Basic electrical concepts question

In the end, it doesn't matter if you consider it pushed or pulled, the equations we use for circuit design don't 'care' if you think current as pushed or pulled.

Using a water analogy. in most cases you can think of increasing voltage aswater getting higher behind a dam -- it has more potential, more pressure. Think of it as a water level behind a dam. When you apply a load, you're giving the water a way to get lower. Like energy in a battery, we drain some of it away when we apply a load. Some times our equations could be used to describe water flow from a dam just as well as they could describe current flow from an electrical power source.

(To my more exacting readers, yes, I know the difference between laminar and turbulent flow and Reynolds numbers)

In this problem, it is more important to get the voltage correct, if this is done, the current (AMPS) is a function of the load and will take care of itself provided that the current capacity of the dc adapter (rated milliamps) is large enough to handle the load.

A 9 volt 600 ma adapter is on the larger side, and if this was the specified original equipment, you would probably be in trouble if you replaced it with a 9 volt 200 ma version, for example.

The other point to consider is to make sure that your polarity (+ - ) is correct with respect to the tip and sleeve of the connector. DC adapters come in both (+ -) and (- +) with respect to connector polarity. The worst thing that will happen if you guess polarity wrong is it will blow up your device (and maybe the adapter too!). The best thing that will happen is that the device simply won't work until you reverse the polarity. Look for a little symbol on the adapter, a semi-circle with a + - next to it to tell you what kind you have.

Finally, there are unregulated adapters and regulated ones. An unregulated 9 vdc adapter might produce an actual open circuit voltage of 15 volts or so. A regulated adapter will give you the rated voltage, open circuit or loaded. Which kind you need depends on the device connected, although you generally will be playing it safer with a regulated adapter.

I don't normally recommend Radio Shack for anything, but they do sell switchable multivotage - switchable polarity, replacement DC adapters with all sorts of connectors.

Beachcomber

Does water flow thru a hose because there is more pressure at the spigot end than at the sprayer, or does it flow because there is less pressure at the sprayer than at the spigot?

In a simile to what you asked - does the water flow to the end because there is 10 psi at the spigot and 0 psi at the end, or because there is 0 psi at the spigot and -10 psi at the end?

Push and pull are qualitative "conventions", mathematical devices used to imply a direction of a force. If you pull on the rope in a westerly direction. and he pushes on the wagon attached to the same rope and also in a westerly direction, the force is in the same direction. So electricity can push, or it can pull, depeneding on how you personally want to look at it.

So to keep the reference frames in order and the same for all engineers and technicians, engineers view electricity as the movement of charge - not electrons or current or the like - charge. Thus, voltage has force derived from charge. Current is flow derived from charge. Power is the amount of charge delivered at a force in a certain period of time.

The voltage should match, the current rating match or exceed. But if you need to ask this I suggest you start reading some basic textbooks!

Wouter van Ooijen

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Also, the current rating should not exceed by too much.

Best regards, Spehro Pefhany

An 220 V appliance, when connected to mains, will draw it's rated current whether the circuitry can handle it or not.By the Ohm's law, I =U/R i.e. current is voltage divided by resistance.If you apply 220 V to a 10 ohm resistance, it will draw 22 Amperes, whether you connect it by a hair-thin conductor or a 2-cm-thick conductor.As you have experienced, if you turn everything on at home, you will blow the fuse. If you search for an adapter, the mA rating is the MAXIMUM current that adapter can 'give'.But, it's a waste to buy a 1000 mA adapter for something that draws only 50 mA.AND the voltage must be identical. The electricity is both pushed and pulled.From the negative side of the battery, the electrons are pushed away, and from the positive they are pulled.

-- Dimitris Tzortzakakis,Greece Visit our website-now with aircondition!

Spehro brings up a good point. Some devices are designed to rely on the current limiting capacity of their associated adapters. (Even though the voltages match).

An adapter with a too large current rating can lead to safety and/or operational problems.

You wouldn't want to connect a small flashlight or cell phone charger to an un-fused car battery connection, for example. The device might work ok most of the time, but if you had a short anywhere it would melt you wiring and possible start a fire. I've seen this happen to car alarms that were incorrectly wired directly to a car battery terminal without a fuse.

When you need clarity in science you should always go back to the concept of energy. Force alone is insufficient--for every force there is an opposite opposing force, so how can you from force alone know the 'direction' of something?

In electrical systems 'pushing' is voltage and 'draw' is current. 'Pushing' is analogous to force and is equally ambiguous about direction. For example, a battery may be said to always try to 'push' current in a particular direction, but this tells you nothing about what direction of current is actually going (think car battery).

So what does this all mean?

