Ohm's Law/Kirchhoff's Law Question

[rstofer]

"When electricity meets an intersection, if its fanning apart, the current divides (fractions), if its fanning together, the current adds.
So theres 2 mathematical functions/calculations able to be used straight away just with having the wires and electricity itself."

Congratulations.  You have just renamed Kirchhoff's Current Law, abbreviated "KCL".  Other beginners, looking for information on this topic, won't find it under "junction theory", but can find it under "KCL" or "Kirchhoff" (note the double "h" in the proper name).  See https://www.allaboutcircuits.com/textbook/direct-current/chpt-6/kirchhoffs-current-law-kcl/ , one of many results for Googling the standard term.

They might also find it under "Nodal Analysis".

[Capernicus]

Profound, if confusing.  How does that produce a logarithm?

Just as consequtive multiplies are a power,  consequtive divides could be a logarythm, as consequtive adds are a multiply, to consequtive subtracts are a divide.      log-> Its the count of consequtive divides it takes to get to 1.  it would be cool to see that done with wire dividers!

IanB - Ive got done work with the FFT before, and it has amplitude and phase as x+y as well, I never understood it but I got it working and used it.  I found all of digital and analogue filtering virtually impossible to understand, so I always just nabbed code instead to get my projects finished.   I have an idea of how I'd go about attempting my own filter tho,  if there was some way to increase the falloff speed of a blur, you could make a simple blur turn into a lowpass filter, as it is without increasing the falloff, its on orders of octaves not harmonics how slow it attenuates the sound. How a capacitor and resistor does it is quite dumbfounding to me just an ordinary bin can do it.

[TimFox]

The base 10 logarithm of 2 is 0.30103...
How do you compute that useful result with your "junctions"?
If it only works for base 2, how do you compute log₂(3)?

[TimFox]

"When electricity meets an intersection, if its fanning apart, the current divides (fractions), if its fanning together, the current adds.
So theres 2 mathematical functions/calculations able to be used straight away just with having the wires and electricity itself."

Congratulations.  You have just renamed Kirchhoff's Current Law, abbreviated "KCL".  Other beginners, looking for information on this topic, won't find it under "junction theory", but can find it under "KCL" or "Kirchhoff" (note the double "h" in the proper name).  See https://www.allaboutcircuits.com/textbook/direct-current/chpt-6/kirchhoffs-current-law-kcl/ , one of many results for Googling the standard term.

They might also find it under "Nodal Analysis".

Yes, "Nodal analysis" is a useful algebraic method to solve KCL on a practical circuit.

[Capernicus]

the integral log2 is all you need to make an ADC.   fractional results aren't as important.

log2(3) should abort on the second divider.    1^2.  (the first divider is 0^2)   (its 2.)
the strange thing about the integer log2, is it actually truncates upwards instead of downwards.

[TimFox]

That's a far less useful result than your original statement:  "If you divide, then divide again, youll have get two consequtive! divides happening,  so it actually computes log as well."
Maybe it works better with punctuation.

[Capernicus]

Got me drawing up another SCHEMMY!!!

adjust that top line of resistors and itll change the sensitivity,   but it has a problem still because when it starts activating it keeps activating all past the threshold.   so has a problem still.  I think if you make the resistors resist less ohms towards the left, it actually make the inverters outputs precedent over each other.  im still learnin'!

Because its just a binary output, it can be heaps convoluted charge and it doesnt matter,  its only an ADC so that makes it easier.

the wire coming in the side is an inverter, that activates when the charge has divided off enough for it to activate.



But then the question is,  can u get the rest of the bits?

[Electro Fan]

Hi:
I ran across this Popular Electronics Magazine (Dec. 1970) quiz while looking for another article. Electronic quizzes were a fairly regular feature in the magazine.

I have enlarged the paragraph above the circuits, so it is legible. Can anyone explain how the formula works? Ohm's Law and ratios?

It mentions that Ohm's Law and Kirchoff's Law applies but does not explain how the formula is derived from them.

My experience with Ohm's law is that if two quantities are given, the third can be determined.
The quiz uses voltages only. It does some squaring, addition and subtraction, but neither resistance nor current is specified.

Answers, too, in case you want to try your hand at manipulating the basic algebraic relationship for the various circuits.

(Not sure if my two images will appear in my post. I cannot view them in the preview option. If they do not appear, what must I do to have them in my post?)

Thanks,
Ron Reeland

(Attachment Link)

Not sure if you are looking for any more help with Ohm’s Law but if you are and you haven’t come across the circle of Ohm’s Law (and the associated triangles) you might find this useful:

http://electrical-mcqs.blogspot.com/2017/12/ohms-law-interview-question.html

[george.b]

I'm no good for computing exact values,

You don’t say… 

Its hardly important at all to me! =)     Alot of the time when I put caps in my circuits the actual farads dont even matter at all!   And when I'm deciding resistances as long as the machine still outputs acceptably, a range of values is acceptable.   I've got my next oscillator on the way, I don't know what the frequency will be exactly but as long as its oscillating that's the main part right?

It's very important, because you're getting bogus results most of the time. I could be mistaken, but I think the truth goes more along the lines of you being scared of doing the math, because that requires a modicum of formal education, even if it's self-education - you'll have to sit down with a book or an article and read through it and actually be taught about it by somebody else, directly or indirectly, which is something I think you're too arrogant to bring yourself to. It seems that, in your mind, you're this unparalleled, misunderstood genius who can solve complicated problems intuitively in his head and with some trivial experimentation using nothing but capacitors and resistors. In reality, you're someone with a seemingly weak maths background, scared of complex numbers because you can't understand them, so you'd rather go around them than just accept and use them, just like you'd rather go around all of the well-established circuit theory and just throw stuff at the wall hoping something will stick (and not explode). You're free to do so, of course, but then don't be surprised when your ideas are met by others (or by reality) with resistance and/or ridicule.

Being afraid of learning will only set you back, you know.

[Ron Reeland]

Hi:
basinstreetdesign posted a reply that the quiz formula is based on the famous Pythagorean Theorem. And he is right. A closer look at the quiz formula reveals that it is the usual A squared = B squared + C squared of the classic example. Thanks, as I did not comprehend that fact. The author pointed this out to a degree when he mentioned the values follow the 3,4,5 solution.

So, I printed up some graph paper and worked out the quiz #1 and quiz #2. My scans of them are shown below. Are they correct?

But I am stumped by quiz#3. I cannot deduce how to form a triangle since there is not an in phase or horizontal base for a resistor. I believe the capacitor's 6 volts is to be subtracted from the unknown inductor voltage to form one leg of a triangle. But I just am getting myself more confused.

Please assist. I am not ready to proceed in the quiz until I am taught how to solve quiz #3. I wonder if I will stumble on the balance of the quizzes.

Thanks,
Ron Reeland

[ledtester]

Well, the introduction of the problem set says this:

(VT)^2 = (VL-VC)^2 + VR^2

In the case of #3, VR = 0, so you just have to solve VT = VL - VC.