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Old 9th March 2006, 06:54 PM   #1
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Default Solving Voltage with multiple resistors.

Hi Everyone, here is another little resistor/voltage puzzle.


Say I have 3 resistors, R1, R2, R3.

I wish to calculate the voltage at the center where all 3 connect together. On the outside of the 3 resistors, I have V1, V2, V3.

What's the formula?

Example:
V1 = 0v, R1 = 22K
V2 = 5V, R2 = 18k
V3 = 10v, R3 = 90k

How do I solve for the voltage of the other side where all 3 resistors short together?

What if I have 4, of 5 resistors?
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Old 9th March 2006, 07:07 PM   #2
lineup is offline lineup  Sweden
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that must be impossible to solve (unless you have worlds's most powerful computer to make a model simulation)
even with only 3 resistors and 3 voltages

do like this
1. hook it up, 3 voltage supplies and 3 resistors you will need
2. measure with voltmeter from 0V to the joining point of resistors

this will give a correct result

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Old 9th March 2006, 07:11 PM   #3
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@lineup: no it isn't imposible to solve. & will wait 30 more minutes before working out the solution.

Here is a hint: You must work out the parallel total resistance first, then plug use it to solve a multiple voltage/current divider.
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Old 9th March 2006, 07:25 PM   #4
moamps is offline moamps  Croatia
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Default Re: Solving Volateg with multiple resistors.

Quote:
Originally posted by Brian Guralnick
What's the formula?
Use Kirchofs law.
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Old 9th March 2006, 07:36 PM   #5
lineup is offline lineup  Sweden
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Quote:
Originally posted by Brian Guralnick
@lineup: no it isn't imposible to solve. & will wait 30 more minutes before working out the solution.

Here is a hint: You must work out the parallel total resistance first, then plug use it to solve a multiple voltage/current divider.

maybe ....

... a simulation spice program will solve it in a moment
they have some formulas programmed
... or is it just repeated corrections in a loop, until a good, satisfying value is reached?
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Old 9th March 2006, 07:38 PM   #6
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Default Re: Re: Solving Volateg with multiple resistors.

Quote:
Originally posted by moamps


Use Kirchofs law.

Bingo!


Here is the formual.

First, solve total resistance, I'll call it Rt.

Rt = 1/ ( (1/R1) + (1/R2) + (1/R3) )

Then plug in the voltages, I'll use Vt for total voltage:

Vt = (Rt*V1/R1) + (Rt*V2/R2) + (Rt*V3/R3)

For more resistors, just increase the R# & V# in the pattern of the above formulai.
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Old 9th March 2006, 08:14 PM   #7
lineup is offline lineup  Sweden
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Is it that easy.
I think, even I,
now know how to solve this sorta puzzle.



Not that often, at least in amplifiers,
we need to find out voltage in 3 resistors crossing point,
with 3 different voltage sources.

But I know one case where this happen.
This is in a sound mixer input.
Often 3,4 or more resistors are connected to mixer Op-Amp inverted input pin.
Each coming from a sound channel.
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Old 9th March 2006, 09:02 PM   #8
Nordic is offline Nordic  South Africa
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http://www.physics.northwestern.edu/Lab/ec_c.pdf

I have found the above link very usefull to get to grips with alot of these calculations... seeing that school was a lifetime ago... it all comes back reasonable easy though, but then again, I always liked science.
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Old 11th March 2006, 09:25 AM   #9
AndrewT is online now AndrewT  Scotland
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Hi,
what result did you calculate for the voltage at the junction.

I got 3.468V.

but your posted formulae gave 8.441V which is clearly wrong.

Tell us, which is correct?
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Old 11th March 2006, 09:37 AM   #10
lineup is offline lineup  Sweden
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Quote:
Originally posted by AndrewT
Hi,
what result did you calculate for the voltage at the junction.
I got 3.468V.

but your posted formulae gave 8.441V which is clearly wrong.
Tell us, which is correct?
I have not tried to use the formula.
When you say 'calculate', AndrewT, what you mean?
You used a simulation? And got 3.468V

Calculate is for me using math, as with a mini calculator.
Anybody with access to a circuit simulation program, will not need any formulas to calculate.

I never use any circuit simulations.
Just my head or my pocket calculator and paper and pen.
So for me a general useful and correct formula would be interesting.

Regards
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