kirchoffs law help

Discussion in 'Homework Help' started by tom6123, Jun 26, 2014.

  1. tom6123

    Thread Starter New Member

    May 31, 2014
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    hello, ive been trying to figure out how to do kirchoffs law.
    i havent got far into it but i keep getting lost here is what i have so far


    R1=10 Ohms
    R2=10 Ohms
    R3= 75 Ohms
    R4 = 33 Ohms
    R5=47 Ohms
    V1= 5v
    V2 = 21V

    what i got so far is:
    Us1= I1(10+33+75)+I2(75)
    US1= I1 118 + I2 75

    Us2=I2 (10+47+75)+ I1 (75)
    Us2=I2 132 +I1 75

    5v=I1 118 + I2 75 (X132)
    21V= I2 132 +I1 75 (X75)
    -------------------------
    660=I1 23760 +I2 9900
    1575= I1 5625 +I2 9900
    -------------------------
    -915 + I1 18135 -915/18135 = -50.5 mA



    5v=I1 118 + I2 75 (X75)
    21V= I2 132 +I1 75 (x180)
    ---------------------------
    375= i1 13500+I2 5625
    3780=I1 13500+ I2 23760
    -----------------------------
    -3405=-18135 -3405/-18135= 187.8mA



    Am i on the right tracks because i think im missing something
     
  2. tshuck

    Well-Known Member

    Oct 18, 2012
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    yes, a schematic would help as listing what resistors there are doesn't tell us how they are connected...

    Kirchhoff has two laws associates with his name: KCL & KVL...
     
  3. tom6123

    Thread Starter New Member

    May 31, 2014
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    [​IMG]
    I dont know if its properally showing the image

    [​IMG]
     
  4. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Ok.

    Step 1.
    Schematic.
    Got it.

    Step 2.
    Part values.
    Got it.

    Step 3.
    What are we trying to find?
     
  5. tom6123

    Thread Starter New Member

    May 31, 2014
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    im trying to find the
    Current through R1 UR1=I1(R1)
    Current through R2 UR2=I2(R2)
    Current through R3 UR3=(I1+I1)+R3
    Current through R4 UR4=I1(R4)
    Current through R5 UR5=I2(R5)
    Voltage through R3
    Power Dissipated in R3

    And i dont know the formulae for the last two.but i dont know if my working out is correct right now.
    i think the power dissipated in r3 is PR3=(I1+I2)UR3
     
  6. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Voltage is always across.
    Current is through.
    So.
    Voltage ACROSS R3 is equal to current THROUGH R3 times resistance of R3:
    V=I*R


    Power at R3 is easy.

    Option 1.
    Voltage across R3 times current through R3.
    P=V*I

    Option 2.
    Voltage across R3 squared divided by R3.
    P=(V^2)/R

    Option 3.
    Current though R3 squared times R3.
    P=(I^2)*R
     
    Last edited: Jun 26, 2014
  7. tom6123

    Thread Starter New Member

    May 31, 2014
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    so are these formulas ( UR1=I1(R1) ) incorrect because the formula is finding the value of the voltage instead of the current
     
  8. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Ok. I am looking at it. If I remember right, Mesh Current Method is KVL. So I solved the problem using Mesh Current Method.

    What you have there is setup for Mesh Current Method, but you made two mistakes. One mistake you made twice, third mistake is the sign of voltages.

    Also there is some weird (X132) and (X75). I have no idea what those are so I did not even count them as mistakes, they are just too weird.
     
  9. tom6123

    Thread Starter New Member

    May 31, 2014
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    the (X 132) and (X75) is where i was using simultaneous equations to solve the current
     
  10. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    It depends.

    If you need voltages, you use V=I*R.

    If you need currents, you use I=V/R.
     
  11. tom6123

    Thread Starter New Member

    May 31, 2014
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    does the simultaneous equations look right
     
  12. shteii01

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    Feb 19, 2010
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    No. They are not.

    Give my a couple of minutes and I will upload schematic with the currents shown. That will show you the error that you made twice when you were setting up the equations.
     
  13. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    Here is your circuit.

    [​IMG]



    The error that you did twice is the voltage across R3. We assume that all currents move clockwise. I1 moves clockwise, I2 move clockwise. But that means that they enter R3 from opposite directions. So the current though R3 is the difference between the two currents.

