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DC Networks and Kirchoff's Laws

  1. Nov 27, 2011 #1
    1. The problem statement, all variables and given/known data

    Calculate the value of the current through the 12V battery shown in figure (attached).

    2. Relevant equations

    V = IR
    I1 = I2 + I3

    3. The attempt at a solution

    I've calculated the current for the left hand loop, see below:

    Loop 1:
    E_1- E_2- V_1- V_2=0
    E_1- E_2= V_1+ V_2
    E_1- E_2=I(R_1+ R_2)
    13-12= I (3+1)
    1=4I
    1/4=I
    I=0.25A

    I understand it's a process of doing the same to the right hand side and then subtracting the two sides? I'm just unsure as to when i should be adding the voltages and resistors and when i should be subtracting them.Is it related to the direction of the current and voltage potential? If anyone can help here it would be greatly appreciated. Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Nov 27, 2011 #2

    gneill

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    Staff: Mentor

    You have two loops which will influence each other in terms of voltage and current. So a single equation for one loop isn't going to capture the interplay.

    What circuit analysis methods have you learned?
     
  4. Nov 27, 2011 #3
    gneill, thanks for your help.

    We've been shown the branch current method.
     
  5. Nov 27, 2011 #4

    gneill

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    Staff: Mentor

    Okay. It looked as though you were attempting to apply Kirchhoff's Voltage Law (KVL) around a loop, but branch currents are good too! That boils down to Kirchhoff's Current Law (KCL), which states that the sum of the currents entering a given node of the circuit must equal the sum of the currents leaving that node.

    Your circuit has three branches and two nodes. For the nodes, label the bottom horizontal conductor as "ground". Label the top horizontal conductor "Node A".

    attachment.php?attachmentid=41323&stc=1&d=1322454721.jpg

    If you happened to known the voltage at node A (with respect to ground), VA, then you could calculate the current through each of the branches. For the leftmost branch, for example, you 'd have I = (VA - 13V)/R1. So how to determine VA? That's where Kirchhoff's Law comes in.

    Assume that there's a voltage VA at node A. Write the equation for the sum of the currents leaving Node A for each branch connected to that node. I gave you the example for the first branch. Solve the resulting equation for VA. Once you know that voltage you can determine the current in each of the branches (including the one you're interested in).
     

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    • Fig1.jpg
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  6. Nov 28, 2011 #5
    Thanks again for your reply.

    For the left hand loop I have:

    +E1 - I3R3-E3-I1R1 = 0
    19 - 3I3 - 12 - I1 = 0

    -3I3 - I1 = 7
    I2 = I1 - I3


    I1 - I2 = 7

    For the right hand loop I have:

    +E2 - I3R3 - E3 - I2R2 = 0
    14 - 3I3 - 12 - 2I2 = 0

    -3I3 - 2I2 = 2
    I1 = I2 + I3

    I1 - 3I2 = 2

    Multiply left hand side by four and add to right hand side ...

    4I1 - 4I2 = 28

    Left hand loop plus right hand loop ...

    5I1 - 7I2 = 30

    How can I know calculate the value of I1 and I2?
     
  7. Nov 28, 2011 #6

    gneill

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    Where did the 19 come from? Doesn't E1 = 13V?
    After you straighten out the voltage of E1, the first two equations that you wrote in bold constitute two equations in two unknowns. You can solve for both I1 and I2 with them. For example, if you subtract one equation from the other you should eliminate I1 and be left with an equation for I2.
     
  8. Nov 28, 2011 #7
    Thanks for that, didn't even spot that.
    So ...

    I1 - I2 = 1

    (I1 - I2) - (I1 - 3I2) = 1 - 2

    2I2 = -1

    I2 = -1 / 2 = -0.5A

    Therefore if I1 - I2 = 1, I1 - -0.5 = 1

    I1 = 0.5

    If I1 = I2 + I3, 0.5 = -0.5 + 1

    So current at E3 would be 1Amp
     
  9. Nov 28, 2011 #8

    gneill

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    One problem I just spotted (should have caught it sooner) is that, according to your assumed current directions you should have I3 = I1 + I2. This is going to change your result, since you've written I1 = I2 + I3.

