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Help about kirchoff's law

  1. Jul 6, 2006 #1
    help about kirchoff's law!!

    :cry: is there a shortcut way to solve the current in a circuit using kirchoffs law?? or... is there a law or other easiest solution that can solve the the current in a circuit?? because my prof. gave us the equation and solution, but it is too long to solve? hope u can help me...

    can you solve this circuit and can you show me the solution on how to get the current using a shortcut way.... if there is?
    I would really appreciate some help. Thank you!

    Attached Files:

    Last edited: Jul 6, 2006
  2. jcsd
  3. Jul 6, 2006 #2
    Do you mean you have a big system of long equations to solve? For that you can use Gaussian elimination.
  4. Jul 6, 2006 #3
    If you take a circuit class, you will learn easier methods for dealing with circuits. However, Kirchoff's laws are fine if you setup them up in Matrix form and solve accordingly. Can you use a calculator? More than likely you can setup a matrix with it, and it's straightfoward... otherwise, it takes a lot of algebraic steps and it is too easy to make a mistake.
  5. Jul 6, 2006 #4
    im not familliar of that matrix form.... but if i used that kind of solution, will i get the final answer? meaning the current.. or there is another steps to use to get the answer??
  6. Jul 7, 2006 #5
    Have you taken linear algebra? Or do you know how to solve a system of linear equations with a matrix?

    I really should not have said "matrix form", it is somewhat of a misnomer.

    When your attachment is approved, I or someone else can walk you through it.

    And yes, it will give you the currents, in fact it will yield all of them.
  7. Jul 7, 2006 #6
    First, let's start with the basics. I don't know how you visualize circuits, but what helps me is to think of it as a system of pipes and water sources; let me explain.

    Picture the voltage source (in this case you have two) as a water pump. It's purpose is to pump water through the pipe it is connected to. Think of a resistor as something in the pipe that slows the water down. So starting from your [itex]20V[/itex] "water pump" picture water being pumped out towards the [itex] 4\Omega[/itex] resistor. The amount of water that is being pumped out is the current. In your diagram you lableled it [itex] I_3 [/itex]. So the amount of water, or [itex] I_3 [/itex] is being pumped from the [itex] 20V [/itex] voltage source. It hits the [itex] 4 \Omega [/itex] resistor and changes speed, which means it changed voltage, NOT CURRENT. So The current accross the [itex] 20V [/itex] is the same accross the [itex] 4 \Omega [/itex]. This is the case for elements of a circuit in series. So lets just call this a rule:

    1) The current accross elements hooked up in series is equal.

    So following along with our water analogy, you hit the [itex] 30 \Omega [/itex] resistor. Again, the current is the same, the voltage is different. Next up we get to the "hard" part. The current comes in and hits the node you labled "b". Now what?

    Well think of it like water. The water will go through all of those pipes right? But you only have so much water to move through them. So water will move to node "c" and travel to "d" and it will move to node "g" and travel to node "f", it will even move down the middle and travel from node "b" accross "e". So this means you are actually going to have 4 different currents.

    You didn't label the current from node "b" to "e" so lets do that now. We will call it [itex] I_0 [/itex].

    So now lets use kirchoff's laws to solve the circuit.

    First, what are they?
    Well you have KVL, and KCL, or...
    KVL = kirchoff's voltage law
    KCL = kirchoff's current law

    Let's start with KCL. KCL something along the lines of, the current that comes into a node must equal the current going out of that node.

    So go back to the circuit and look at node "J" for a second. Remember we labeled this current [itex] I_3 [/itex]. Well you have [itex] I_3 [/itex] coming into the node from the [itex] 30\Omega [/itex] resistor side, and you also have [itex] I_3 [/itex] leaving a side. So your expression for it would be:

    [itex] I_3 = I_3 [/itex].

    This isn't very useful, as it doesn't yield any new information. So you really don't want to chose a node like that. Instead choose a node where more than one current is entering. And you happen to have two of them. Do you see why?

