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Effect of the supply internal resistance

Discussion in 'Electronic Basics' started by Rich, Feb 7, 2009.

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  1. Rich

    Rich Guest

    I'm looking at voltage dividers. Trying to learn what I probably knew one

    Say you have a battery supplying 10v, and it has in internal resistance of 1

    You connect it to a voltage divider network R1 and R2. R1 is 90 Ohms in
    series with R2 which is 10 Ohms.

    Vcc connects to one end of R1, Ground connects to one end of R2.

    No load on the tappinp point.

    There is 10V from tap to GND. There is 90V to Vcc.

    Okay I get this. But the battery connects between Vcc and GND and that has
    in internal resistance of 1 Ohm.

    This makes it look like R1 and R2 are also connected together not only at
    the tap point, but at their other ends.

    Which of course produces a kind of obserdity because that makes irt look
    like there is the same voltage across both R1 and R2.

    Of course, as far as the battey is concerned, it's 10V across it does not
    short through it's own 1 ohm internal resistance.

    But anyway, when is that 1 Ohm internal resaistance going to be significant
    in any way? I know there will be a voltage drop across it when current flows
    and the battery terminal voltage will drop.

    But why exactly to you pretty much ignore that internal resistance when
    calculating voltages in a two resistor divider network?

    But I think in times past for some reason, because of low internal
    resistance of the supply, Vcc and GND have been taken in some way as being
    connected. I've forgotten now what that was.
  2. Rich

    Rich Guest

    1V & 9V. :c)
  3. Tom Biasi

    Tom Biasi Guest

    The battery's internal resistance is in series with its ideal voltage. The
    internal resistance is always part of the circuit.
    If the load resistance is much larger than the internal resistance then it
    won't matter much, if not , its another resistor in your circuit.
    Draw the circuit being fed with your supply voltage and a resistor in

  4. Rich

    Rich Guest

    | V out
    | |
    | |
    | |
    R2 R1
    | |
    | |
    |--------B-------- I R ------------------|
    X Y

    B= ideal battery with zero internal resistance.

    Lets say the battery voltage is 10.1V open circuit.
    1 Ohm Internal Resistence (I R).
    R1 = 90 Ohms, R2 = 10 Ohms.

    0.1 A of current flows.

    10V between X and Y.

    9V across R1. 1V across R2.

    0.1V across I R. (V = 1A * 0.1 Ohm = 0.1 V.)

    Wait, there is 10V across X and Y, yet if B represents an ideal battery with
    zero resistance, there should be 10V across I R!
  5. Rich

    Rich Guest

    The current is 0.1A a. So, using V= I * R voltage across I R is, 0.1A * 1
    Ohm = 1V.

    I know that the actual voltage across I R *is* 0.1V

    But looking it another way there is supposed to be 10V across X and Y. Which
    is effectively 10V across I R.
  6. Rich

    Rich Guest

    I think I've figured it.

    Youve got to draw a center line down the circuit.


    Looking at the above circuit, because B is ideal battery and zero
    resistance, it looks like R1 and R2 are practically in parallel because I R
    (Internal resistance), is 1 Ohm.

    But you must look at the left hand side of IR and then the right hand side.

    Add up the voltageson either side and see what IR sees (Z & Y).

    | V out
    | -
    R1 9V
    | +
    I R ------------------|

    | V out
    | +
    R2 1V
    | -
    |--------B-------- I R
    X - + Z
    10.1 V

    We see that there are two voltages as far as I R is concerned. And they are
    opposing each other.

    On the right, I R sees + 9V

    On the left, I R sees 10.1V - 1V = 9.1V.

    So, there is *not* 10V across I R. There is 9.1V - 9V = 0.1V

    So, there is 0.1V across X and Y. There is 1 Ohm between these points. Yet
    *because of battery B* you cannot say that R1 and R2 are practically in
    parallel. Despite the low impedance of the suply source.

    And I think this sort of stuff is going to explain Op Amps.
  7. Rich

    Rich Guest

    I think I've figured it.
    In fact there is something special about point Vout here. It's like a
    reference point. Like a ground.
  8. Ned

    Ned Guest

    Here is the way I learned to draw circuits:
    And simplify them in College:

    R(int)=1 ohm
    +| |
    --- |
    | B | \
    |10v| R1 / 90 ohm
    |___| \
    | /
    -| |_____________ V(out)
    | |
    | \ /\
    | / |
    | R2 \ 10 ohm |
    | / |
    | | \/
    |__________________|____________ 0 volts

    V(out) = (output / input) * V(source)

    output is I * R2
    Input is I *[(int) + R1 + R2]

    then V(out) = (10 ohms / 101 ohms) * 10 volts
    v(out) = 0.9901 volts

    Your equivelent source impeadance will be R2 parallel [R1 + R(int)]

    R eq = product / sum
    = 10 ohms * 91 ohms / (101 ohms)
    = 9.01 ohms

    There: at V(out) the voltage will be 0.9901 volts with an equivelent source
    resistance of 9.01 ohms

  9. Shaun

    Shaun Guest

    That's some wierd email address, and if I click it, Outlook express pops up.

    link didn't work

  10. Rich

    Rich Guest

    I'm slightly wong.

    There is a point along the circuit where positive meets negative.

    Going one way is like going into negative, going the other way like being
    into positive. Of course anywhere along the circuit one part can be
    considered more positive, or negative relative to another.

