saps
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electricity etc
I know very little about electricity and have several questions.
-Firstly, why do battery packaces list the voltages of thier product but not the amerage??
-Secondly do resistors actually use a material (eg. tungsten) to decrease electric charges?
- Another question that i have is: how can i get a capacitator to release its full charge in a fraction of a second?
I know that these questions may seem dumb. And i also realize that i should probably be asking a differnt source for such simple questions, But if
anybody could respond it would be greatly apreciated.
-Saps
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12AX7
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| Quote: | Originally posted by saps
I know very little about electricity and have several questions.
-Firstly, why do battery packaces list the voltages of thier product but not the amerage??
|
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 
| Quote: |
-Secondly do resistors actually use a material (eg. tungsten) to decrease electric charges? |
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.
| Quote: |
- Another question that i have is: how can i get a capacitator to release its full charge in a fraction of a second? |
The discharge rate of a capacitor depends on the resistance and inductance of the circuit. There is resistance to everything: the wires, the load,
even (or in some cases, especially!) the capacitor itself. Inductance is the inverse of capacitance, like how in mechanical terms, springiness is the
opposite of mass (indeed both systems oscillate back and forth, and capacitance can be likened to an electrical "weight" on the circuit).
The simplest way to discharge a capacitor is to short its terminals. This minimizes resistance, maximizing current.
Let's say you have a charged capacitor sitting on a table... it has 10V on its terminals. Let's say it has 1000µF = 1mF storing that
energy. Now we connect a one ohm resistor across the terminals, at t=0. We know the capacitor has 10V on the terminals, so what, there must be
10V/1ohm = 10 amperes flowing. Because a capacitor is a "massive" thing, current and voltage stay reasonably constant for a pretty good
period of time. Well, not very long, but we're looking at a short slice of time after all.
The next thing that happens is, now that we're drawing 10A from that capacitor, electrons are moving around on its plates. Charge (Q) is
dropping, which means voltage is dropping, as Q = V*C. Since time is a big part of this, let's differentiate that: dQ/dt = C * dV/dt. dQ/dt,
the rate of change of charge, is otherwise known as.....current! One ampere is one coulomb per second. dQ/dt *poof* turns into <I>i</I>.
Alrighty, so i = 10A and C = 0.001F. dV/dt, the rate at which voltage is falling, equals 10/0.001 = 10kV/s, or more convieniently, 10V/ms.
So very roughly... the 1000uF capacitor charged to 10V will discharge through a 1 ohm resistor in 0.001 second.
But note that the resistor would have to draw 10A the whole while. EVEN THOUGH voltage is dropping. This doesn't happen, because resistance
would have to drop, for some reason, at the same time! (Constant current sources and sinks are easy to make, but that requires an active device like
a transistor.) What actually happens is, over a short period of time, you get a small voltage drop, which decreases current draw, decreasing voltage
drop slightly, and so on and so forth. Eventually, you develop an equation that shows <U>exponential decay</U>. Voltage starts at 10V,
drops quickly to half, quarter, eighth, and so on it decays. The time constant, R * C, is the point in time when 63% of the charge (voltage) has
decayed. A practical "the cap is empty, jim!" point is 5RC, where the charge is reduced to 99% of the initial value. A 100uF capacitor
charged to 500V, discharged with a 1kohm resistor, will be discharged to an insignificant 5V in a half second.
FYI: all this talk of resistance (particularly internal resistance), time constants and whatnot makes another interesting consequence: when
discharging capacitors into one-another, charge *is* conserved. We know this from the loop law. However, with each exchange, you lose exactly half
the *energy* transferred from the charged cap. Can you prove why?
Tim
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DDTea
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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??
"In the end the proud scientist or philosopher who cannot be bothered to make his thought accessible has no choice but to retire to the heights in
which dwell the Great Misunderstood and the Great Ignored, there to rail in Olympic superiority at the folly of mankind." - Reginald Kapp.
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sparkgap
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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 (~_~)
"What's UTFSE? I keep hearing about it, but I can't be arsed to search for the answer..."
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12AX7
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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
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saps
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12AX7, So a D-cell would be 9 watts because1.5 volts *6 =9 ??
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12AX7
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No, a D cell contains 1.5 * 6 = 9 watt-<B>hours</B>. Volts times amps equals watts, volts times amphours equals watthours.
