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3 batteries, 5 resistors using KCL & KVL

  1. Mar 19, 2012 #1
    1. The problem statement, all variables and given/known data

    1. Using Kirchhoff’s Laws, find all the currents flowing in the circuit.

    2. Hence, determine the voltage across each resistor and show that all loops comply with KVL


    2. Relevant equations

    KVL and KCL



    3. The attempt at a solution

    Please see the pictures attached
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Mar 19, 2012 #2

    gneill

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    Staff: Mentor

    Hi amy1983, Welcome to Physics Forums.

    For your Loop 2 equation be careful about the orientation of the voltage sources; as you go around the loop in the direction of mesh current I2, both give potential rises.

    Take another look at your Loop 3 equation. Again, pay attention to the voltage source polarity with respect to the current loop direction. Also check that you haven't inadvertently counted some voltage drops twice...
     
  4. Mar 19, 2012 #3
    It's a bit confusing because you've taken clockwise mesh currents as negative. Also that loop 2 is not correct. As Gneill said, you've made an error with values of the voltage sources in that loop.
     
  5. Mar 19, 2012 #4
    A good program to use for Nodal Analysis and Mesh Analysis is Pspice, for these kinds of circuits your answers can be checked very quickly.
     
  6. Mar 19, 2012 #5
    Hi gneill / NewtonianAlch - thank you kindly for your welcome and for your suggestions, I will review this PM.
     
  7. Mar 21, 2012 #6
    Hi g/neill / NewtonianAlch, I typed it all up and came up with the following according to your suggestions:

    At node V_1,
    I_1+I_2-I_3=0
    ⇒I_1+I_2=I_3
    ⇒(15-V_1)/(1 k)+(10-V_1)/(2 k)=(V_1-V_2)/(1 k)
    ⇒(30-2V_1 )+(10-V_1 )=(2V_1-2V_2)
    ⇒40-3V_1=2V_1-2V_2
    ⇒5V_1-2V_2=40
    At node〖 V〗_2,
    I_3-I_4-I_5=0
    ⇒I_3=I_4+I_5
    ⇒(V_1-V_2)/(1 k)=(5-V_2)/(1 k)+V_2/(2 k)
    ⇒2(V_1-V_2 )=(10-2V_2 )+V_2
    ⇒2V_1-V_2=10
    Solving the two equations;
    V_1=-20V &V_2=30V
    I_1=(15-V_1)/(1 k)
    ⇒I_1=(15-(-20))/(1 k)
    ⇒I_1=35/(1 k)
    ⇒I_1=35mA
    I_2=(10-V_1)/(2 k)
    ⇒I_2=(10-(-20))/(2 k)
    ⇒I_2=30/(2 k)
    ⇒I_2=15mA
    I_3=(V_1-V_2)/(1 k)
    ⇒I_3=((-20)-30)/(1 k)
    ⇒I_3=(-50)/(1 k)
    ⇒I_3=-50mA
    I_4=(5-V_2)/(1 k)
    ⇒I_4=(5-30)/(1 k)
    ⇒I_4=(-25)/(1 k)
    ⇒I_4=-25mA
    I_5=V_2/(2 k)
    ⇒I_5=30/(2 k)
    ⇒I_5=15mA




    Using KVL,
    E_1-E_(R_1k )+E_(R_2k )-E_2=0
    Where,
    E_(R_1k )=I_1×R_1k
    〖⇒E〗_(R_1k )=35×〖10〗^(-3)×1×〖10〗^3
    〖⇒E〗_(R_1k )=35V

    E_(R_2k )=I_2×R_2k
    〖⇒E〗_(R_2k )=15×〖10〗^(-3)×2×〖10〗^3
    〖⇒E〗_(R_2k )=30V
    ⇒E_1-E_(R_1k )+E_(R_2k )-E_2=0
    ⇒15-35+30-10=0
     
  8. Mar 21, 2012 #7

    gneill

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    Staff: Mentor

    Okay, looks good for Node V1.
    Oops. Check the polarity of the E3 source. Is it going to make the potential difference across R4 larger or smaller?
     
