Two loop circuit with 3 voltage sources

Discussion in 'Homework Help' started by mzsyed85, Jan 1, 2009.

  1. mzsyed85

    Thread Starter Member

    Jan 1, 2009
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    Hey guys, I hope you're all well and Happy New Year.
    I have some exams coming up, but I am having some trouble commanding KVL and KCL effectively. I understand how to use them when there are two loops and two sources, but not so confident when it comes to three sources... I can't get my head around how KCL operates at a node when theres a voltage supply just before it... Could somebody please help me derive the correct equations so I can find the voltage and current through each resistor? I appreciate any help you guys can give me.

    in circuit a, R1=R2=R4= 1 kΩ, R3 = 2 kΩ
    in circuit b, R1=R2=R3=R4 = 1 kΩ
     
    Last edited: Jan 1, 2009
  2. vvkannan

    Active Member

    Aug 9, 2008
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    hello mzsyed85

    assume current direction as usual say clockwise
    now travelling in the direction of current if we move from negative terminal to positive terminal of battery take it as positive . In the direction of current if we travel across a resistor there is a drop and take it as negative and this can be done irrespective of the number of sources.
    for circuit 1 loop 1 let the current be I1.
    10-I1(1k)+5 - (I1-I2)1k = 0.
    similarly for loop 2 but the drop across common resistor is considered as
    (I2 -I1)1k.

    similarly for loop 1 of circuit 2
    10-I1(1K) - 10 - (I1 - I2)2k = 0
    and similarly for loop 2
    -(I2 - I1)2k + 10 - I2(1k) + 5 - I2(1k) = 0.
     
    Last edited: Jan 2, 2009
  3. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    The trick is to recognize that when you encounter a series circuit such at the one you have posted, you can change the order of the series elements.

    hgmjr
     
  4. mzsyed85

    Thread Starter Member

    Jan 1, 2009
    10
    0
    thanks for that! I really appreciate it... I think one of the things that is confusing me is whether adding a voltage supply affects the current - i mean, shouldnt the current and voltage in R2 and R4 be different in both circuits? and how do you derive the current for R3 (in both circuits)?
     
  5. hgmjr

    Moderator

    Jan 28, 2005
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    Can you rephrase your question?

    hgmjr
     
  6. mzsyed85

    Thread Starter Member

    Jan 1, 2009
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    I apologise that my question was kinda ambiguous. What I meant was that is the magnitude of the currents and voltages going through R2 and R4 the same? And how do you derive the current going through R3?
     
  7. hgmjr

    Moderator

    Jan 28, 2005
    9,030
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    Assuming your question is concerning the first of the two circuits then it is correct to assume that the current in R2 and R4 are the same. It is also correct to assume that the voltage drop across R2 and R4 are the same since the are both have the same resistance.

    Once you have calculated the voltage across R3 in the first circuit using KVL, then the current in R3 of the first circuit is voltage drop across the resistor divided by the resistance.

    hgmjr
     
  8. mzsyed85

    Thread Starter Member

    Jan 1, 2009
    10
    0
    Thanks guys, this is making more sense as a go along. I just wanted to say my background is in human biology so I am very much a novice in this stuff. That said...
    I am having some trouble using the equations that you guys gave me above - please check if my algebra is wrong...

    for circuit 2: (with the 10, 10, and 5 voltage supply)

    Loop 1:
    -10-I1(1K) - 10 - (I1 - I2)2k = 0
    20 = -I1(1K) -I1(2K) + I2(2K)
    20 = -I1(3K) + I2(2K)

    Loop 2
    -(I2 - I1)2K + 10 - I2(1K) + 5 - I2(1K) = 0
    15 - I2(2K) + I1(2K) - I2(1K) - I2(1K)
    15 - 12(4K) + I1(2K)
    15 = I2(4K) - I1(2K)

    using substitution,
    I1 = -6.875 mA, I2 = .625 mA

    ANSWER SHEET - doesnt say which one is which, but lists the currents as 3.75, 5.625, and 1.875 mA AND 3.75, 5.625, 3.75, and 5.625 V

    can someone please tell me where I am going wrong...:confused:
     
  9. silvrstring

    Active Member

    Mar 27, 2008
    159
    0
    You need to correct Loop 1

    When I use a clockwise current loop for I1, I start off:

    -10 + 1k(I1) + 10 + 2k(I1) - 2k(I2) = 0
     
  10. mzsyed85

    Thread Starter Member

    Jan 1, 2009
    10
    0
    if I use your equation, I get I1 = (2/3)I2 => 1.875 mA
    which matches one of the answers but doesnt give me the right voltages, so I still must be going wrong somewhere. I again apologise for my inexperience. Could someone please show me how they would work out the voltage and current across each resistor... The answers from the answer sheet are posted above
     
  11. mik3

    Senior Member

    Feb 4, 2008
    4,846
    63
  12. mik3

    Senior Member

    Feb 4, 2008
    4,846
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    This is wrong the correct is 10-I1(1K) - 10 - (I1 - I2)2k = 0.
    Be careful.
     
  13. silvrstring

    Active Member

    Mar 27, 2008
    159
    0
    mzsyed85,

    Let's start over. We've gone over the first circuit, so I have attached a step-by-step solution for you to better understand it.

    The second attachment is similar, but it is NOT the same as your second circuit--it is an example of the same circuit, but with different values. Go through it, and then you should be able to solve your second circuit.

    Take care,
    silvrstring
     
  14. mzsyed85

    Thread Starter Member

    Jan 1, 2009
    10
    0
    Dear all
    Thank you so much for helping me. Not only do my answers match, but I now understand how to tackle this problem. I can see now that I was overcomplicating the loop - I thought before that adding a second voltage supply would somehow change the magnitude of the current for the entire loop.. I guess it does in a way, but the way you tackle the problem remains unchanged. Thank you vvkannan for initially helping me with derive the equations, thank you hgmjr for clarifying it a bit, thank you Mik3 for correcting the equation I had, and Silvrstring, thank you so much for actually s-p-e-l-l-i-n-g out the problem for me - I actually feel embarassed that you actually went out of your way to do that for me. I really really appreciate it, thank you guys so much
     
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