Equivalent resistance as seen by the source

Discussion in 'Homework Help' started by SilverKing, Feb 3, 2014.

  1. SilverKing

    Thread Starter Member

    Feb 2, 2014
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    Hi everyone,


    The problem as shown in the attachment is to "Find the equivalent resistance as seen by the source VS (The resistance between a and b)"

    The line "as seen by the source VS" is disturbing :S

    So, is it only R1 or R1 and R8 and R7?
     
  2. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    Calculate including ALL the parallel and series resistor network values and reduce the resistance to a single value.
    Is this Homework, if yes, what do you calculate the equivalent resistance to be.?
    E
     
    Last edited: Feb 3, 2014
  3. LvW

    Active Member

    Jun 13, 2013
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    Silver King,
    I think, your start (first attempt) as scetched on the right figure looks good.
    However, two questions:
    1.) "The line "as seen by the source VS" is disturbing :S"
    What does this mean?

    2.) What is the meaning of your question in red (right figure)?
     
  4. SilverKing

    Thread Starter Member

    Feb 2, 2014
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    ericgibbs:
    I didn't get your question, but if you're asking about the equivalent resistance, I'd say (depending on what you said) that it's equal to [(R2+R3+R4//R5)+R6+(R1//R7//R8)]
    This is not a homework, but consider it like one

    LvW:
    1) I don't know if it's same when question says "Find the equivalent resistance" and "Find the equivalent resistance as seen by the source"

    2) My question in red says: "Can I assume those points a and b?"
     
  5. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi SK,
    Ref your equ: [(R2+R3+R4//R5)+R6+(R1//R7//R8)]

    Recheck the section marked in Red.
    E
     
  6. SilverKing

    Thread Starter Member

    Feb 2, 2014
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    ericgibbs:
    So basically when I face a question like doesn't change the fact that I've to calculate all the resistance in the circuit?

    and for R1//R7//R8, after second thought, it looks like that same current is passing through R8 and R7 which means that they're in series and their sum is in parallel with R1, am I right?
     
  7. studiot

    AAC Fanatic!

    Nov 9, 2007
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    This is not a problem it just works like this:

    Take any two nodes in a circuit and draw a line ab through them.

    The line divides the circuit into two halves, like a diameter divides a circle into two semicircles.

    The important point is the the two halves of the circuit only connect at nodes a and b, they do not connect anywhere else.

    For linear circuits there are three theorems all of similar form.
    These say we may replace one half of the circuit withone or two components, depending upon circumstance.

    Let us say we are going to replace the right hand side (RHS)

    1) If there are no sources of electricity (batteries generators etc) in that half then we may replace the entire half circuit with a single resistance (impedence).

    This is called the equivalent resistance looking from the left hand half of the circuit
    The formulae for parallel and series resistors are simple examples of this.


    2) If there are sources of electricity the we may replace the entire half circuit with a sinngle constant current source in parallel with a single resistance (impedence).

    This is called Norton's theorem.

    3) If there are sources of electricity we may replace the entire half with a single voltage source in series with a single resistance (impedence).

    This is called Thevenin's theorem.

    Note that we may do either of 2 or 3 so we have a way of providing
    an equivalent source
     
  8. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi SK,
    Have you arrived at a final solution for the equivalent resistance.? if so, what is its value.?
    E
     
    Last edited: Feb 3, 2014
  9. SilverKing

    Thread Starter Member

    Feb 2, 2014
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    studiot:
    First of all, I must say that this problem is under Series and Parallel resistances only, there should no use of any network theorem

    In our case, we have a voltage source, can I assume that the equivalent resistance that voltage source sees (between points a and b) are the resistors that have same volts as the source? the ones in parallel with the source?

    P.S. would you take a look at the attachment below? is the current division (which is upon it I calculate the total resistance) correct?

    ericgibbs:
    If my calculations in #6 are correct, then it would be equal to 50 ohms//R1=9.6 ohms
     
  10. studiot

    AAC Fanatic!

