Deriving the Initial Equations
We derive the initial equations by noting that for the two configurations to appear the same externally, it is necessary for the externally measured resistances, RXY, RXZ, and RYZ, to be the same for either configuration. Therefore, we can determine these and set them equal to each other. This gives us our three initial simultaneous equations:
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| (1) RXY = R1 + R2 = | (RA + RB) × RC | = | RA×RC + RB×RC |
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| (RA + RB) + RC | RA + RB + RC |
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| (2) RXZ = R1 + R3 = | (RA + RC) × RB | = | RA×RB + RB×RC |
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| (RA + RC) + RB | RA + RB + RC |
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| (3) RYZ = R2 + R3 = | (RB + RC) × RA | = | RA×RB + RA×RC |
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| (RB + RC) + RA | RA + RB + RC |
Solving For R1, R2, and R3
From the basic equations above, it looks much easier to solve for the numbered resistors, so we'll do that first. That in turn may make it easier to solve these equations in the other direction. We'll isolate R1 by subtracting Equation (3) from Equation (1) to get R1 - R3, and then add Equation (2) to that result. The expressions for R2 and R3 can be derived the same way, and will be quite similar.
| (1) - (3) | = | R1 + R2 - (R2 + R3) |
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| | = | R1 - R3 |
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| (1) - (3) + (2) | = | R1 - R3 + R1 + R3 |
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| | = | 2R1 |
Applying these to the lettered resistors, we get:
| R1 - R3 | = | RA×RC + RB×RC | - | RA×RB + RA×RC |
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| RA + RB + RC | RA + RB + RC |
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| | = | RB×RC - RA×RB | |
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| RA + RB + RC |
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| 2R1 | = | RB×RC - RA×RB | + | RA×RB + RB×RC |
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| RA + RB + RC | RA + RB + RC |
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| | = | 2(RB×RC) | |
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| RA + RB + RC |
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| R1 | = | RB×RC | |
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| RA + RB + RC |
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| R2 | = | RA×RC | |
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| RA + RB + RC |
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| R3 | = | RA×RB |
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| RA + RB + RC | |
Solving for RA, RB, and RC
To solve these expressions for the lettered resistors, we first note that the equation for R1 above contains only a single instance of RA. Therefore we will rearrange that equation and solve it for RA. Then we will substitute that value for RA in the denominator of the equation for R2, and simplify the result as much as possible. This will give us simplified relationships that we can more easily apply to these expressions.
| R1 | = | RB × RC |
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| RA + RB + RC |
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| R1(RA + RB + RC) | = | RB × RC |
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| RA + RB + RC | = | RB × RC | |
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| R1 |
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| RA | = | RB × RC | - RB - RC |
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| R1 |
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| R2 | = | RA × RC |
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| RA + RB + RC |
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| R2(RA + RB + RC) | = | RA × RC |
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| R2( | RB × RC | - RB - RC + RB + RC) | = | RA × RC |
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| R1 |
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| R2 × RB × RC | = | R1 × RA × RC |
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| [ R3 × RC = ] R2 × RB | = | R1 × RA |
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| RB | = | R1 × RA | |
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| R2 |
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| RC | = | R1 × RA | |
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| R3 |
Now we can replace RB and RC with very simple expressions involving RA, so that we will be able to solve for RA in terms of only numbered resistors. RB and RC can then be found in the same way, and will have similar expressions.
| RA | = | RB × RC | - RB - RC |
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| R1 |
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| | R1 × RA | | R1 × RA | |
| × |
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| RA | = | R2 | | R3 | - | R1 × RA | - | R1 × RA |
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| R1 | R2 | R3 |
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| RA | = | R1 × RA × R1 × RA | - | R1 × RA | - | R1 × RA |
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| R1 × R2 × R3 | R2 | R3 |
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| 1 | = | R1 × RA | - | R1 | - | R1 | |
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| R2 × R3 | R2 | R3 |
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| R1 × RA | = | 1 | + | R1 | + | R1 | |
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| R2 × R3 | R2 | R3 |
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| R1 × RA | = | R2 × R3 | + | R1 × R3 | + | R1 × R2 | |
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| RA | = | R2 × R3 | + | R3 | + | R2 | |
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| R1 |
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| RB | = | R1 × R3 | + | R1 | + | R3 | |
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| R2 |
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| RC | = | R1 × R2 | + | R1 | + | R2 | |
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