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Explanation of Amps

Discussion in 'Electronic Basics' started by tuxtlequino, Jan 16, 2005.

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  1. tuxtlequino

    tuxtlequino Guest

    I am trying to learn more about electronic, but for some reason I
    cannot fully Amps with the examples the book provided. I memorized
    Ohm's law, and the formulas necesary to come up with the total
    resistance in a circuit (either parallel or in a serie), but there is
    still more questions that the book does not answer. Could somebody help
    me with that please?

    Here are my questions:

    If E=IR can we just then increase the resistance and get as many volts
    as we want?

    The schematics in the book explains the number of volts (for example a
    9 volt battery), but they never mention the amps in the battery, how
    can we then figure out how much resistance do we need?

    If I get a LED that says in the package 2 Volts, and .005amp. I
    understand that I need to divide 7/.005 to figure out how much
    resistance I need. But why .005? Where did it came from, why, what is
    the resistance of the LED?

    Now, if I have a system of 9 volts (I bought an electric kit with a
    schematic explaning how it works!), and then there is a resistor for
    4.7MOhms, what is my amperage? Would I use the same formula(e=ir) to
    figure out? Does the use of capacitors change my amperage (I have seen
    them in my circuit!)?
  2. If you have a current source (that provides as many volts as it takes
    to push a fixed current through a resistance), then, yes.

    That formula makes a lot more sense if you rearrange it to show that
    R=E/I, or that ohms means 'volts per ampere'. A 100 ohm resistor is
    one that requires 100 volts across it to force 1 ampere through it.
    At any current or voltage, the ratio of volts divided by amperes is
    always 100. That is what 100 ohms means.
    Divide the volts by the resistance connected across it to find the
    amperes passing through that resistance.
    An LED is not a resistance, because the ratio of volts across it to
    the current through it does not stay constant, regardless of the
    voltage and current. It is a non ohmic device. It is specified to
    drop about 2 volts at one current, .005 amperes. You need some other
    formula than ohm's law to figure out the voltage current pairs for
    other currents. If you want ot run this LED at this current from a 9
    volt source, you might put a resistor in series with it to burn up all
    the extra voltage while 5 milliamps passes through both the LED and
    resistor. So, by ohm's law, you need a resistor that drops (9-2)= 7
    volts while .005 amperes passes through it and the LED. 7/.005= 1400
    If you can say that the circuit puts 9 volts across the resistor, then
    ohm's law tells you how much current that voltage will push through
    that resistance.
    9V / 4.7M = 1.9 microamperes.

    But if you put a capacitor in series with that resistor and connect
    the battery across that pair, the situation is very different. The
    rule that relates voltage across a capacitor ot the current through it
    is I=C*dv/dt) or current in amperes equals capacitance in farads times
    the rate of change of voltage in volts per second. Time is involved
    in this relationship, but was missing from ohm's law. So the current
    will depend on how long the voltage has been applied.
  3. If you already know the I and you already know the R, then it allows you to
    compute the voltage which must be present across the two terminals in order to
    make that particular current through that particular resistance.

    However, it is the more usual case in simple circuits that you have a given
    voltage and you either want to calculate the expected current, given some
    particular resistance... or else you want to calculate the desired resistor to
    use in order to achieve some given current value.

    Your next question gets to this...
    The battery supplies whatever current is required by the circuit. Most circuits
    will specify the desired voltage (say, 9 volts) and will have some particular
    effective resistance (not usually specified, though.) When you connect up the
    circuit, the battery presents a particular voltage to the circuit and the
    current simply is what it is, based on what the circuit requires.

    If you hook up a 1k Ohm resistor to your 9V battery, 9 milliamps will flow. If
    you hook up a 10k Ohm resistor to your 9V battery, 0.9 milliamps will flow (or
    900 microamps, as you prefer to imagine it.) The amount of current flowing
    depends on the "load" resistance. But the voltage at the terminals remains the
    same, with a battery -- 9 volts, in this case.

    Batteries actually prefer it if the circuits require less current. They last
    longer and they provide the correct voltage for longer, as well. If the battery
    were an "ideal battery" it would provide 9 volts no matter what current is, but
    real batteries are limited.

    A more useful specification for a battery, other than its voltage, is the total
    amount of energy it holds. Batteries don't really "hold amps" in them, they
    hold potential energy. As the energy is consumed, by way of the voltage and
    amps required over time, they gradually expire. While that takes place, their
    ability to maintain the specified voltage degrades somewhat and their ability to
    supply higher levels of current suffers. But it is the circuit that determines
    the "amps" that the battery needs to supply it. Not something inside the

    Still, batteries *do* have limitations in current. A car battery has a very
    high current rating and is able to supply very high currents into very tiny
    resistances, while still maintaining their voltage. A small hearing aid battery
    (a so-called button battery), on the other hand, may not be able to supply more
    than a milliamp or two and still hold their proper voltage, too. Something will
    suffer, if the circuit demands more than some tiny current limitation for those
    button batteries.

