| Brian Guralnick |
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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| lineup |
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
:D |
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| 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. |
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| moamps |
| quote: | Originally posted by Brian Guralnick
What's the formula? |
Use Kirchofs law. |
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| lineup |
| 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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| Brian Guralnick |
| 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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| lineup |
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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| Nordic |
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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| 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? |
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| lineup |
| 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
:cool: |
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| AndrewT |
Hi,
No, I cannot use simulators.
I just used a hand calculator and some simultaneous equations. It took about 4 minutes.
Equate the currents into the node.
Set unknown voltage =V then equate each resistor current to (Vx-V)/Rx etc. |
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| moamps |
Vx=V1-I1*R1
Vx=V2-I2*R2
Vx=V3-I3*R3
I1+I2+I3=0
for V1=0, V2=5, V3=10, R1=22, R2=18, R3=90
Vx=3,465V |
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