Need help with KCL and KVL

Discussion in 'Homework Help' started by BiscuitDuke, Apr 23, 2014.

  1. BiscuitDuke

    Thread Starter New Member

    Apr 23, 2014
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    Hey I need help using KCL and KVL for this circuit. I was able to solve it using nodal analysis but using loops on this one is pretty difficult. Is there a way to get rid of that 100kohm or will I have to account for that loop? Would it be useful to have a loop that goes around the outside of the circuit? Thanks for your help
     
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  2. studiot

    AAC Fanatic!

    Nov 9, 2007
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    4 unknown voltage drops, 8 loops to choose from is that not enough?

    What have you done so far?
     
  3. BiscuitDuke

    Thread Starter New Member

    Apr 23, 2014
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    Loop 1 with R1 & R3:
    V1 + R1*I1 + R3*I3 = 0

    Loop 2 with R2 and R3
    V2 + R2*I2 + R3*I3 = 0

    Loop 3 with R1 R2 R4
    R4*I4 - R1*I1 - R2*I2

    I1+I2 = I3

    I started solving this and it got out of hand so I figured I did something wrong
     
  4. d3ck

    New Member

    Jul 14, 2013
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  5. studiot

    AAC Fanatic!

    Nov 9, 2007
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    d3ck

    Please be aware that Maxwells mesh current method is not the same as KVL.

    biscuitduke

    so you have 3 equations and 4 unknowns, what about the other 5 I mentioned?
     
  6. d3ck

    New Member

    Jul 14, 2013
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    When I saw the circuit, the first thing comes in my mind to solve this is by using mesh analysis.
    Thanks for correcting me studiot.
     
  7. BiscuitDuke

    Thread Starter New Member

    Apr 23, 2014
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    Maybe I'm not seeing these 8 loops. I just see 5 loops.

    loop 1 with V1 R1 and R3
    loop 2 with V2 R2 and R3
    Loop 3 with V1 R1 R2 and V2
    loop 4 with R1 R2 and R4
    loop 5 with V1 V2 and R4.

    Am i missing any?

    My main problem comes with the 100kohm. would I have to do something like R1(I1 - I4) if the loops i choose go in opposite direction?
     
  8. WBahn

    Moderator

    Mar 31, 2012
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    I only see seven loops. If I start with the meshes and call them M1, M2, and M3, then I see:

    L1 = M1
    L2 = M2
    L3 = M3
    L4 = M1+M2
    L5 = M1+M3
    L6 = M2+M3
    L7 = M1+M2+M3

    Did you miscount (easy to do) or did I miss one (easy to do)?
     
  9. WBahn

    Moderator

    Mar 31, 2012
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    Part of the problem is that you aren't clearly defining your currents. What I I2? It may be reasonable for me to infer that it is the current through R2, but I have no idea what direction I2 is flowing through R2 and it is unreasonable to expect us to reverse engineer your equations to figure out what you intended. Instead, annotate your drawing with all of the currents, using arrows to indicate the direction that current is flowing is that current is positive.

    This will also help you keep things straight because then you don't have to wonder if you add or subtract a term because you know that the voltage across the resistor is (symbolically) positive on the side that the current is entering.
     
  10. studiot

    AAC Fanatic!

    Nov 9, 2007
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    I have drawn out the loops in attachment1.
    WBahn was correct (as usual) there are only 7.
    I realised when drawing that my loop 8 was just my loop 6 in reverse.

    WBahn was also correct to properly label your diagram.

    I have done this in attachment2.

    As you can see there are four nodes so there are four equations and six unknowns.
    But because D is not a proper node, they are not independent.

    So you need some results from KVL and or ohms law to complete the set.
     
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