electricity etc

Batteries (cells rather; a battery, as you might imagine from the word, is a number of cells connected together, a battery of cells that is to say) do not have an amperage rating, because they supply energy, not power. Thus you can only define current for a period of time, measured in ampere-hours (Ah, or mAh for 1/1000th). If a battery can supply five amperes for three hours, it has a capacity of 5 * 3 = 15Ah. You can also draw one amp for 15 hours.

Since a battery's voltage is constant, you can calculate energy capacity from voltage and Ah. (Volts times amp-hours = watt-hours, 1 watt-hour is 3600 watt-seconds, otherwise known as a joule.)

Now to answer your question, batteries sold don't have an Ah rating, for who knows why. You can assume AAAs are somewhere around 500mAh to 1Ah, on up to D cells which are in the 4-6Ah range.

If you meant amp rating as actual instantaneous current capacity, that is determined by the internal resistance of the cell. This is a fault condition, so I don't recommend you test it

Huh??

I don't know what you mean by charges, but resistors do use particular resistive materials to drop the voltage. (FYI, charges are not dropped, charges are electrons and the same number of electrons going in a system must come out!)

Most resistors use either carbon, metal or oxides. Originally, carbon powder was glued together and called carbon composition. Noisy as hell. They also used resistance wire, first iron, then alloys formulated to have a low temperature coefficient (resistance usually rises, quite dramatically, when a metal is heated). Wirewound resistors are still used today because, being solid wire... they change very little with time so can be made very stable. They are also tolerant of heat so can dissipate lots.

The average small resistor these days is carbon or metal film (vapor deposited on a ceramic core, IIRC, and etched in a spiral pattern to obtain the proper resistance). Power resistors range from thick film to metal oxide (often in a gray to white cemented case, hence the term "sand resistors" to wirewound.

When it comes to dealing with electricity, there are a few things that always screw me up/keep me from understanding it as well as I'd like. So I have a dumb question too

I understand that electricity is "the flow of charge;" but when we talk about current, is it the flow of positive charge or negative charge (i.e., electrons)?? What is doing work in the system--positive or negative charge??

It's the electrons that move around in an electrical current, good Samosa. Protons are not (usually) used to carry charges from one place to another.

sparky (~_~)

Yep. The whole confusion, however, started when people starting noticing the utterly horrid mess of minus sign after minus sign, not to mention occassionally forgetting them between steps! So "conventional current flow" was created. Current flows from anode to cathode, whether or not electrons are actually emitted from the cathode, or positrons somehow manage to emit from the anode and locate the cathode exactly. (Which could still be, I mean...when's the last time you physically saw an electron? When you experimentally test the direction, how do you know that isn't accounted for? Damned quantum mechanics! )

Then along came semiconductor physics. To visualize the flowing charges while keeping in step with conventional current flow, they called the absence of electrons "holes"...

Tim

It's useful for us to define our units, that's what had me confused for a long time. Once I understood the units, I had a better idea of where I was going...

1 coulomb (C) = the quantity of charge transferred in one second by a current of 1 Ampere

1 Ampere (A) = 1 Coulomb/Second--a measurement of current

1 Volt (V) = 1 Joule/Coulomb--a measurement of electric potential (one needs a potential difference for current to flow--just as water needs to be at a higher point in order to fall)

1 Joule (J) = a measurement of energy (from what I've been told, a single match contains about 1 joule of energy)

1 Ohm (capital greek letter 'Omega' = a measurement of resistance (electrical resistance, not Iraqi resistance )

1 Watt (W) = 1 Joule/second--a measurement of work

1 Newton (N) = 1 kg*m/s^2 -- a measurement of force; specifically, the amount of force required to accelerate a 1 kg mass 1 m/s^2, according to the equation F = m*a .

There is a nice equation that relates Voltage, Current, and Resistance (this will be a bastardization of it for you purists, but it's easier to remember):

V = IR

V = Voltage (V), I = Current (A), R = Resistance (Ohms)

So as you see from this equation, for a given voltage, if resistance increases, current decreases! That makes sense, right? Memorize that equation.

Current is conventionally talked about in terms of positive charges moving from A to B. However, now we know that most of the time it has been negative electrons moving from B to A.

Sometimes it's negative ions moving from B to A while positive ions move from A to B at the same time.

It all comes to the same thing, like how -1 * -1 = 1 * 1 = 1, which is why it took so long for us to realise how it really works. If a charge moves in an electric field, it does work.

(I might note that when a DC arc forms the positive end gets hotter because of bombardment from electrons. Otherwise current flow is usually symmetrical.)

Saps: you still seem confused about electricity. Keep asking those noob questions!

For typical fluorescent bulbs, you have two tungsten filaments at either end with mercury vapor inbetween. Each filament is rated maybe 40V. When you flip the switch, current goes through the ballast, down one filament, through the starter and down the other filament (and vice-versa, since this is AC). The ballast is nothing more than a resistor, or since that wastes power, an inductor is used. The starter device is essentially a neon lamp with a bimetallic strip inside: when the lamp and strip are cold, the main tube insulates (it isn't ionized yet) and current flows through the starter. This causes a discharge which heats the bimetallic strip to the point that it shorts the contacts, putting line voltage (less ballast drop) across the filaments, heating them up and allowing emission. The bimetallic strip in the starter cools down, which would allow it to arc again (it normally arcs erratically as it heats and cools like this), but now that the main tube is ionized, the total voltage is lower (more drop across the ballast) so it just relaxes and stays out of it.
Inductive kick (from the erratic closing of the starter) may be a factor, but if it were, the filaments wouldn't be necessary. Also, how can any significant voltage appear across the main tube if the starter tube discharge has a lower starting voltage?

But back to the fluo you have... that could be anything. Inverter fluoros ("cold cathode" obviously needs a high voltage spike to ignite the tube, either from the inverter (with an open-circuit tube, an unregluated inverter's voltage rises dramatically) or a seperate ignitor circuit (like photoflash tubes, except continuous rather than pulsed discharge).

So to answer your question, RTFM. (If you can find a datasheet for that tube....there ya go!)

Tim

I'm pretty sure it does have a major effect on the starting of the tube. No inductive surge, no starting of the lamp. There is oscillation involved I think, but I don't know the mechanism off the top of my head... I have to get back to you on that
I have a reprint of an old Navy electronics manual somewhere around here that explains things pretty well.

In the old-style set up, when you turn on power, current flows through the inductive ballast and starter. As explained, the starter then goes open circuit, forcing the current from the inductor into the tube with that inductive kick.

The first few flashes from the tube are from the inductive kick and you'll notice dead tubes keep flickering.

The ballast's main purpose is to limit current.

In short: yes, you need the ballast. RTFM might be a good idea too if you're going to try wiring up something yourself.

A gaggle of electrons = ? Coulombs

European 230 volt TL-tube lighting system

zoomer, shouldn't it have been an Avogadro's number (~6×10²³ electrons for a coulomb of charge?

sparky (~_~)

Coincidently, I fished one of those avagardros out of my pool this morning. I do have to correct my earlier typo, a coulomb works out to 6.241x10^18 electrons, not 6.41... .

Z