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Question regarding introductory circuit analysis

  1. Nov 19, 2009 #1
    Hello! I have been working on this problem for around 1.5 hours, and decided I simply need help.

    1. The problem statement, all variables and given/known data
    "Using branch-current analysis, find the magnitude and direction of the current through each resistor for the network.
    rP0RL.jpg

    2. Relevant equations
    I tried using kirchoff's current law? :D


    3. The attempt at a solution

    I tried using various attempts at the loop, first trying using the 10V voltage source as positive, then minus, then same with the 12v, but I just can't seem to get it quite right.

    When I eventually did (-10 -4I1(3.06) +3I2(3.25)+12), I got -.4900. Am I perhaps supposed to convert these Voltage sources to current sources? I am totally lost here.

    Thanks!!
     
  2. jcsd
  3. Nov 19, 2009 #2

    djeitnstine

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    If you are doing branch currents, or mesh analysis, You're using KVL to solve the system not KCL.

    The most common mistake most students make is use the wrong convention. You need perform KVL in a clockwise fashion (it helps to label your mesh currents accordingly). Moreover, you label the voltages across each resistor as if the current was flowing in a counter clockwise fashion. You leave the voltage sources alone.

    Using this convention you should get the correct result.
     
  4. Nov 19, 2009 #3

    berkeman

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    When the problem says use "branch current analysis", that suggests the KCL to me. I could be wrong, though.

    I totally did not follow your attempt at an equation. To use the KCL, you need to label the top node V1 (or whatever), and write an equation that shows the sum of all currents out of that top node equals zero. If there is a voltage source in a branch, that will add to or subtract from the current through the associated resistor.

    Can you please show us the KCL equation for the top node?
     
  5. Nov 19, 2009 #4

    djeitnstine

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    That's the thing, that statement and his equations imply mesh current analysis which needs KVL. What he does is imagine a single current in one branch and applies KVL to each component in a single loop and write the voltage of that component in terms of the branch current.

    Hence the use of I1 and I2 in his equation. But as you said the equation is very incorrect and needs revising.
     
  6. Nov 22, 2009 #5
    You could try using the Superposition theorem by first considering one source , then the other, separately, then adding or subtracting the currents as needed.
     
  7. Nov 22, 2009 #6
    Thanks for the replies everyone! It is beginning to make more sense to me :)

    What I now have is:

    -10v+(4ohms)I1+(3ohms)I3-12v=0
    12v-(3ohms)I3)-(12ohms)I2=0
    Then I1+I2=I3

    My question is, are my current numberings right? Is it that the left branch adds with the far right branch to make the middle branch? Or is it that it is left+middle=right? Oy!

    Thanks again for the replies!! :)

    EDIT: I am also pretty confused as to which side of R3 is positive and which is minus :( Help?

    EDIT again: Since my first equation didn't work, I tried having the right branch be I1 being the left branch, I2 being the middle, and I3 being the right.

    I got:

    4I1+3I2+0=22
    0--3I2-12I3=-12
    I1+I2-I3=0

    Yet I still get a wrong answer *sigh*

    Can anyone possibly give me a few hints as to how I am supposed to set up my formulas? I am simply out of ideas... :( By the way, the book tells me to use third-order determinants, if that makes any difference :(

    The book is telling me: "IR1 = 3.06A, IR2=3.25 and IR3=0.19A". But my teacher posted the answers as well, and she said it was "IR1 =I1 = 3.06 A, IR2 = I2 = 0.19 A
    IR3 =I3 = 3.25 A"
     
    Last edited: Nov 22, 2009
  8. Nov 22, 2009 #7
    Hi Mohdoo,
    I wrote a full solution a few days ago and posted it, but this was deleted as it broke the PF homework policy, which is fair enough I suppose.

    The method I used was node-datum analysis. I have a feeling there are a multitude of different names for the same forms of analysis. This one uses KCL to get the solution.

    You identify each node in the circuit. As I understand so far a node is any point on the circuit where the current and voltage will be the same. Treating wires as ideal (zero resistance) this will mean all wires that touch each other directly form a node.

    Label each node A, B, C, D etc.
    Write down the voltage at each node, Ie, Va = 10V, Vb = 0V, etc. If unknown, leave unknown. Realize that at a grounding point, the voltage is zero. The voltages of the nodes on either side of a voltage source / sink will have a difference equal to the voltage level of that source / sink. Ie, in your question, the node on the bottom will have a voltage of 0V (grounded), while the node directly above the 10V (but below the 4ohm resistor) will have a voltage of -10V.

    Do KCL at each node.
    Current In = Current Out
    Realize that as V=IR, I = V/R. The voltage drop across each component, divided by the resistance of that component, will equal the current passing through it. Va - Vb = difference in voltage between nodes Va and Vb.
    You'll get an equation at each node like (Va - Vb)/Rc + (Vc - Vb)/Rc + (Vd - Vb)/Vd = 0.
    You'll have to make assumptions about current direction and whether a voltage source is a source or a sink. If your current comes out negative, that means you've assumed the wrong direction, so you just swap it over at the end.
    Hope that's clear.
    Also I did get the same results your teacher posted.
     
  9. Nov 22, 2009 #8

    vk6kro

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    The question said use branch current analysis.

    Mark the junction at the top of the 3 ohm resistor as V.
    Going away from this along the top rail, mark I1 going left and I2 going right.

    Now, using branch current analysis, what is the current in the 3 ohm resistor?
    Mark it on the diagram.

    Now, there are 3 vertical paths and they all add up to V volts relative to the bottom rail.

    So, all these things that add up to V must equal each other?
    You don't have to find V, so there are only two unknowns here, I1 and I2.

    When you solve the equations, put the values in the original equations and they should all equal the same V. I have my answers, so I can compare them with you if you post yours first. The book answers are rounded.
     
    Last edited: Nov 22, 2009
  10. Nov 22, 2009 #9

    djeitnstine

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    Hi again Mohdoo, reading your first set of equations, it seems you have used the exact opposite convention as I have stated in post 2.

    Again,
    The convention these books use for labeling is that you label your resistors in such a way that current flows counter clock wise. However, you use KVL in a clock wise manner.

    Let the branch current on the left be [tex] I_{M1} [/tex] and the one on the right be [tex] I_{M2} [/tex]

    So hint #1 your first equation look like [tex] 10V + 4(I_{M1}) + 3(I_{M1} - I_{M2}) + 12V =0 [/tex]

    Correct labeling should be apparent from this.
     
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