Resistance Across Cubic NetworkDate: 12/18/2003 at 23:50:34 From: Jack Subject: How to solve a cube of resistors for total resistance I'm trying to compute the resistance between opposing corners of the following cube: A------B Each of the twelve edges of the cube is /| /| a resistor of value R. D------C | | E----|-F What is the resistance between points |/ |/ A and G? H------G How can I do this? Date: 12/19/2003 at 00:01:16 From: Doctor Douglas Subject: Re: How to solve a cube of resistors for total resistance Hi, Jack. Thanks for writing to the Math Forum. If all of the resistors are identical, then there is a quick way to find the equivalent resistance of the cube network. To do this, you simply insert additional wires between points that you know to be at the same voltage. A------B Each of the twelve edges of the cube is /| /| a resistor of value R. D------C | | E----|-F What is the resistance between points |/ |/ A and G? H------G In this network, the points B, D, and E are equivalent, i.e., all of them have the same voltage, namely exactly one resistance drop down from the voltage at point A. Thus, we may connect these three points together by wires BD, DE, and BE without changing anything. Similarly, the points F, H, and C are equivalent, and we may connect them together with wires. Now imagine shrinking these wire loops down to a single point. This collapses together the points BDE and CFH. The circuit now looks like this (when flattened out): /-R-\ --R-- /--R--\ -R- / \ /---R---\ / \ o---A---R--(BCD)----R--(CFH)--R--G---o \ / \---R---/ \ / --R-- \--R--/ -R- Thus, this cube network is equivalent to the series combination of three things: a set of 3 resistors in parallel, a set of 6 resistors in parallel, and another set of 3 resistors in parallel. Thus the total equivalent resistance of the cube from A to G is R_total = (R/3) + (R/6) + (R/3) = R(1/3 + 1/6 + 1/3) = 5R/6. Note that the key idea is to notice that you can connect equivalent voltages in a circuit by a wire without changing the current flow, so that the overall properties of the circuit remain unchanged. This is possible when all of the resistors are the same. When the resistors are not all the same, one may have to resort to Kirchoff's Laws, which keep track of the currents in the loops and the voltages at the nodes of any circuit network. You are probably familiar with the formulas for series and parallel resistances: R[series] = R1 + R2 and R[parallel] = 1/[1/R1 + 1/R2]. These formulas are shortcuts for replacing a small network of resistors with an equivalent effective resistance. Kirchhoff's laws are used to derive these transformations, which are useful because the series and parallel combinations occur often in real circuits. Another transformation that is sometimes useful, but less commonly seen, is the delta-wye transformation: X--Rxy--Y X Y \ / \ / Rxz Ryz <----> Rx Ry \ / \ / Z | Rz "delta" | "wye" Z The formulas that relate these resistances are Rxy = (RxRy + RyRz + RzRx)/Rz "wye to delta" Ryz = (RxRy + RyRz + RzRx)/Rx Rxz = (RxRy + RyRz + RzRx)/Ry Rx = RxyRxz/(Rxy + Ryz + Rxz) "delta to wye" Ry = RxyRyz/(Rxy + Ryz + Rxz) Rz = RxzRyz/(Rxy + Ryz + Rxz) Sometimes using these equations are useful for simplifying a circuit without going through all of the equations from Kirchhoff's Laws. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/ |
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