electric-circuitselectrical-resistancehomework-and-exercises
Actually , I'm having trouble in solving for the net resistance of the below network of resistors.
Can anyone help me finding the equivalent resistance between points a & b.
My answer is $2/3r$ .
I've done it by first solving for the two resistances in the middle , Then coverting the circuit into a Wheatstone Bridge.
PS- SORRY for the bad diagram
Think about current flow.
If we take each individual resistor and determine the current for the applied voltage, we get: $$I_T=\frac {V}{R_1} +\frac {V}{R_2} + ...$$
Dividing everything by the voltage give us:
$$\frac {I_T}{V}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$
Which is the same as:
$$\frac {1}{R_{eq}}=\frac {1}{R_1} +\frac {1}{R_2} + ...$$
Since there is more current flowing in all the resistors than through just one resistor, then the equivalent resistance must be less than the individual resistors.
The general idea is to find and exploit symmetries in the network. A symmetry means that if you change something about the problem, it remains the same. Generalized method for dealing with circuit involving symmetry? links to a basic introduction to a formal procedure for identifying the symmetries of the network, and the effect these symmetries have on the relation between the inputs and outputs of the circuit. However, in many cases the symmetries are most easily identified visually from a diagram of the circuit.
For example, the following infinite ladder of resistors looks exactly the same if you add another "unit" at the front. This suggests a method of finding the total resistance $R_\infty$ which is the same as $R$ in series with $R$ || $R_\infty$ :
$$R_\infty=R+\frac{R_\infty R}{R_\infty+R}$$
This is a quadratic equation which can be solved to find $R_\infty$.
In your cube problem, resistors $a$, $b$ and $c$ are in equivalent positions - ie if you rotate the cube about an axis through AB you can replace $a\to b\to c\to a$ without making any difference to the resistance between A and B. This symmetry means that the points marked $\alpha$ are all at the same potential, as are those marked $\beta$.
Without affecting the circuit we can connect wires between the points marked $\alpha$ - and likewise between those marked $\beta$ - because no current will flow through them. The cube is then equivalent to the following series of parallel resistors :
http://www.rfcafe.com/miscellany/factoids/kirts-cogitations-256.htm
Best Answer
My answer is r. 2 resistances on the vertical line are redundant as no current flows via that route due to the symmetry of the problem.
Consider this circuit joined to the source with A joined to the positive terminal, then current will equally split along two possible routes from A. Due to symmetry of the problem, the electric potential energy above the vertical line is same as the point below it. So no current flows via that part of the circuit. So that section of wire can be removed, yet giving same equivalent resistance. In our simplified circuit, 2 resistors of resistance 2r are in parallel giving equivalent resistance r.