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Multi-loop circuit

  1. Jan 18, 2015 #1
    Ω1. The problem statement, all variables and given/known data
    In the following circuit, what is the current across the R2 resistor?
    V1 = 1.5 V, V2 = 1.5 V, R1 = 470 Ω, R2 = 560 Ω

    333x1z5.png

    2. Relevant equations
    Kirchhoff's laws

    3. The attempt at a solution
    I attempted to create a system of equations with the intention of finding R2, as was demonstrated in an example in class. Here are my equations:

    Node A: I2 = I1 + I3
    Loop 1: 1.5 V - I1 * 470 Ω - I2 * 560 Ω = 0
    Loop 2: I2 * 560 Ω - 1.5 V = 0

    V1 - I1 * R1 - I2 * R2 = 0 ⇒I1 * R1 = V - I2 * R2 ⇒ (V1 - I2 * R2) / R1 = (1.5 V - I2 * 560 Ω) / 470 Ω = I1

    I've now expressed I1 in terms of I2. So I can write:

    I2 = [(1.5 V - I2 * 560 Ω) / 470 Ω] + I3

    However, I realized that I don't have enough equations to use to solve this system of equations for I2. In class we created a third loop that went around the whole outside. However, would that be possible in this case, as that isn't a path of current? And even if I were to add a third loop, I can't see how that'd help my cause.
     
    Last edited: Jan 19, 2015
  2. jcsd
  3. Jan 19, 2015 #2

    CWatters

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    Take a step back and inspect the circuit again. Don't write any equations just ask yourself...

    What is the voltage across R2?
    What is the voltage on both sides of R1?
     
  4. Jan 19, 2015 #3
    The voltage drop across R2 is I2 * 560 Ω.
    The voltage before R1 is 1.5 V and after is 1.5 V - I1 * 470 Ω.
    Right?
     
  5. Jan 19, 2015 #4

    CWatters

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    That's an equation. I said don't write any equations ;-)

    What does a voltage source like V2 do?
     
  6. Jan 19, 2015 #5
    It creates a potential difference of 1.5 V.
     
  7. Jan 19, 2015 #6

    CWatters

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    V2 is connected in parallel with R2 so what's the voltage across R2?
     
  8. Jan 19, 2015 #7
    -1.5 V ?
     
  9. Jan 19, 2015 #8

    CWatters

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    +1.5V eg same as V2

    Sorry - I have to go out for a few hours.
     
  10. Jan 19, 2015 #9

    CWatters

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    Back again.

    So what's the voltage on either side of R1?
     
  11. Jan 19, 2015 #10
    I think I may have solved it. I got I2 = 7.5 mA.
     
  12. Jan 19, 2015 #11

    CWatters

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    That's not what I make it. Can you show your working.
     
  13. Jan 19, 2015 #12
    Req1 = 470 Ω + 560 Ω = 1030 Ω
    I1 = 5 V / 1030 Ω = .00485 A
    I3 = 1.5 V / 560 Ω = .00268 A
    I2 = I1 + I3 = .00485 A + .00268 A = 7.5 mA
     
  14. Jan 19, 2015 #13

    CWatters

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    Ah no that's somewhat mixed up. You can't work out I1 like that because V2 dictates the voltage across R2.

    The first thing to notice about the circuit is that V2 is in parallel with R2. This means that the voltage across R2 must = V2 = 1.5V (as mentioned above). Then if the voltage across R2 is 1.5V the current through R2 must be what (according to Ohms law)?

    Edit: Where do you get 5V from? Problem statement says both V1 and V2 are 1.5V. Is that a typo?
     
  15. Jan 19, 2015 #14
    I2 = V / R = 1.5 V / 560 Ω = 2.7 mA

    How does the first loop work then? V1 adds 1.5 V and R2 drops 1.5 V. What about R1?

    A typo indeed. Nice catch.
     
  16. Jan 19, 2015 #15

    CWatters

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    Indeed, hence my earlier question.. So what's the voltage on either side of R1? or if you prefer..What's the voltage drop across R1?
     
  17. Jan 19, 2015 #16
    I = 1.5 V / (470 Ω + 560 Ω) = 1.5 V / 1030 Ω = 0.001456 A
    ΔV = 0.001456 A * 470 Ω = 680 mV
     
  18. Jan 19, 2015 #17

    CWatters

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    It's not as complicated. You were closer with..

    That implies both sides of R1 are at 1.5V and there is no voltage drop across R1. What does that mean for I1?
     
  19. Jan 19, 2015 #18
    Hmm. No current?
     
  20. Jan 19, 2015 #19

    CWatters

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    I forgot to confirm that this....
    is correct.
     
  21. Jan 19, 2015 #20

    CWatters

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    light bulb moment.jpg

    In effect we have just written KVL for the outside loop, solved it for i1 and found it to be zero.
     
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