    The voltage rating may be the same, but bigger batteries are usually able to
    handle more current for longer times, because they store more energy and because
    they are designed for higher current requirements. For example, a D-cell has
    the ability to supply much higher current than a AAA-cell does, while still
    holding a proper 1.5 volts. If the circuit requires too much for the AAA, for
    example, the AAA voltage itself may suffer *and* the battery may also warm up
    and the total lifetime of use will also suffer. In the same situation, the D
    cell may be perfectly fine supplying the higher current, not getting warm at
    all, and holding its proper voltage the entire time.
    The LED "requires" about 2V to light up. With .005 amps (5 milliamps), the LED
    will be reasonably visible. Since your 9V battery has too much voltage for the
    LED to operate properly, you need to "throw away" part of that. To do so, you
    take the difference of what you have in the battery and what you need at the LED
    to compute the part to throw away. This is the 7 volts, as you said. Now,
    since you also want about 5 milliamps in the LED at the same time, and since the
    resistor is to be in series with the LED, whatever current the LED requires must
    come through the resistor. So the current through the resistor must be equal to
    the desired current through the LED. (Otherwise, the charge would have to "pile
    up" somewhere.) If you compute your resistor to be 7V/.005, or 1400 Ohms, and
    use that value then it will be true that the voltage taken away by the resistor
    (opposing the battery voltage) will be 7 volts *if* it turns out that 5
    milliamps occurs. But if the LED takes a little less than 2 volts then the
    actual current through the resistor will be a little more and if the LED takes a
    little more than 2 volts, then the actual current will be a little less.
    If the resistor is hooked directly across the two battery terminals, then the
    current through the resistor will be 9V/4.7Meg or slightly less than 2
    Yes, after some algebra to solve for I.
    It makes your current time-dependent. That's too complex for you to worry
    about, just yet.

  4. Miles Harris

    Miles Harris Guest

    I appreciate where you're coming from on this one, John. Also, JK's
    explanation was outstanding in detail and clarity.
    However, it's confusing to introduce a newbie to the concept of the
    current source. It doesn't exist in the real world as we know. No one
    - to my knowledge - has mentioned the essential missing ingredient
    here: the battery's _internal resistance_. This is what the OP really
    needs to be informed about, but I'm not apparently very good at
    expounding the basics, or so I'm told. Can someone please elucidate in
    a language the OP will understand?
  5. I don't like to try to explain everything in a single post, but try to
    guess what the op is ready to soak up in one dose. If he asks further
    questions, I will try to guess what he is ready for, then. I didn't
    think that the question of source resistance was needed at this place
    in his education, but I will be watching for the opportunity to inject
    it. If your judgment differs, by all means jump in with both feet.
  6. tuxtlequino

    tuxtlequino Guest

    Wow guys, thank you very much for explaining everything so quickly and
    in such a maner that it finally makes sense. Everybody apported
    something new to help me understand Amp's. Part of my problem
    understanding Amps had to do with the common knowledge out there. Most
    of what I heard was as a little child being warned by his father, or
    asking a silly question to an electrician. I heard many times that what
    kills is not Voltage, but Current.

    I also remember about how much static electricity we can accoumulate
    and discharge (I used to get shocked everytime at my College's
    library). So, through common sense I reached the conclussion that a
    certain amount of Amps was storaged at the batteries(or cells). When I
    first started to read the book in basic electricity and electronics my
    common knowledge made everything harder to understand. Now I know that
    everything has a certain resistance (and I supposed that is why a
    lighting can kill us, since we offer resistance). And knowing that
    batteries have current ratings explain why a car batery is more
    dangerous than another one.

    More questions will come as I learn more, but thank you for helping me
    with my questions about Amperes.

    I think I know that what Milles was trying to explain is that since
    every batery has a different resistance, they will have a certain
    amount of amp's (although not significant enough!). What I will like to
    know is how does the electrical companies carry electricity from far
    places. I mean, they have to fight all the resistance from the long
    cables, so the voltage has to be greater, and therefore we will have a
    very good amount of Amps in the lines. Does our AC electricity comes
    with Amp's?, or they are somehow wasted at the transformers?

    Sorry about the silly questions. I am a computer programer, so if you
    have questions about your computer I can help, but my knowledge in
    electricity is very limited.
  7. There is a way of thinking about this that makes sense. If you touch
    a source of voltage (with respect to the Earth, let us say) but you
    are not also touching the Earth, your body potential rises to match
    the potential of what you are touching, but, since there is no path
    for current to leave your body through a second point, you will not
    receive a shock.

