Electric circuit

1. Aug 30, 2005

Craig113

Hello!

I have a few questions that need answers. They are just simple thoughts.

First of, its is said that the formula P = UI can applied to the energy being transformed every second over any component. But the formula P = (U^2)/R and the P= (I^2)R only can be applied over resistors.

But I think that the two formulas can I fact be applied to any component as well, as long as we know the resistance R of that component there really is no difference. The only difference may be that not all of the energy goes to waste as heat in other components that are not resistors.
Am I right?

Secondly, when one connects a spot of a circuit to earth that spot and only that spot gets the electrical potential zero. It does not affect other parts of the circuit that are located behind that point of grounding. But from that point and forward a positive test charge will not get affected of any electrical field that is located behind the spot, it will only be affected by new electrical fields located forward counting from the spot.
Why is that?

2. Aug 30, 2005

rbj

i use $$v$$ for voltage.

the instantaneous power entering a two-terminal electrical component is

$$p(t) = v(t) i(t)$$

that power is, of course, the time derivative of the energy going in. now that energy can be stored (as it is in inductors and capacitors) or it could be dissapated (as it is in resistors and other "memoryless" devices such as diodes). the problem is that the resistance is not constant with applied voltage except for the ideal resistor (which doesn't exist in real life but we pretend it does when we do circuits).

anyway, the equation above is the first principle. it's always correct. equations like

$$p(t) = v^2(t)/R = i^2(t) R$$

only apply when the resistance is known and reasonably constant.

3. Aug 30, 2005

Craig113

Okay, so you are telling me that the resistance in other devices than resistors is not reasonably constant when applying different voltages? Or even the same voltage and therefore the only formula describing the power correctly is the first.

Well what about the second question?
Can you spread some light over this problem?
If one connect any a point/spot of a circuit to ground then only that particular spot get the potential zero. The spot does not affect other part of the circuit. All it does is to “neutralize” any potential that a test charge may had before the spot.
How does this work? One could think that the spot that has contact with the ground could make the whole circuit or a part of if free of any voltage.

4. Aug 30, 2005

LarrrSDonald

I don't exactly know what he is saying (though I'm sure he's right) but the resistance over anything that has any degree of inductance, coils, capacitors, etc isn't even constant over time with constant voltage. A capacitor will at first have almost no resistance and rapidly rise to infinite resistance as it charges. A coil will initially have infinite resistance and then rapidly drop to none as it adjusts to the changing current (both cases ideal of course, but even in reallity they by no means at all give constant resistance - that's their whole function). When voltage is dropped a capacitor will (when treated as a black box) have negative resistance, if you drop it to 0 it will still be charged and have a voltage potential between the sides, provide current (briefly) if shorted, etc. So no, not especially constant..

Don't quite understand your second question, but perhaps someone else does.

[EDIT] Ok, so in case you'd like to further specify you're second question, I now (sort of) get why I'm not understanding it. You talk about "behind" the earth point, there is nothing behind it. If there is, it's not an earth point - it's assumed to be tied to earth, i.e. +-0 v with enough of it there to suck up any voltage and/or current going into it. This is obviously in an ideal system, but also holds true for most non-broken real systems. The theory will fail a bit if you, say, connect +5v straight to GND. In an ideal system, this is one of those "irresistable force meets immovable object" situations, a +5 ideal provides +5 no matter what and an ideal GND pulls anything to 0. In real life, you have a short and GND is usually "more ideal" then your +5 and the +5 will give first (not so irresistable as the GND is immovable). Provided your circuit is all nice and balanced, the +5 will drop evenly over your resistances (or rather trickily over things with inductance over time, but at a given moment it still holds sort of true) winding up at 0 at GND.

I'm pretty sure that's not the answer you're looking for, but perhaps you can get a better sense of why I'm not sure what you're asking.