    Nov 9, 2007
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    (1) Is not the replacement of 2 parallel resistors by one equivalent what you are talking about? I just thought you might like a wider view since I don't know why you are doing this.

    (2) Definitely not, you need to take on board the first part of what I said even if you don't go as far as Norton & Thevenin.
    The voltage source 'sees' either the original arrangement of resistors or something equivalent.
    The something equivalent is any arrangement that will draw the same current from the voltage source.
    There are an infinite number of possible equivalent arrangements.
    The simplest one is a single resistor for which current is the same (by Ohm's law) as that drawn by the original arrangement.
     
  11. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi SK,
    I dont agree with your answer, but this may be due to me misreading the resistor values from your blurred drawing.
    Can you confirm the values?
    E
     
  12. WBahn

    Moderator

    Mar 31, 2012
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    Maybe this will help you get the basic concept involved.

    The source is hooked up to this mishmash of resistors. It connects to them at points 'a' and 'b' (the two terminals of the source). We can replace the mishmash with a single resistor and, if chosen properly, the source can't tell that anything has changed. As far as it's concerned, the single resistor and the mishmash are equivalent.

    However, this is NOT to say that that same resistor is always the equivalent resistance of that particular mishmash. Imagine removing that source from the circuit and instead putting a different source that is connected between a different pair of points. It's the same mishmash, but we can't replace it with the same resistor that we did before. We need to choose a different value in order to make it so that the new source can't tell the difference.

    While we may say things like, "that the source VC sees," what we really mean is more like, "that something connected between points a and b sees after removing the source VS." Both pieces are important -- knowing that we are talking about the source VC tells us which points we are interested it and it also tells us that we should find an equivalent that does NOT include the source VS itself.
     
  13. SilverKing

    Thread Starter Member

    Feb 2, 2014
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    ericgibbs:
    R1= 12 R2=16 R3=25 R4=50 R5=30 R6=7 R7=4 R8=16

    studiot, WBahn:
    After getting through what you said, I see that I'm starting to get the idea
    Thanks
     
  14. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi,
    This image shows the method I used, a couple of days ago, I misread the R2 value as 15R.
    It should be easy for you to substitute the correct value in the equations.

    Your answer should be 6R as opposed to mine of 5.98R

    Do you agree.?
    E
     
  15. The Electrician

    AAC Fanatic!

    Oct 9, 2007
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    You've got the right end of R6 connected to the right end of R1, but in the OP's original diagram the right end of R6 is connected to the junction of R7 and R8.
     
  16. ericgibbs

    AAC Fanatic!

    Jan 29, 2010
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    hi TE,
    Woops, I will relook at the OP's original image.

    Thanks for the heads up.;)

    E

    EDIT:

    I see it now, duh!!!
     
    Last edited: Feb 6, 2014
  17. JoeJester

    AAC Fanatic!

    Apr 26, 2005
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    Question for the OP:

    Do you think you would have trouble analyzing the question if the circuit looked like this diagram?
     
  18. SilverKing

    Thread Starter Member

    Feb 2, 2014
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    JoeJester:
    No, I don't think so.

    After getting through in the subject, I've seen that my problem was with identifying the nodes.
    Please take a look at the attachment, the nodes which connects only two elements are just called "nodes",aren't they?
    And the nodes which connects more than two elements are called "essential nodes", right?

    So, taking in my mind that any two elements connected by a "node" are in series, and any two elements have two essential nodes in common are in parallel; the only two elements that have a and b (two essential nodes) in common are the voltage source VS and and R1 (R1=12 ohms).


    How do you see my approach?
     
  19. JoeJester

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    Apr 26, 2005
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    There is nothing wrong with the way you did the node identification. Did you happen to notice that my drawing is your problem ... only redrawn for clarity?
     
  20. SilverKing

    Thread Starter Member

    Feb 2, 2014
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    Of course I did notice that, just wanna make sure if my approach is correct or not.

    So, If I have a question like this (to find Req), I'll take the nodes in priority, which results - in this case - not the combination of all the resistors in the circuit, but only between a and b, which is 12 ohm, right?
     
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