    Your second error was the sign that you assigned to the voltage sources when you put them into the equation. Notice how I1 enters battery V1 from the negative terminal so you have:
    Voltage across R1+Voltage across R3+Voltage across R4 - 5=0
    Voltage across R1+Voltage across R3+Voltage across R4 = 5 volts
    Here you got the equation right, the sum of voltages is equal to +5 volts.
    But the next equation you got wrong. Notice that I2 enters battery V2 at the positive terminal:
    Voltage across R5 + Voltage across R3 + Voltage across R2 + 21=0
    So the next step is:
    Voltage across R5 + Voltage across R3 + Voltage across R2 = -21 volts
    The correct equation is equal to -21 volts. You had +21 volts which is wrong.
     
    Last edited: Jun 26, 2014
  14. tom6123

    Thread Starter New Member

    May 31, 2014
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    so to work out all the questions i can use:

    5v/r1
    21v/r2
    i dont know how to work out r3
    5v/r4
    21v/r5

    to get all of my currents through those resistors?

    also i tried to find the voltages using the current and resistor, would these be correct?

    UR1=I1(R1) -50.5 mA X 10ohm= -0.505V
    UR2=I2(R2) 187.8 mA X 10 Ohm 1.878V
    UR3=(I1+I1)R3 ( -50.5 mA + 187.8 mA) 75ohm=10.2975
    UR4=I1(R4) -50.5 mA X 33Ohm= -1.6665V
    UR5=I2(R5) 187.8 mA X 47Ohm= 8.8266V

    Are these correct because if i add the I1 voltages up its more than my voltage sources
     
    Last edited: Jun 30, 2014
  15. Jony130

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    Feb 17, 2009
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  16. tom6123

    Thread Starter New Member

    May 31, 2014
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    well i applied the value of I1 and I2 back into the formula and got this

    (-50.5mA x 180) + (187.8mA X 75) =4.995
    and i also did (-50.5mA X 75) + (187.8mA X 132) = 21.0021
     
  17. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    KVL states: The directed sum of the electrical potential differences (voltage) around any closed network is zero.

    So. Let us do sums!

    UR1+UR3+UR4-5=0
    UR1+UR3+UR4=5
    -0.505+10.297+(-1.666)=8.126
    In my universe 8.126 is not equal to 5. Do you live in some other universe?

    UR2+UR3+UR5-21=0
    UR2+UR3+UR5=21
    1.878+10.297+8.826=21.001
    This one is close.
    You will probably call me petty for saying this, but... it has to be said: since this is system of simultaneous equations, they both HAS TO BE RIGHT, the fact that second equation came out sort of "right" is not enough.
     
  18. tom6123

    Thread Starter New Member

    May 31, 2014
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    ok, i found my problem, it was in the simultaneous equations, i was multiplying by 180 when i was meant to be multiplying by 118, but now my currents are I1=-91.95mA , I2=211.3mA

    now i can get the voltage to 4.9974V
    and the second volt to 20.99535v

    but how would i be able to find the current through the resistors using these calculations?
     
    Last edited: Jun 30, 2014
  19. shteii01

    AAC Fanatic!

    Feb 19, 2010
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    [​IMG]

    Using above notation (I1 and I2 clockwise) I found:
    I1=-0.091 A=-91 mA
    I2=-0.211 A=-211 mA

    The minus sign indicates the direction of the current.
    Example:
    I assumed that I1 will be going through R1 from left to right. So. If I keep my assumption, then the current through R1 is -91 mA. However, I can change my mind, now that I have actual numerical value for I1, I can change its direction which in turn changes the sign of I1. So I say that I1 goes through R1 from right to left, therefore the current through R1 is +91 mA.

    To answer your question about the current.
    The current in R1 and R4 is I1.
    The current in R2 and R5 is I2.
    The current in R3 is the I1-I2 or I2-I1, notice that numerically they are the same. The difference is the sign, the sign indicates the direction. If the sign is positive, then your assumption about the direction of the current is correct. If the sign is negative, then your assumption about the direction is wrong, if you drew the arrow to indicate the direction of the current, simply turn the arrow around and remove the minus sign.
     
  20. tom6123

    Thread Starter New Member

    May 31, 2014
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    so am i right in saying that the current in the diagram above is correct, as the direction of that current isnt in the negative numbers

    ando so the current through r1 and r4 is just -91.95mA do i have to apply some calculations to it?
     
    Last edited: Jun 30, 2014
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