[DOUBLE EDIT] Perhaps I just explained to myself what you're saying.. Yes, a GND point will make that part (up until anything with resistance and a +v on the other side somewhere) free of voltage. In reallity, GND isn't ideal but very close to it. It's considered a point that is large and can absorb any amount of voltage/current (such as pounding a stake into the ground and connection to it) large enough that it will not effect it's potential. Now, in low power circuits GND is usually the negative pole on the power supply, the opposite point to that providing the voltage. Thus, anything put out by the +v will by need be possible to absorb by the -. Any point connected to it with a degree of resistance between it an + (if there is none you have the afforementioned problem, one is just putting out and one is just sucking up until one gives - a short) will put all voltage over this point (it's the fastest way back to gnd) and the rest of the circuit past it will be potential-less baring other sources of voltage.

This is, of course, assuming resistors or resistance only devices - a device containing inductance or capacitance will have to have those considered. If you wind up, for instance, with a capacitor between two ground points that was previously charged you also have a short of sorts - it's discharging itself straight into GND. Any component storing charge must be considered differently although in real life they'll probably (in low v/a devices, high is another matter entirely that I have little experience but much respect for) discharge peacefully until the GND is truly +-0v.

I have no idea about what implementation if any this is for, but I should perhaps mention that the current will rise rather quickly when grounding random points in a circuit so it's by no means something one can do willy nelly with sensitive devices. If a point is grounded the current through the remaining circuit will rise accordingly (it's the new gnd, so everything is loaded over the "new" circuit) possibly way over it's limits (resistors glowing redish, capacitors exploding, ICs suddently losing their magic smoke, etc) so if this is application based rather then theoretical be careful.

Last edited: Aug 30, 2005
5. Aug 30, 2005

Craig113

All I am saying is that, when a part of a circuit gets connected to earth, only that part of the conductor gets the potential zero.

That part that is connected to earth is what I call the "spot". And behind or in front of that "spot" is meant parts of the circuit that is front or behind the spot of circuit connected to earth, How can this be so hard to understand?

My question is, why does only that spot get affected by the earth and not other parts of the circuit? I mean they are connected as well to earth indirectly

6. Aug 30, 2005

LarrrSDonald

<shrug> Color me stupid if you feel like it, I just wasn't sure what you were saying. I wasn't trying to say that it was poorly explained, just that I wasn't sure what you were saying, reason unknown.

The indirectly connected parts of the circuit aren't affected because there is resistance between them and gnd. Think of it this way: you pour a tons of electrons into one place and they flow towards gnd where they get absorbed by it (since it's always +-0). If you have a resistor, it will flow through it. GND is essentially an "ultimate" cathode, a place where any current will flow through. Anything with resistance will keep a degree of voltage on the side away from earth. If it wasn't so (i.e. if ohms law was wrong) then electricity wouldn't work at all like it does - all voltage in any circuit would immediatly drain to gnd, be it earth or a "normal" cathode.

If you're aiming for a more physical explanation of why electrons "pile up" in this way, dropping based on resistance from + to gnd, that's more theoretical then I can handle. I suppose I've always figured it was kind of intuitive, you have two equal resistors between +5 and gnd then the point between them will have 2.5v potential, or half the "power" available. Being an elecronics freak rather then a physicist I can't honestly explain it further I'm afraid.

7. Aug 31, 2005

phillip256

To have a working circuit there must be a source and a ground. Current will flow from the source to ground (in standard theory). Now lets say you have a circuit with; a source, a beginning, a spot, an end, and a ground, as shown below.

source (+25V) ----> beginning ----> spot ----> end ----> ground

Under normal operation the current will flow from the source to ground, dissapating its energy through out the whole circuit.

If earth (ground) is connected to "spot" than the current will flow from the source through beginning to the new ground at spot, dissapating its energy through beginning circuit. Because end circuit is located between two equally potential grounds there is no source, or energy, to dissapate through that part of the circuit (unless a new source is added).

As a side note, the energy dissapated through beginning will be more with the new ground than there was when the circuit was in normal operation.