2nd battery confusing analysis

Discussion in 'Homework Help' started by dandeeny, Sep 16, 2014.

  1. dandeeny

    Thread Starter New Member

    Jul 28, 2014
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    This thread was originally posted to the wrong Forum. Perhaps it will get better feedback here.

    I'm a mature newcomer to electronics and regard All About Circuits as a tremendous resource. I've studied Vol 1, Chapters 1 to 7 and tried to put the knowledge to use by analysing an ohmeter circuit (the schematic hopefully, accompanies this help request) which was presented as part of the instructions manual for a multimeter construction project.

    I am at loss how to incorporate the second (9 volt) battery into the circuit simplification proceedure as described in All About Circuits, Chapter 7. The treatment there deals only with series/parallel resistors in its circuit analysis. This second battery only becomes involved when the resistance range selector switch is in the X 10000 position. The second battery is brought into play as a series connection with the 3v battery but separated from it by the 195k resistor (R18 in the attached schematic). This separation further complicates the issue for me. Its presence disallows treatment of the power source as a single 12v battery.

    Can anyone help me in analysing this circuit ?

    Dand
     
  2. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Here you have I redraw the diagram for you. Are you now able to analyze the circuit?
    4.PNG
     
    Last edited: Sep 16, 2014
  3. dandeeny

    Thread Starter New Member

    Jul 28, 2014
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    Many thanks for that Jony130. It certainly looks much simpler. I shall have to study it more closely and see if I can analyse it now. I can see immediately that I'm going to have a problem at the 9V-3V battery junction point and how to incorporate it into my analysis. I'm wondering how R18 and R41 can be combined into a single resistance. THIS IS GOING TO BOTHER ME. Thanks again Jony31. I will keep you informed.

    Dand
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Simply, some circuit cannot be solved by combined a component into a single resistance.
     
  5. dandeeny

    Thread Starter New Member

    Jul 28, 2014
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    Jony130,

    I've thought about your new schematic and to me (admittedly a complete newcomer) its not the same circuit as the original. You've repositioned R18 and changed its polarity. The core of my problem remains the interface of the two batteries and the voltage at this node. I'm not able to figure it out.

    Dand
     
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    This is why you need to learn how to use some more advanced analysis techniques.

    For example if I use a Thevenin's theorem I can find this equivalent circuit

    Rth = Rexternal + R18||R23 = Rexternal + 35.9k Ohm

    And Vth = 3V + 9V*R23/(R18 + R23) = 3V + 1.659V = 4.569V

    5.PNG
     
    Last edited: Sep 16, 2014
  7. dandeeny

    Thread Starter New Member

    Jul 28, 2014
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    Thanks again Jony130. I've moved on to studying Chapt 10 - Network analysis.
     
  8. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    I agree with Jony in that a couple source transformations will leave you with just one source to deal with.
    You can also use superposition as an alternate method.

    Here's a step by step procedure using the source transformation methods of Thevenin and Norton...


    Start by converting the 9v source and series 195k resistor
    into a current source and parallel resistor. The current
    source is ix=9/195000 amps and the parallel resistor is
    Rp=195000 ohms. Now we have a current source ix, a parallel
    resistor Rp, and these two are now in parallel with the 44k
    resistor. The parallel combination of Rp and 44k is:
    Rx=195000*44000/(195000+44000)=8580000/239 ohms.
    Now we have one current source ix and one resistor Rx
    in parallel. Converting this to a voltage source in series
    with a resistor, we have the voltage source:
    Vx=Rx*ix=396/239 volts and the resistor is Rx=8580000/239 ohms.

    So the network comprised of the 9v battery, the 195k resistor,
    and the 44k resistor can all be reduced to a single voltage source
    of 396/239 volts in series with a single resistor of 8580000/239 ohms.
    Replacing those three elements with the new voltage source and
    new resistor, we see that we are left with the two new elements in
    series with the 3v source and so now the new source and 3v source
    can be combined to form a single voltage source, and the new resistor
    can be combined with the External Resistor to form a single resistor.
    This leaves us with only one source in the circuit so it makes it
    easier to analyze.
     
  9. dandeeny

    Thread Starter New Member

    Jul 28, 2014
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    MrAl, many thanks for your comprehensive and very clearly presented treatise. However, it will require time and further reading of Chapt 10 for me to understand. I immediately have a problem with your conversion of the 9V source/series 195k resistor into a source and parallel resistor, using, it seems, Ohms law, the whole, apparently, divorced from the fact that there is a 3v source and other components connected further along the line...! Believe me, I shall study your presentation during my further reading of Chapt 10.

    Thanks again for your time and trouble.

    Dand
     
  10. MrAl

    Well-Known Member

    Jun 17, 2014
    2,440
    492
    Hi,

    It comes from one of two theories which read in simple form:

    1. A voltage source in series with a resistor is equivalent to a current source in parallel with a resistor.
    2. A current source in parallel with a resistor is equivalent to a voltage source in series with a resistor.

    So you see if we have either a voltage source in series with a resistor or a current source in parallel with a resistor we can immediately convert to the other form in order to simplify the network. In the case of your circuit, we were able to convert to a current source first and then back to a voltage source and then we were able to combine that with other elements so the whole thing got a lot simpler.

    Note that we dont have to know anything else about the network when we want to convert either, we just need knowledge about the source and resistor. If you want to you can call it "divorced circuit analysis" :)

    Example 1:
    A 5v source in series with a 5 ohm resistor converts to a 1 amp source in parallel with a 5 ohm resistor.
    Here the new source value is 5v/5ohms=1 amp.

    Example 2:
    A 2 amp current source in parallel with a 10 ohm resistor converts to a 20v source in series with a 10 ohm resistor.
    Here the new source value is 2 amps times 10 ohms=20 volts.

    So to convert the voltage source to a current source divide the voltage by the resistance.
    To convert the current source to a voltage source multiply the current times the resistance.
     
    dandeeny likes this.
  11. dandeeny

    Thread Starter New Member

    Jul 28, 2014
    22
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    MrAl - what I've just read is (to me) breathtakingly interesting. I just dont have the time to think about it immediately... But believe me, I will and will be back to you very soon.

    Many thanks.
    Dand
     
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