff's Laws » Go on... » Experiment 15 - Resistor Cube II

Experiment 15 - The Resistor Cube II

Consider the schematic illustration of the resistor cube...

Illustration resistor cube.

  • Because of the symmetry, the currents are I3 = I4 = I7 = 1/3 Itotal (KCL!).
  • The points C, D and E are therefore on the same potential, as UAC = UAD = UAE (Ohm's law!).
  • This applies analogously to the currents (I5, I10, I12) and the potential at the points F, G and H

Now you can connect the points of the same potential with each other. No current flows through the wire, as no voltage applies (KVL!). The additional wires have no impact on the circuit!

The additional wires are colored in green. The circuit diagram can therefore be redrawn...

 

Illustration resistor cube with fictitious connections.

Circuit diagram resistor cube with fictitious connections.

This new circuit is now composed of a series circuit of three parallel circuits of resistors. This can be easily calculated...

Rtotal = 1/3 R + 1/6 R + 1/3 R
          = 5/6 R

However, symmetries of a circuit can not always be identified as easily as with a cube. Therefore, general circuits are calculated using the so-called network analysis...

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