    Here's the figure again with the assumed currents:

    attachment.php?attachmentid=41345&stc=1&d=1322543216.jpg
     

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  10. Nov 28, 2011 #9
    Ok thanks for seeing that. So,

    If I2 = -0.5A

    I1 - I2 = 1

    Therefore I1 = 0.5

    I3 = I1 + I2

    I3 = 0.5 + -0.5

    I3 = 0A

    Current at E3 would be 0A?
     
  11. Nov 28, 2011 #10

    gneill

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    You'll have to go back and rework the loop equations where you've used the current sum relationship. The current through E3 is not zero.
     
  12. Nov 29, 2011 #11
    Thanks for all your help to date. I've come up with a final figure of 0.95A for I3. Does this seem right?
     
  13. Nov 29, 2011 #12

    gneill

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    Unfortunately, no :cry: Something must have gone awry in the math.

    How about one more attempt, from the beginning? I'll get things started, using the equations that you've written before:

    attachment.php?attachmentid=41345&stc=1&d=1322543216.jpg

    KCL for top node: I3 = I1 + I2 ............(1)

    KVL for left-hand loop:

    13 - 3*I3 -12 - 1*I1 = 0

    Replacing I3 using (1):
    4*I1 + 3*I2 = 1 ............(2)

    KVL for right-hand loop:

    14 - 3*I3 - 12 - 2*I2 = 0

    and using (1):
    3*I1 + 5*I2 = 2 ............(3)

    You should be able to solve (2) and (3) for I1 and I2, then use (1) to find I3.





    I
     
  14. Nov 29, 2011 #13
    Thanks for all your help.

    So ...

    4I1 + 3I2 = 1 1
    3I1 + 5I2 = 2 2

    Multiply 1 by 3
    12I1 + 9I2 = 3

    Multiply 2 by 4
    12I1 + 20I2 = 8


    (12I1 + 9I2) - (12I1 + 20I2) = 3 - 8
    -11I2 = -5
    I2 = -5 / 11
    I2 = 0.45


    4I1 + 3I2 = 1
    4I1 + 3x0.45 = 1
    4I1 + 1.35 = 1
    4I1 = 1 - 1.35
    = -0.35
    I1 = -0.35 / 4
    I1 = -0.0875


    I3 = I1 + I2
    = -0.0875 + 0.45
    = 0.3625
    = 0.4A?
     
  15. Nov 29, 2011 #14

    gneill

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    Your method is good. You might want to keep a few more decimal places in intermediate values to prevent rounding errors from creeping into the final result.

    Bravo.
     
  16. Nov 29, 2011 #15
    Brilliant thanks for all your help.
     
  17. Jan 13, 2012 #16
    I think your I2 should be
    -0.4545 which changes your I1 to to 0.8409 giving I3 = 0.39 A.
    This would affect your answer when working out the power dissipated on each resistor wouldn't it?
     
  18. Mar 22, 2012 #17
    If i wanted to work out the power dissipated in each resistor. would i use P=I^2*R or P=I*V ??
     
  19. Mar 22, 2012 #18

    gneill

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    Yes :smile:

    Both are applicable depending upon what quantities you have to work with. There's also V2/R that'll work if you happen to have the voltage across the resistor.
     
  20. Mar 22, 2012 #19
    ok so if i use P=I2R

    with the folowing values for I:

    I1 = -0.0875
    I2 = 0.45
    I3 = 0.4

    i get:

    PR1 = -0.08752 * 1 = -0.00766 Watts

    PR2 = 0.452 * 2 = 0.405 Watts

    PR3 = 0.42 * 3 = 0.48 watts


    but if i use the other formula P=I*V

    i get:

    PR1 = 0.0875 * 13 = 1.1375 watts

    PR2 = 0.45 * 14 = 6.3 watts

    PR3 = 0.4 * 12 = 4.8 watts

    which is correct?
     
  21. Mar 22, 2012 #20

    gneill

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    As was pointed out by charger9198 in a previous post, there seems to be some issue with the calculated values for the currents. I've just taken a look at the circuit and I can see that the values are not those given above :frown:

    You should recalculate the currents before moving forward.
    Well the second set is definitely incorrect because the voltage you need to use is the potential that appears across each resistance. This will NOT be the values of the individual cells in the branches; it will be the voltage drop due to the current passing through a given resistor.
     
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