Equivalent Resistance Help

Discussion in 'General Electronics Chat' started by Aadrik, Jun 13, 2016.

  1. Aadrik

    Thread Starter New Member

    Dec 15, 2014
    4
    0
    Me and a friend are trying to brush up on our circuit skills and have run into a snag trying to find an equivalent resistance for a given circuit. I am not really looking for an answer as much as I am looking for the process to continue getting this circuit down to a single equivalent resistor. Thank you for any help you guys can provide pointing me in the right direction.
     
  2. AnalogKid

    Distinguished Member

    Aug 1, 2013
    4,544
    1,251
    The circuit is drawn badly to be intentionally confusing. Redraw it with R3 as the crossbar of an H. This should make more clear what is in parallel with what and what is in series with what. Two Thevenin equivalents and a little algebra, and you're there.

    ak
     
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,394
    1,606
    Funny you state the voltage but not the resistances...

    I don't see any simple way to reduce the circuit. I would use three current loops to get three equations and work it down from there.
     
  4. WBahn

    Moderator

    Mar 31, 2012
    17,760
    4,800
    I don't see any resistor that is in series or parallel with any other resistor.

    There are a number of ways to tackle it. One way (which is often the intended way for a question like this) is to use a delta-wye transform.

    Another way is to use nodal or mesh analysis to find the total current as a function of the applied voltage and then divide the two to get the equivalent resistance.

    One method that will NOT work (in general) is to use Thevenin equivalents since the equivalent circuit is NOT equivalent as far as anything inside the equivalent is concerned. This can easily be seen by imagining a sixth resistor of low value (relative to any of the other resistors) placed directly across the voltage source terminals. While this resistor would clearly dominate the equivalent resistance of the network as seen by the voltage source, it would have absolutely zero effect on the Thevenin equivalent circuit seen by any of the other resistors. You would need to do a Thevenin equivalent to find the current (and hence the power) in each of the resistors (treated separately as a load) and then use the total power plus the actual applied voltage to determine the total equivalent resistance.
     
    anhnha likes this.
  5. WBahn

    Moderator

    Mar 31, 2012
    17,760
    4,800
    Oh, by the way, this is a good case where setting some bounds can be very useful.

    Imagine that R3 is very large relative to the other resistors. If it were infinite (i.e., an open circuit), then you would have

    Req = (R1 + R2) || (R4 + R5)

    On the other hand, imagine it is very small relative to the other resistors. If it were zero (i.e., a short circuit), then you would have

    Req = (R1 || R4) + (R2 || R5)

    So you KNOW that the actual value has to be somewhere between the two. You also can probably get a feel for whether it should be closer to the upper or the lower bound depending on how big R3 is relative to the other resistors.

    Even if you don't have actual values, you can take your final symbolic result and let R3 go to infinity and see if you get the first equation and then let R3 go to zero and see if you get the second. If you pass both of these tests, then you are almost guaranteed that you have the right solution.
     
    ErnieM and atferrari like this.
  6. EM Fields

    Member

    Jun 8, 2016
    177
    30
    Resistor reduction 2.png
     
    Last edited: Jun 17, 2016
    Mohamed Bensalah likes this.
  7. WBahn

    Moderator

    Mar 31, 2012
    17,760
    4,800
    How do you figure that R1 and R2 are in series? Or that R4 and R5 are in series?

    Why do you think it is reasonable to assume that all resistors are equal value? The normal way of indicating that would be to just designate all five resistors with "R".
     
  8. Alec_t

    AAC Fanatic!

    Sep 17, 2013
    5,804
    1,105
    I think that's a typo. Swap R2 and R4.
     
  9. WBahn

    Moderator

    Mar 31, 2012
    17,760
    4,800
    I don't follow. Are you saying I made the typo, or EM Fields.

    His diagram shows R1 in series with R2 and it shows R4 in series with R5.

    In the original circuit, there are no two components that are in series. In order for two components to be in series, the node connecting them must ONLY connect those TWO components. Every single node in that circuit connects THREE components.
     
  10. EM Fields

    Member

    Jun 8, 2016
    177
    30
    They're obviously connected in series because they're connected head - to - tail instead of head -to - head and tail - to - tail.

    It's reasonable to assume that they're equal valued because the OP only posted reference designations. Were they to be of different values, then one would assume that those values would be posted by the OP.

    In the end, it makes no difference since if the procedure is followed, for the topology invoked, the single resistor solution will always out.
     
  11. EM Fields

    Member

    Jun 8, 2016
    177
    30
    Nonsense. Blow up the the OP's thumbnail to view the actual circuit.
     
  12. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,283
    326
    The very first diagram in your series of diagrams is not the same as the one given by the TS.

    You have R3 connected between the (junction of R1 and R4) and the (junction of R2 and R5). That's not how it is in post #1.

    Also, your diagram is finding the resistance as measured across R3, which is not what the TS must find.
     
    anhnha likes this.
  13. ErnieM

    AAC Fanatic!

    Apr 24, 2011
    7,394
    1,606
    I just noticed(after reading WBahns comment) that Wikipedia has this exact problem demonstrated in the article on Y-delta transformation.

    Sorry no link, can't make them on this device.
     
  14. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,283
    326
  15. EM Fields

    Member

    Jun 8, 2016
    177
    30
    OK, then, what's your solution?
     
  16. WBahn

    Moderator

    Mar 31, 2012
    17,760
    4,800
    Huh?

    Do you know what it means for two components to be in series?

    They are in series if and only if whatever current that flows in one MUST flow in the other.

    Wrong. Because only reference designators were provided it is only reasonable to assume that the values are unknown -- and that includes having no knowledge about any relationships between any of them.

    Consider any of the many very common schematic diagrams that often give only reference designators, such as a voltage divider or any of the classic simple op-amp circuits. It is not reasonable to assume that the different resistors are the same in any of those situations.

    But it can only work out if you actually work with the same circuit! You can't just go and take five resistors and hook them up in a completely different way and expect the result for your different circuit to be a valid solution to the original circuit.

    Here is the original circuit with the nodes highlighted:
    Edit_2016-06-13_1.png
    In order for R1 and R2 to be in series, then it MUST be true that whatever current is flowing in R1 would HAVE to then flow in R2. Look at the diagram! The current flowing in R1 is SPLIT between R2 and R3.

    Look at that blue node. It connects R1, R2, and R3. Now look at YOUR diagram. You have a node that connects JUST R1 and R2. Where is the connection to R3?
     
    Mohamed Bensalah likes this.
  17. EM Fields

    Member

    Jun 8, 2016
    177
    30
    Not true, in all cases.
     
  18. BR-549

    Well-Known Member

    Sep 22, 2013
    2,004
    394
  19. WBahn

    Moderator

    Mar 31, 2012
    17,760
    4,800
    How can you possibly claim that your diagram is the same as the one in the first post? You don't have a single node that matches the original circuit. Not one! Zilch, zero, nada!
     
  20. EM Fields

    Member

    Jun 8, 2016
    177
    30
    Actually, the left end of R3 is connected to the junction of R1 and R2 and its right end connected junction of R3 and R4 .
     
Loading...