well i know this isnt the place but

[turboedbug]

I need to vent! Well i am off in steamboat springs colorado for electrical school and wholy crap are series and parralell circuits are hard! 13 people had one hell of a time with two questions......and yess that was the whole class. That wasnt even all the homework........wow what did i get my self into

[Marc]

There's no better place in this neighborhood. Hopefully the light will come on soon, hang in there.

Practice helps build confidence, try this (for decimal answers don't go past the second decimal point (it's too dumb to accept .3333 when it expects .33):
http://www.wisc-online.com/objects/View ... ID=dce8904

Count yourself lucky if the course curriculum doesn't include problems which require Kirchoff's Law and/or Thevinin's Theorem to solve...



... resistance AB is 5/6th Ω - the hard part is proving why.

[Piledriver]

For a little more "light reading":
http://jacquesricher.com/NEETS/

[turboedbug]

Well i have the series and parallel circuits down. I can figure voltage and amps now. Our instructor threw us a curve ball. The exact cube that marc posted but with 3 ohm resistors. The instructor said he has only seen 14 do it in the last 14 years. Anyone wanna lend a hand?

[turboedbug]

Thats 4 years......for some reason my stupid phone wont let me edit it.

[Marc]

The first thing most folks want to do when confronted with this problem is to redraw the circuit in two dimensions (I know I did)...don't waste your time, it won't help
Kirchoff's law says that the current leaving any point (node) must be equal to the sum of the currents going into - and vise-versa.
I already gave you the answer to the cube with 1Ω resistors, let me show you how that one's solved. Once you have it figured out, the 3Ω version will be a breeze. The key to it is in assuming a current through the circuit to give you a starting point for calculations. ANY current would do, but since we know the first thing it'll do is split three ways, 3A is a good choice to make the math simple.
Assume, then, that 3A is entering the upper-left node. Since the three resistors attached to it are of equal value, 1/3 of the current must go through each (so 1A through each) ...that's a bit oversimplified since the values of the other resistors would also affect the "split" but thankfully they're all the same so we don't need to think that hard, we can simply trust that the symmetry of the circuit will keep things simple and move forward (that's less obvious when drawn in 2D, another reason why it's futile to do so). When 1A through any one of them gets to the next node it has to split off through two paths, so .5A each through them. As the .5A currents meet up at the nodes on the way to the lower-right "exit", they add back up to the original 3A....make sense? That's Kirchoff's Law in action.


Now that you know the current through each resistor you can calculate the voltage dropped across it. Follow any of the three possible routes from the "in" to the "out" and you'll see that the voltage drops all add up the same (the first and last in the path drop the same voltage, and the middle one drops half of that) - that makes sense, since it's essentially three parallel branches and the voltage drop across any branch must be the same as that of the total circuit.
Now you have the current and the voltage drop, and can calculate the resistance of the circuit.


You could substitute resistors of any value, so long as they're all the same the total circuit resistance will always be 5/6 of the value of a single resistor. The instructor will probably be more impressed if you show your work, but if some smartass some day again tries to stump you with this problem, take three seconds to multiply the single resistor value by .833333 and blow their mind. When they ask how you figured it out, just say "it's intuitively obvious, isn't it?

Will you ever encounter a practical application for this knowledge? Very doubtful, but like me you'll remember it for life - if nothing else, it drives home Kirchoff's Law.

[turboedbug]

Yeah he wants amps, ohms and voltage at each resistor. Ur way is simplified. The way we were tought were to start at end of circuit and work back to the battery and then backtrack it once we get everything simplified. I am going to give it a shot tomorrow after i get some rest. He swamps us with so much work that we only get 5 hours of sleep a day. So i am beat

[Marc]

The only simplification I offer is in the recognition of the elegant solution. I know I've said some dumb-ass stuff around here but I honestly am a certified super-genius - once in a while it comes in handy.

It's not unlike when you're asked to give the sum of all the numbers from 1 to 100, and realize that the problem is simply 50 x 101 = 5050. There are 50 pairs of numbers, each pair adding up to 101 - 1+100, 2+99, etc. through 50+51.

I cannot imagine how someone other than a mathematics savant could solve the resistor cube any other way...other than grabbing a handful of precision resistors and a soldering iron, anyway. I'd be really impressed by the work of those 14 guys if they managed this the "hard way".

[turboedbug]

It was actualy 4 guys......my phone is stupid. I dont have my laptop with me

[Marc]

I know I said it was a waste of time to redraw the circuit, but I did it anyway in case it'll make it easier for you to follow. Note how the individual currents entering each node are equal to the current leaving it.

If all resistors are 3Ω, any current path you choose from top to bottom goes through two resistors which each drop 3V and one that drops 1.5V, for 7.5V total.
If the voltage across the circuit is 7.5 and the total current through it is 3A, the circuit's total resistance is 2.5Ω...(as promised, 5/6 of the single resistor value).

[madmike]

I was sent to Radio school in the Army some 35+ yrs ago and learned what a 'Tube' does after 8 wks ,shipped out to Germany only to find out permanet duty station had test equpment,If the green light comes on its good ,,Red,, it's bad ,why did I need to know what a 'Cathro' was Ohm's law is still handy when wiring a VW

[Ol'fogasaurus]

Mike my "radio" was in the late 50's so you know what kind of shape I am in on this. I think Marc's answer is wrong as there is no ground shown The ground leg does all the work and that is about what I seem to remember... the hard way.

[Marc]

Mike my "radio" was in the late 50's so you know what kind of shape I am in on this. I think Marc's answer is wrong as there is no ground shown The ground leg does all the work and that is about what I seem to remember... the hard way.

Surely you can't be serious. The circuit is completed by the ohmmeter, no "ground" is needed. I'm aware of your fixation on grounds, based upon other posts you've made, but a statement like "the ground leg does all the work" is patently ridiculous.

[Ol'fogasaurus]

Marc, there was a smiley face there and I was pulling your chain... lightly . The ground thing is what I have been told many times (I think it only applies to 12 volt in this manner. I'm dating myself but remember the big who-ha many years ago when some American cars and some European cars were positive ground, as I understood it that was the reason they did it).

[Ol'fogasaurus]

http://community.cartalk.com/discussion ... cal-system

http://answers.yahoo.com/question/index ... 519AAO5PZZ

I know it doesn’t mean much but I think the second it what was driving things *pun may or many not have been intended ) for a while.

Lee