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#1 |
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diyAudio Member
Join Date: Mar 2002
Location: Montreal, Quebec, Canada
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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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_______ Brian |
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#2 |
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diyAudio Member
Join Date: Dec 2005
Location: the north
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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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lineup |
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#3 |
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diyAudio Member
Join Date: Mar 2002
Location: Montreal, Quebec, Canada
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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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_______ Brian |
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#4 | |
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diyAudio Member
Join Date: Oct 2002
Location: Croatia
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Quote:
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#5 | |
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diyAudio Member
Join Date: Dec 2005
Location: the north
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Quote:
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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lineup |
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#6 | |
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diyAudio Member
Join Date: Mar 2002
Location: Montreal, Quebec, Canada
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Quote:
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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_______ Brian |
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#7 |
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diyAudio Member
Join Date: Dec 2005
Location: the north
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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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lineup |
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#8 |
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diyAudio Member
Join Date: Sep 2005
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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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#9 |
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diyAudio Member
Join Date: Jul 2004
Location: Scottish Borders
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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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regards Andrew T. |
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#10 | |
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diyAudio Member
Join Date: Dec 2005
Location: the north
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Quote:
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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lineup |
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