Our R's are computable,but...

...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?

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

W.

Reply to
wilby

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

Solve the simultaneous equations.

Reply to
krw

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.

--
http://www.wescottdesign.com
Reply to
Tim Wescott

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

Reply to
J.A. Legris

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

Reply to
Ralph Barone

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

Reply to
Rich Grise

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.

--
Dave Platt                                    AE6EO
Friends of Jade Warrior home page:  http://www.radagast.org/jade-warrior
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Reply to
Dave Platt

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??
--
John G
Reply to
John G

The resisters are not all the same.

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

Reply to
Clifford Heath

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

--
John G
Reply to
John G

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

Reply to
Uwe Hercksen

There is no solution for the above numbers...

Reply to
PeterD

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

George H.

Reply to
George Herold

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.

Reply to
Jeff Johnson

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?)

--
Thanks,
Fred.
Reply to
Fred Bartoli

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

Reply to
Robert Baer

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

Reply to
Robert Baer

Thanks.

Reply to
Robert Baer

Thanks.

Reply to
Robert Baer

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