# Our R's are computable,but...

• posted

...how? Let || represent "in parallel with". The equations are: 200 = R1||R2||R3||R4||R5||R6||R7 250 = R1||R3||R4||R5||R6||R7 300 = R1||R2||R4||R5||R6||R7 600 = R1||R2||R3||R5||R6||R7 1000 = R1||R2||R3||R4||R6||R7 2000 = R1||R2||R3||R4||R5||R7 5000 = R1||R2||R3||R4||R5||R6

How does one go about solving for the resistor values?

• posted

Hmmm. You have 7 equations and 7 unknowns. I guess some simple, but drawn out, algebra should do it.

W.

• posted

1/200 = 1/R1 + 1/R2 + 1/R3 + 1/...

Solve the simultaneous equations.

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As in

1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5 + 1/R6 + 1/R7 = 1/200 1/R1 + 1/R3 + 1/R4 + 1/R5 + 1/R6 + 1/R7 = 1/250 etc.

Then solve using any simple linear algebra solver. That's not too drawn out at all.

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• posted

Not as drawn out as you might expect:

Observe that the first equation includes all 7 resistors while the other 6 equations are each missing 1 resistor. So, subtracting eq'n 1 minus eq'n 2:

1/200 =3D 1/R1 + 1/R2 + 1/R3 + 1/R4 + 1/R5 + 1/R6 + 1/R7 1/250 =3D 1/R1 + 1/R3 + 1/R4 + 1/R5 + 1/R6 + 1/R7

---------------------------------------------------------------------------

1/200 - 1/250 =3D 1/R2

Similarly:

1/200 - 1/300 =3D 1/R3 1/200 - 1/600 =3D 1/R4 1/200 - 1/1000 =3D 1/R5 1/200 - 1/2000 =3D 1/R6 1/200 - 1/5000 =3D 1/R7

-- Joe

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Solve for conductance (7 equations and 7 unknowns as above), then invert to get resistance.

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7 equations in 7 unknowns.

You could do it algebraically if you have all day, but I suspect that somebody that knows how to do matrix algebra on a computer could probably answer it in minutes.

Cheers! Rich

• posted

First, transform these equations into admittance form. Define

A1 = 1/R1, A2 = 1/R2, and so forth.

Then,

1/200 = A1 + A2 + A3 + A4 + A5 + A6 + A7 1/250 = A1 + A3 + A4 + A5 + A6 + A7 1/300 = A1 + A2 + + A4 + A5 + A6 + A7

etc.

This leaves you with 7 linear equations, in 7 unknowns.

Solve for the unknowns A1 through A7. You can do this manually by a process of reduction (adding and subtracting combinations of these equations in order to eliminate unknowns), or by a matrix inversion and multiplication process. Any good text on linear algebra would show the procedure.

In this case, reduction is trivial. For example, to calculate A2, you would just subtract the second equation from the first, leaving you with

A2 = 1/200 - 1/250 Do the same trick for the subsequent equations (subtract each from the first) to calculate A3 through A7

A3 = 1/200 - 1/300 A4 = 1/200 - 1/600

and do forth.

Plug all of these values into the first equation, simplify, and you have the value of A1.

Then, invert each value A1 through A7 to calculate R1 through R7, and you're done.

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Dave Platt                                    AE6EO
• posted

Dave Platt submitted this idea :

What am I missing here??

7 resistors to make 200 ohms = 1400 ohms each 6 resistors to make 250 ohms = 1500 ohms each etc. etc. no computations just a little mental arithmetic. Or did I miss somthing??
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John G```
• posted

The resisters are not all the same.

J.A Legris already pointed out the simple way to solve it.

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The OP did not say that. And without some clues otherwise the answer is indefineable.

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John G```
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Dave Platt schrieb:

Hello,

but the problem is that A1 is a negative admittance of -113/10000, R1 is a negative resistor of -10000/113.

Bye

• posted

There is no solution for the above numbers...

• posted

If everything is in parallel it might be better to write the equations in terms of conductances. Then everything just adds.

George H.

• posted

X = A || B 1/X = 1/A + 1/B

So,

1/200 = 1/r1 + 1/r2 + 1/r3 + 1/r4 + 1/r5 + + 1/r6 + 1/r7 1/250 = 1/r1 + 0/r2 + 1/r3 + 1/r4 + 1/r5 + + 1/r6 + 1/r7 1/300 = 1/r1 + 1/r2 + 0/r3 + 1/r4 + 1/r5 + + 1/r6 + 1/r7 1/600 = 1/r1 + 1/r2 + 1/r3 + 0/r4 + 1/r5 + + 1/r6 + 1/r7 1/1000 = 1/r1 + 1/r2 + 0/r3 + 1/r4 + 0/r5 + + 1/r6 + 1/r7 1/2000 = 1/r1 + 1/r2 + 0/r3 + 1/r4 + 1/r5 + + 0/r6 + 1/r7 1/5000 = 1/r1 + 1/r2 + 0/r3 + 1/r4 + 1/r5 + + 1/r6 + 0/r7

or

B = A * R

The values you get are

1.0e+003 *

-0.0709 1.0000 0.6000 0.3000 0.2500 0.2222 0.2174

Which gives you a negative resistance.

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Robert Baer a écrit :

How simple... Just work with conductances and you have a simple linear system to solve.

Getting back to resistances is left as an exercise to the student. (did I really manage to tell something that stupid?)

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Thanks,
Fred.```
• posted

No shit, Dick Tracy. And that method is......................

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That is a start..i guess i will have to find one of those somewhere and download it.

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Thanks.

• posted

Thanks.

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