    Grab the water faucet with the other hand, and that potential will
    drive current through you on its way to ground, and you will be
    shocked by the current passing through you. But it was the voltage
    difference between the two things you touched that pushed that current
    through you.
    Wrap yourself in metal, and the lightning will mostly choose that
    material and avoid you almost completely.
    A car battery does not produce enough voltage to push a dangerous
    current through your skin resistance (though it will give your tongue
    an unpleasant sensation). But it is dangerous because of the energy
    it can release into a low resistance load that will get hot enough to
    burn you. You can weld metal with a car battery.
    It will have a short circuit current limit that occurs when its
    internal voltage is all used up across its internal resistance.
    About half of all the electrical energy that is generated is lost
    heating the resistance of the distribution grid. They jack the
    voltage up very high, since doubling the voltage allows the same power
    (volts times amperes) to be delivered with half the amperes. And half
    the amperes drops half the voltage across a given piece of wire
    resistance. This reduces the wire losses to 1/4. Etc. But there are
    practical limits to how high they can raise the voltage. Stand under
    any high tension transmission line, and you can hear the crackle of
    corona discharge taking place, draining energy from the wires.
    But increasing.
  8. Andrew Holme

    Andrew Holme Guest

    Are you sure about that, John? I find that hard to believe. Googling for a
    figure, I found this:

    It says:

    "Seven percent of the energy is lost in transmission," said George David,
    chairman and chief executive of United Technologies, which is based in
    Hartford. "The solution is to put power generation much closer to where the
    electricity is consumed."
  9. Andrew Holme

    Andrew Holme Guest

  10. I hope you are right. I heard the 50% figure a long time ago and
    don't know where to look it up.
  11. Bob

    Bob Guest

    Wow guys, thank you very much for explaining everything so quickly and
    Yes!! This is a great way to learn. You can only learn so much from books.
    Experimentation is the key to a long, successful, career in electronics.

    As a great philosopher once said:

    "Give a man a fire and he'll be warm for a day. Set a man afire and he'll be
    warm for the rest of his life."

  12. Miles Harris

    Miles Harris Guest

    Thanks, Bob. I can always use a bit of humor to lighten an otherwise
    dull day. :)
  13. John Fields

    John Fields Guest

    Yes, it does. an _extremely_ common example is a high voltage being
    fed, through a high value of resistance, to a load which requires a
    constant current. An arguably less common example is a current
    regulating diode.
  14. Miles Harris

    Miles Harris Guest

    I'm talking about a _perfect_ current source, Junior. Surely even
    *you* must have worked that out.

    That "someone" will 'correct' me even if I don't make any mistake, it
    seems. :-/
  15. John Fields

    John Fields Guest

    Well, maroon, in the context of the thread, everybody knows there's no
    perfect current source, so when you don't explicitly state that that's
    what you're talking about, the default becomes real current sources
    which, like lead-acid batteries and diodes, we can add to the
    inventory of devices you've demonstrated you know little about.
  16. Miles Harris

    Miles Harris Guest

    Wrong again, Junior. It was John Popelish that introduced the current
    source in his first response on this thread. It was clear that he was
    also talking about a perfect current source:

    Nil points for observation and comprehension again, Jr.
  17. John Fields

    John Fields Guest

    As far as I can tell, JP was expounding on the _concept_ of a current
    source, which does assume ideal conditions, much like the
    impossible-to-find massless lever, weightless string, and frictionless

    Moreover, John clearly defined his terms in order to, ostensibly,
    remove any ambiguity from his argument, so his meaning was perfectly

    You, on the other hand, with your sloppy:

    "However, it's confusing to introduce a newbie to the concept of the
    current source. It doesn't exist in the real world as we know."

    Are claiming that the current source (any current source) doesn't
    exist in the real world. You seem to be reasonably adept with the
    language, and if you had meant otherwise I find it difficult to
    believe that you would have deliberately omitted the 'perfect' from:

    "However, it's confusing to introduce a newbie to the concept of the
    perfect current source."

    and in the process botched what you claim your intent was so
    perfectly. But perhaps it was only an oversight?
  18. Miles Harris

    Miles Harris Guest

    So you retract your earlier mis-statement? That's a welcome change.
    It certainly was. And yet again you failed to pick up on it, Junior.
    Are you actually bunking off high school.... or _grade_ school? Don't
    they teach English any more in American schools? I saw something on
    the news tonight that indicated Darwin and his work was to be dropped
    from the curriculum so I guess anything's possible these days in the
    good ol' US of A.. :-(
    I won't bother to request an apology for obvious reasons...
  19. John Fields

    John Fields Guest

    I'll retract nothing, and it wasn't a mis-statement. A welcome change
    from you would be something a little more substantial than footwork,
    but I guess you're afraid to try it 'cause you might get nailed again?
  20. keith

    keith Guest

    ....zipless ****. Though Miles' parents apparently mastered it.
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