    In the circuit above the center point, which is kind of a reference point is
    within R1. It's 40.5 Ohms into R1 on one side. Other side is 40.5 Ohms.
    Adding resistances from the center point either way adds up to 50.5 Ohms, so
    total is 101 Ohms. True center point, which I suppose is a reference point
    is total resistance divided by 2.

    So, in my circuit Vout, is not the center point.

    And the voltage across I R is + 5.05V on one side, + 4.95V on the other, so
    there is 0.1V across I R (internal resistance of battery).

    Yet between point X and Y, you would measure 10V, and for a moment it looks
    there is 10V across I R, without closer analysis.
  11. Rich

    Rich Guest

    40.5 Ohms and 49.5 Ohms.
  12. Shaun

    Shaun Guest

    The supply internal resistance will give you a voltage drop, lowering the
    source voltage available at the battery that depends on the current you
    voltage divider and load require. V(battery) = V(10 volts) - I *
    R(internal). I is the total circuit current with your voltage divider and
    whatever is connected to it. V(10 volts) is your open circuit voltage.

  13. Shaun

    Shaun Guest

    NO! NO! NO!

    You have the battery and R(internal), This is your source. Then you have
    your voltage divider and you want the voltage across the 10 ohm resistor.
    See Neds description of how to work out the circuit, it is the proper method
    that is taught in schools and books. You don't use an imaginary reference
    point, you're over complicating the circuit, the only reference point to use
    is battery neg (common).

  14. Rich

    Rich Guest

    One think I'm shaky about is this concept of reference and ground.
    Especially when I try to grasp Op Amps.

    Of course, I know that there is 0.1V across 1R (the internal resistance) not
    10 Volts.

    10.1 Volts at Ebat is referenced to ground. We can draw 1R as connected to
    that point. Of course that one side of 1R is 10.1V above ground is not that
    significant, if it is above ground is a fact that does not assist with the
    understanding of how to calculate the voltage across 1R.

    What is significant is the voltage across 1R. In fact what is significant is
    what 1R sees as being it's source.The following circuit shows what that
    there across the interal resistance of the battery. It's usefullness is in
    showing more clearly how only 0.1V PD apears across 1R. It ignores any kind
    of reference point to ground, although in practice you may need to know the
    voltage across 1R and ground.

    +---------+--E2 = 0.1V
    | |
    [10R] R2 |
    | |
    | |
    [90R] R1 |
    | [1R]
    +--Ebat |
    |+ |
    [BAT] |
    | |
    | |
    | |
  15. Rich

    Rich Guest

    BAT is suppossed to be a device with zero resistance. So, you can
    erroniously conclude there is 10V across 1R, thinking one part of 1R is at
    ground. The voltage stated is a voltage *reference to ground*. One end of 1R
    is + 10V above ground or the negative terminal that's for sure. But in fact
    the other side of 1R is *not* connected to ground at all through a zero
    resistance, so there is not + 10V across R1. This shows how careful one must
    be when associating a zero resaistance value to some ideal component. I must
    be careful in doing that.

    +-----------+--- 10V
    | |
    | [R1]
    [1R| |
    | |
    | |
    [BAT] [R2]
    | |
    | |
  16. Rich

    Rich Guest

    But, when I drew the circuit, BAT and 1R is an equivalent circuit for a
    battery. Consisting of a perfect battery (BAT) and an internal resistance

    Face value you want to say, hey if BAT is a perfect battery with zero
    resistance, there's 10V across 1R. Of course that is not true. It's just one
    of these issues you get when dealing with perfect components in series with
    resistance. 10v is just a statement that that point is 10V above a refrence
    point. And there really is not a zero resistance across BAT. It's mixing a
    fiction with reality. The equivalent circuit of a batttery is a fiction.
  17. Rich

    Rich Guest

    Really, the issue has been what to make of that fact that in the circuit as
    drawn, R1 is at a potential of + 10v and it's other side is "seemingly*
    connected through BAT to GND.

    That can cause you to wonder how to explain why there is not really a 10V
    potential across R1. Not that one is unable to calculate the actual voltage
    across R1, which I can.

    Not really, just trying to see where the error would be in thinking there
    was 10V across R1. That I don't think would be an uncommon thing to do.
    Yes, I've noticed that in my ham radio experience.
  18. Jon Kirwan

    Jon Kirwan Guest

    Shaun, if you are using a newsgroup aware program (as I do) it should
    regonize the address as a newsgroup (usenet) ID. All I need to do is
    double click it and a search takes place looking for the article, for
    example. First within my existing local database, then next online
    with my newsgroup server via NNTP (one of many protocols, this one
    called "network news transfer protocol," if memory serves.) If found
    online, the article is downloaded and stored locally, automatically,
    and then displayed for me. (It's possible that the server I use no
    longer stores the article, though.)

    For example, the message ID for your own message (the one I'm
    responding to) was found by looking at the message headers for your
    message as:

    Message-ID: <MSljl.14757$>

    Every newsgroup message that gets posted, I think, has one of these
    fields present in the headers. I don't know how it is generated or
    who does the generation such that they are unique, but there must be
    an RFC (request for comment), or more than one, that documents it.

    If I double-click the message ID above, for example, your message
    immediately shows in my newsgroup browser -- flipping me away from
    this message I'm composing. Then I just click back to the tab for
    this message and continue writing.

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