Electronics math

[Jim147]

It's too late in the night but I'm working on something in my head.

100k resistor, 11.5 VDC output, what would the input need to be?

If you don't have it in front of you don't spend too long looking. I just have this question bugging me when I should be asleep.

I should have a chance tomorrow afternoon to pull out my books to get a formula, but I just hate it when I have things like this bugging me.

I'm thinking it's going to be around 35 or 135 VDC.

jim

[230RN]

Your question, as stated, is not answerable in terms of voltage input.

It can only be answered, as stated, in terms of the current input.

A current of 0.000115 amperes (115 microamperes) will develop 11.5 volts across a 100,000 ohm resistor.

Put another way, only 0.000115 amperes will flow through a 100,000 ohm resistor with 11.5 volts across it.

Current is directly proportional to the voltage impressed across a resistor and inversely proportional to its resistance.

I = E / R.... where I = current, E = voltage, R = resistance.*

current = 11.5 volts / 100,000

Just like a water pipe.  Water flow is directly proportional to the pressure across the pipe and inversely proprtional to the pipe's resistance to the flow.

Other forms, by transposition of the terms, are:

I = E / R

E = I * R

R = E / I

That relationship is common to a lot of things besides water flow through a pipe or electric flow through a resistance. 

However, "reading into" your question, I think what you need is a 200,000 ohm resistor in series with your 100,000 ohm resistor to drop 11.5 volts across your 100,00 ohm resistor with 35 volts across the whole circuit, and a 1.074 megohm resistor to pick off 11.5 volts from a 135 volt input.  By simple proportions in a voltage divider circuit.  If my arithmetic is correct:

Case 1:  What resistor is required to drop 23.5 volts from 35 volts if there already is a 100K resistor in the circuit?  (35-11.5 = 23.5)

unknown resistor = 100,000 * 23.5 / 11.5 = 200,000 ohms

(Note that in this instance, since 11.5 volts is almost exactly 1/3 of 35 volts, the 200,000 ohm additional resistance to make 300,000 ohms total should be pretty intuitive.)

Case 2: What resistor is required to drop 123.5 volts from 135 volts if there already is a 100K resistor in the circuit?  (135 - 11.5 = 123.5)

unknown resistor = 100,000 * 123.5 / 11.5 = 1.074 megohms.


Terry, 230RN

* I believe there are new symbols for current and voltage, but the relationship remains the same.  Been a while.

[RoadKingLarry]

1.21 gigawatts

Or what 230RN said.

[230RN]

^ The power ratings of the resistors is a different question :)

Actually, I forgot that one answer would be 11.5 volts.  Duh on me.

Up all night myself, hashing out something else that's personal.  Had to correct something in it.  Good diversion for me, though.

Terry, 230RN

[Nick1911]

Wow Terry - well said.

Jim - I don't think the question is answerable with the information given.

It seems that you're trying to drop voltage across a resistor.  If this is the case, we have to know what the resistance of the load is before we can calculate that.

[230RN]

^

Jim - I don't think the question is answerable with the information given.

You are correct, as I noted.  The trouble with answering some questions is that you have to "read into" what the querent really meant, which I tried to do.

In any case, adding a load across the 100 K resistor changes the voltage divider relationships.

Using the 35 volt input as an example, adding a "load" of 50K across the 100K resistor changes  this portion of the circuit, by the resistors-in-parallel formula:

1/total resistance = 1 / resistor1 + 1 / resistor2 + ... 1 / resistorN,  AKA "The Law of Reciprocals" *

With that 50 K added load resistor in parallel with the 100K resistor, the total resistance of that part of the circuit is now only 33,333 ohms, thus changing the voltage relationship noted in the prior post in this way:

Case 3: What new voltage will result across the two resistors in parallel (which total 33,333 ohms), if the series resistor is still 200,000 ohms, and still with an input voltage of 35 volts?

(33,333 / 233,333) * 35 volts =  5 volts

Again, by the proportions of the resistances in the voltage divider.

It's a lot easier to see this with diagrams, but my scanner isn't working.

Terry, 230RN
________________________
*  For the special case of only two resistors in parallel, the formula can be the "product over the sum"  formula for two resistors in parallel, or

Total resistance for two resistors in parallel = (R1 * R2) / ( R1 + R2)

Remind me to tell you about how the law of reciprocity also applies to calculating the total intelligence of a group.

[Brad Johnson]

Three.  Three shall be the number.  Count'est not to four lest thy be blown to tiny bits...

Brad

[Jim147]

There is another resistor in there. And a pair of shunt coils.

This thing is almost as old as Terry.

I'm used to inverter boards not exciter motors.

Thanks for the help. I got some testing to do in the morning.

jim

[never_retreat]

Wow that brings back nightmares. I think I would have to dig out a book then proceed to beat myself in the head to remember that stuff.
In other words.

[230RN]

^ I bet that fascinating thread title drew you right in, right?

[Jim147]

I know I didn't ask a very good question but I was getting a little flustered with this machine.

It uses two exciter motors. One putting out 35 VDC and on putting out 125VDC.

I had no wiring diagram and a two inch diameter wire loom running about ten feet.

It was missing voltage to a relay coil. I was trying to figure out where it was getting its power.

It surprised me that it came off the 135 VDC feed. But it made sense once I could trace and see the whole circuit. It had a crack in one of the copper bars running between the shunt coils.

It's working now.

jim

[never_retreat]

^ I bet that fascinating thread title drew you right in, right?

Dam skippy.

[230RN]

Well, you work with that stuff every day for years, and it's kinda hard to forget it even two decades later.

For resistors in series, you just add 'em up.  Simple.  Every resistor you add in series increases the resistance of the total network.

For resistors in parallel, you have to add up the reciprocals. Not so simple.  Every resistor you add in parallel provides another path for the current to flow, thereby lowering the total resistance of the network.

Glad you got it working, Jim147!

[grislyatoms]

Reminds me of first year d.c. circuits. Instructor soldered together a cube of resistors and had us calculate series-parallel resistance from various points. It was rather maddening.

Sheesh, I don't even remember the color codes, now.

[230RN]

For the diagonally-opposite corners (the typical way the problem is posed) with 1-ohm resistors the total resistance is  0.8333... ohms.

The thing to do, see, when someone throws that at you, is to groan, say "Wait a minute," retire to your desk and pretend to jot a couple of numbers down, scribble out a cube with Rs on the edges, then crumple the paper, shove it in your pocket, come back to them and say, "I'm not sure, but I think it should be about point-eight-three-three ohms."  Then you scratch your head, look a little perplexed and doubtful, and add, "If my arithmetic is correct."

Then you drift back to your desk among all the adulating stares, making sure to jot more numbers in the air with your finger, muttering things like, "Carry three," and sit down. 

Then add, as if to yourself, "Yeah, I think that's right."

Terry, 230RN

REFS:

http://www.physics.ucsb.edu/~lecturedemonstrations/Composer/Pages/64.42.html
^ (The simple way)

http://www.radioelectronicschool.net/files/downloads/resistor_cube_problem.pdf
^ (The hard way)

MORE:

http://www.google.com/search?hl=en&source=hp&biw=1037&bih=603&q=CUBE+OF+RESISTORS&aq=f&aqi=&aql=&oq=

[CNYCacher]

[230RN]

Tried it.

Ran out of resistors.

And solder.

Hope nobody asks about a three-dimensional grid.