  9. Mar 21, 2012 #8
    At node V_1,
    I_1+I_2-I_3=0
    ⇒I_1+I_2=I_3
    ⇒(15-V_1)/(1 k)+(10-V_1)/(2 k)=(V_1-V_2)/(1 k)
    ⇒(30-2V_1 )+(10-V_1 )=(2V_1-2V_2)
    ⇒40-3V_1=2V_1-2V_2
    ⇒5V_1-2V_2=40
    At node〖 V〗_2,
    I_3-I_4-I_5=0
    ⇒I_3=I_4+I_5
    ⇒(V_1-V_2)/(1 k)=(5+V_2)/(1 k)+V_2/(2 k)
    ⇒2(V_1-V_2 )=(10+2V_2 )+V_2
    ⇒2V_1-2V_2=10
    Solving the two equations;
    V_1=-20V &V_2=30V
    I_1=(15-V_1)/(1 k)
    ⇒I_1=(15-(-20))/(1 k)
    ⇒I_1=35/(1 k)
    ⇒I_1=35mA
    I_2=(10-V_1)/(2 k)
    ⇒I_2=(10-(-20))/(2 k)
    ⇒I_2=30/(2 k)
    ⇒I_2=15mA
    I_3=(V_1-V_2)/(1 k)
    ⇒I_3=((-20)-30)/(1 k)
    ⇒I_3=(-50)/(1 k)
    ⇒I_3=-50mA
    I_4=(5+V_2)/(1 k)
    ⇒I_4=(5+30)/(1 k)
    ⇒I_4=35/(1 k)
    ⇒I_4=35mA
    I_5=V_2/(2 k)
    ⇒I_5=30/(2 k)
    ⇒I_5=15mA

    Using KVL,
    E_1-E_(R_1k )+E_(R_2k )-E_2=0
    Where,
    E_(R_1k )=I_1×R_1k
    〖⇒E〗_(R_1k )=35×〖10〗^(-3)×1×〖10〗^3
    〖⇒E〗_(R_1k )=35V

    E_(R_2k )=I_2×R_2k
    〖⇒E〗_(R_2k )=15×〖10〗^(-3)×2×〖10〗^3
    〖⇒E〗_(R_2k )=30V
    ⇒E_1-E_(R_1k )+E_(R_2k )-E_2=0
    ⇒15-35+30-10=0



    Hi gneill, Thank you for your advice. Does the remainder of the equation look okay?
     
  10. Mar 21, 2012 #9

    gneill

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    Staff: Mentor

    Oops again. I think you added the V_2's from the right hand side to the left hand side rather than subtracting them from both sides. But you're getting there...
     
  11. Mar 21, 2012 #10
    I_3-I_4-I_5=0
    ⇒I_3=I_4+I_5
    ⇒(V_1-V_2)/(1 k)=(5+V_2)/(1 k)+V_2/(2 k)
    ⇒2(V_1-V_2 )=(10+2V_2 )+V_2
    ⇒2V_1-2V_2=10V_2+2V_2
    ⇒2V_1=10V_2
    ⇒2V_1-V_2=10

    I know it's bad, but I am finding this electrical element difficult.
    :/ Worst case senario if I put up what I have at least it shows I made the effort!
     
  12. Mar 21, 2012 #11

    gneill

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    Staff: Mentor

    You seem to be having difficulty expanding and then collecting terms. On the LHS you've got: 2*V1 - 2*V2. On the RHS you've got: 10 + 2*V2 + V2 = 10 + 3*V2. So:

    2*V1 - 2*V2 = 10 + 3*V2

    Move the 3*V2 from the RHS to the LHS by subtracting 3*V2 from both sides. Then proceed.
    I know it can be tricky, and much of circuit analysis involves a good deal of finicky algebra. But you are indeed showing good effort and insight for what you need to do to get the result. As they say, the devil is in the details :smile:
     
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