Things do not store watts. Period. They store joules. (Capacitors too, E = 1/2CV^2.) Watts are the usage of energy and depend on time. You can
spend 1J in a microsecond (for a power dissipation of 1/0.000001 = 1MW) or spend it over an hour (for a power dissipation of 1/(60*60) = 2.8mW).
Tim
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DDTea
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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.
"In the end the proud scientist or philosopher who cannot be bothered to make his thought accessible has no choice but to retire to the heights in
which dwell the Great Misunderstood and the Great Ignored, there to rail in Olympic superiority at the folly of mankind." - Reginald Kapp.
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Simon
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| Quote: |
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??
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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!
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saps
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I have another question:
I am thinking about buying a Philips 4 watt Germicidal Sterilamp which is essentially just a flourescent lightbulb that creates large amounts of UV-c
rays for sterilization. One of the internet sites i might consider buying the product from says that a "Starter IS Required," but do i need
a ballast too??
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Pyridinium
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If I'm not mistaken, a ballast gives the inductive surge that starts the lamp, then it regulates the current once the gas discharge is underway.
A starter just gives the surge, but it doesn't regulate the current. I think the modern "starter" is actually a ballast / starter in
one unit.
I could be way off on this...
Edit: also, when you buy disposable batteries they don't usually have an amp-hour (Ah) rating, but many storage batteries do. Example: I had a
small gel cell, 12 volts, 7 Ah. I don't think I ever really got 7 Ah out of it, more like 5 or 6.
I have a marine battery for my HV inductor experiments that is, I think, 50 Ah. It has been deep-discharged multiple times and still holds a charge.
:-D
[Edited on 3-7-2005 by Pyridinium]
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12AX7
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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
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Pyridinium
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| Quote: | Originally posted by 12AX7
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? |
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.
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Simon
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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.
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zoomer
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A gaggle of electrons = ? Coulombs
Samosa,
Your definition of coulomb is technically correct, but incomplete. A coulomb is a constant that describes a specific number of electrons, about
6.41x10^18 of them. Push that many electrons through a component in one second and you have one amp.
Z
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Lambda
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European 230 volt TL-tube lighting system
A TL-tube has filaments that act as PTC (positive temperature coeficient). When they glow, the resistance becomes high paralell to the tube and when
disconected to induce a relatively high inductive voltage (even above 1000 volts) via the choke coil. Even when the neon/mercury vapor starter is
triggered again, this parralel filament resistance will still be to high to quench the inductive voltage completely. Heated up Mercury vapor in a say
1.18 m TL-tube ignites at roughly about 400 volts and has an arc voltage of several deca volts up to about 80 volts. This arc voltage must allways be
lower than the voltage needed to ignite the neon/mercury vapor starter again (at least 110 volts). The choke coil forms an inductive ballast, soaking
about 150 volts or more. If different methodes are used to spike ignition of the TL-tube, then a capacitor may be used instead of a choke coil to soak
the voltage not used over the TL-tube. The ignition coil may then be very small. These electronic regulated and ignited tubes use less current and can
be opperated at 30.000 Hz instead of 50 Hz and have no visible strobe effects.
Tl-tubes are filled with argon or krypton in combination with mercury vapor at a low pressure. The filaments are covered with a barium oxide (or
other) layer to stimulate better electron emission. They are kept warm by the small current that still runs through the capacitor parralell to the
neon/mercury vapor starter tube (a consequence of). The capacitor quenches high frequency radio distortion emitted by the sparking of the bi-metall
switch and resonates with the ignition coil. A starting TL-tube gives quit a lot of radio distortion due to the oscillating nature of the circuitry.
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sparkgap
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zoomer, shouldn't it have been an Avogadro's number (~6×10<sup>23</sup> electrons for a coulomb of charge?
sparky (~_~)
"What's UTFSE? I keep hearing about it, but I can't be arsed to search for the answer..."
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12AX7
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Nah, Avagadro's is for underground critters. Matter of fact, that's why it takes 26.4 ampere-hours per mole of electrons.
Tim
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zoomer
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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
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moose sniffer
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I have a question: When I rub a balloon on fur it becomes charged and i'm able to pick up small things. Why cant i pick small things up in the
same way with my Van Der Graff Generator??
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12AX7
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Since when? As I recall a VDG works quite well indeed for attraction. At least until the other object gets charged by the metal surface.
Tim
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kickflip_333
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i have seen people get keys to attract toward one from at least a foot away.
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