1. To understand how an electrical system actually works you must define both voltage and current together (which nails down energy).

1. Energy is always 'pushed' in the sense that you always have a source of energy.

2. Electricity is a form of energy and so is likewise always 'pushed'.

1. 'Pull' is not a useful concept except as a synonym for current.

cheers dave y.

Gosh. If I push a kid on a tricycle, I've applied a force. He and the trike move, and he and I know the direction of motion. The kid and the trike resisted my force with an equal and opposite force, but there was no doubt that the kid and the trike would move in the direction of the force I applied.

Yeah, don't I remember something about force being this vector thing, you know, magnitude and direction stuff. Seems to me energy is a scalar property.

I guess we'll just have to use whatever best describes the problem, huh?

That's one possiblity. Another is he was on a hill rolling backwards and you are applying a braking force that is slowing him down, but he's still going backwards.

And while he is rolling he has kinetic energy but your braking force is actually reducing his energy. It's all about energy.

dave y.

in article snipped-for-privacy@mb-m07.wmconnect.com, tony at snipped-for-privacy@wmconnect.comremoove wrote on 12/31/03 1:32 PM:

What makes special relativity so weird and wonderful is that the components of momentum and energy form a single four-vector that is invariant under relativistic transformation.

Bill

No disrespect to anyone, but hasn't everyone here taken at least a grade school course in general science or introductory physics?

So far as I know, every one of these course teaches the concept that a source of electricity, like pressure with water, supplies the voltage (electromotive force) that results in the flow of current through any load.

So what the heck is this discussion of whether or not current is sucked or pushed? Seems pretty stupid to me.

That said, I wish everyone here a very Happy, Healthy, and Prosperous New Year!

Harry C.

Really? I can describe all of the motion using f = ma and never refer to energy. Given a kid rolling downhill, I can apply sufficient force to make him stop and go back up the hill. All the while he and his tricycle are applying an equal force to me, but it is he and his trike that are slowing and reversing direction. It doesn't matter how high he is on a hill (potential energy) or how fast he's rolling down it (kinetic energy), I can apply enough force to reverse his direction. It's just f = ma; we don't have to consider energy.

Recall that you said "Force alone is insufficient--for every force there is an opposite opposing force, so how can you from force alone know the 'direction' of something?" Easily: f = ma. In the two cases I've described so far, the forces were indeed opposite and equal, but that fact had no bearing on the ultimate direction of the kid and his tricycle. I would argue that in the cases we have discussed, f = ma suffices and we never have to let energy enter the picture. As long as I apply enough force to make a = f/m greater in magnitude than the effects of gravity, the kid is bound to stop and start moving in the other direction. If I don't apply enough force, again determined from f = ma, where now the "a" is not greater than the effects of gravity, the kid will continue down the hill. I can draw those conclusions without reference to energy.

As you say, 'f=ma' will solve these problems and give the right answer, so of course I'm not arguing against Newtons Second Law. But if you look at advanced mechanics theory you'll find the use of energy equations since the old 'f=ma', although correct, can be difficult to apply in more complicated situations. For reference you might want to look up the methods of Lagrange and Hamilton.

The point I was trying to make is that when describing a system you always need a pair of variables to describe what is happening-- some books call these 'across' and 'thru' variables, the product of which is power (or energy). Common pairings are force/velocity and voltage/current. For example when you create a state space description of a system the state variables are always the set of these pairs (or some linear combination of them). That's because they relate to the kinetic and potential energy in the system.

The constitutive laws such as f=ma or e=L*di/dt establish the relationship between the 'across' and 'thru' variables, so when you work with these formulas you are using both. To try to talk about what a system is doing by using only one of those variables is not possible, you need both.

Keeping the mind set of thinking energy comes into play when you consider things like a gearbox, where at first glance you seem to enjoy magically increased torque--hey, just feed back a little of the output torque to the input and you have perpetual motion! Thinking energy keeps you on the right path.

dave y.

in article snipped-for-privacy@4ax.com, dave y. at snipped-for-privacy@ameritech.net wrote on 1/1/04 5:57 PM:

Keep in mind that the concept of energy and its conservation for mechanical systems is derivable from f=ma. How do you tell whic principle is more fundamental when you have two that can do the same thing?

Bill

Remember I was really only saying that you can't talk usefully about 'f' alone or 'a' alone in the 'f=ma' equation.

But to your question here, I would personally think energy is a more fundamental concept. Wouldn't you?

in article snipped-for-privacy@4ax.com, dave y. at snipped-for-privacy@ameritech.net wrote on 1/1/04 9:27 PM:

NO!

Bill

So what is the more fundamental concept if not energy? I've given my reasons why I think force alone is not sufficient.