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Calculating total resistance of a combination circuit

  1. May 15, 2017 #1
    combination circuit.png 1. The problem statement, all variables and given/known data
    Find the total resistance of the circuit
    2. Relevant equations
    Parallel resistance: 1/Rtotal = 1/R1 + 1/R2 + 1/R3 + ..., Series resistance: R1 + R2 + R3...
    3. The attempt at a solution
    The 1470 ohm resistor is parallel to the 50 ohm resistor and the 1000 ohm resistor, but there is a junction between them, which confuses me on how to add the 1470 ohm resistor to each of them... that probably didn't make sense but it's the best way I could word it. Once that is solved I know how to do the rest.
     
    Last edited: May 15, 2017
  2. jcsd
  3. May 15, 2017 #2
    Try adding a picture.
     
  4. May 15, 2017 #3
    (facepalm) oops... there it is
     
  5. May 15, 2017 #4

    vela

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    None of the resistors are in series or in parallel with any other.
     
  6. May 15, 2017 #5
    How would i go about adding up the resistances then?
     
  7. May 15, 2017 #6

    vela

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    Connect the combination to a voltage source V and calculate the current I it draws. The equivalent resistance is V/I.
     
  8. May 15, 2017 #7
    But to find the current that is drawn from the battery (6V) i need to know the total resistance of the circuit...
     
  9. May 15, 2017 #8

    vela

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    You're just going to have to write down equations using Kirchoff's laws and solve the resulting system of equations.
     
  10. May 15, 2017 #9

    scottdave

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  11. May 16, 2017 #10
    I would think that there is no point to this. You need the total resistance to calculate the current, then use I=V/R, and then go back to use R=V/I. That's just a lot of unnecessary extra steps that you would take after you find the total resistance.

    To solve this you would need to either use kirchoff's laws or kirchoff's laws with the delta y transform.
     
  12. May 16, 2017 #11

    scottdave

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    At first, I thought @vela was joking (like actually get a battery and some resistors.. Ha). But I think what was meant is to create a circuit on paper, then go through the loop analysis. Once you find the current delivered by the test voltage, then R = Vsource/Isource.
     
  13. May 17, 2017 #12

    vela

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    Really? There's no other way to find the current I? Like perhaps the way you mention below...

     
  14. May 17, 2017 #13
    Right, but to find the current that way wouldn't you have to find the total resistance first?
     
  15. May 17, 2017 #14

    vela

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    I can't help but feel you're trolling here.
     
  16. May 17, 2017 #15

    gneill

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    No. Solve the circuit, and you'll find all the currents even if you only really require one of them.
     
  17. May 17, 2017 #16
    Ah ok.
     
  18. May 17, 2017 #17

    Janus

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    One method is the loop-equation method. Draw out the possible current loops for the circuit, write an equation for each loop, and then solve the simultaneous equations to find each loop current. The sum of these give you the total current for any given source voltage, and from that you can determine the total resistance of the circuit. Here's the circuit redrawn with the loop currents marked:
    bridge.gif
     
  19. May 17, 2017 #18

    davenn

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    just out of personal curiosity for my learning

    can you do that with just the three loops you have done or should it really be for all loops ?
    as there is a 4th loop
     
  20. May 17, 2017 #19

    vela

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    You'd find that only three of the loop equations are independent. The equations from the other loops would be linear combinations of the three.
     
  21. May 17, 2017 #20

    ehild

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    Independent loops are needed, and there are three in this case. Assign current to each loops, so that there is one or two loop currents flowing through each resistor. If it is possible, the loops are separate. @davenn's construction is too complicated. See the next one:
    upload_2017-5-18_6-25-8.png

    The currents through the resistors are: I1, through 100Ω, I2, through 330Ω, I2-I1, through 220Ω, I2-I3, through 50Ω, and I3-I1, through 1000Ω. The generator current is I1. Anti-clockwise direction is positive. Write Kirchhoff's Loop Rule for each loops, that makes three equations for the three currents, eliminate I2 and I3, solve for I1. The total resistance is R=V/I1.
     
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