    What are they?

    Lets go back to KCL for a second. The general expression that I use to remember KCL is:
    [tex] \sum i_{IN} = \sum i_{OUT} [/tex]

    That may or may not help you.

    So back to the nodes. How many do you have? Yup... you have two:
    Node B and Node E.

    Lets look at Node E. What are the currents coming into it?

    (You didn't label [itex] I_0 [/itex], so lets say it is traveling "up" towards the [itex] 20V [/itex] source. So back to Node E. You have the following:

    [itex] I_3 [/itex] comes in.
    [itex] I_1 [/itex] leaves.
    [itex] I_0 [/itex] leaves.
    [itex] I_2 [/itex] leaves.

    [tex] \sum i_{IN} = \sum i_{OUT} [/tex]

    So the expression would be:

    [itex] I_3 +I_2 = I_1 + I_0 [/itex].

    Now you show me your steps for the other node.

    Now to KVL. KVL says something along the lines of, as you loop around the circuit the voltage drop accross elements totals to 0. I remember it like this:

    [tex] \sum v_i = 0 [/tex]

    What is difficult here, is getting the polarity straight. It would be much easier to show this in person, but I'll try to explain. Lets just go through the example, and hopefully you will pick it up. First, lets define something though, that will hopefully make sense later.

    A) If we are looping along in the direction of the current, a resistor will have a POSITIVE polarity.

    B) If we are looping along in the opposite direction of the defined current, a resistor will have a NEGATIVE polarity.

    C) If we loop into a voltage source, the polarity is given by which sign we touch, either (+) or (-).

    So lets start our first loop. Start from the [itex] 20V [/itex] source. We will start our loop just behind the element, and travel in the direction that we defined [itex] I_3 [/itex]. So we first touch the (+) side of the voltage source, so our expression thus far is:
    [tex] 20V + \cdot \cdot \cdot = 0 [/tex].

    Next we reach the [itex] 4 \Omega [/itex] resistor and because of our rules we defined above, we know that when we reach a resistor when we are following the direction of the current we get a postive polarity, so our new expression is:

    [tex] 20V + (4\Omega)(I_3) \cdot \cdot \cdot = 0 [/tex]. Remember that we are summing up voltages, so from Ohm's law we have [itex] V=IR [/itex].

    We reach the [itex] 30 \Omega [/tex] and get do the same:

    [tex] 20V + (4\Omega)(I_3) + (30\Omega)(I_3) + \cdot \cdot \cdot = 0 [/tex]

    Now we keep looping and we get to the split. We can loop in three directions, down the middle, to the left, and to the right. Lets go down the middle.

    We first get to the [itex] 30V [/itex] source, so because we hit the (+) side first we will continue with our expression and have:

    [tex] 20V + (4\Omega)(I_3) +(30\Omega)(I_3) + 30V + \cdot \cdot \cdot = 0 [/tex].

    Now we get to the [itex] 2 \Omega [/itex] resistor. Remember that the current changed here, and we lableled it [itex] I_0 [/itex]. So we are not dealing with the [itex] I_3 [/itex] current anymore, so our expression would be:

    [tex] 20V + (4\Omega)(I_3) +(30\Omega)(I_3) + 30V + (2\Omega)(I_0)= 0 [/tex]

    So now you have two expressions:
    [itex] I_3 +I_2 = I_1 + I_0 [/itex]
    [tex] 20V + (4\Omega)(I_3) +(30\Omega)(I_3) + 30V + (2\Omega)(I_0)= 0 [/tex]

    The two expressions have 4 unknowns, so you can't solve these yet. You are going to need some more expressions, so use KCL and KVL to finish it up and post your work on here.
  8. Jul 8, 2006 #7
    A good suggestion. However, Cramer's rule is much easier.
  9. Jul 12, 2006 #8
    can you tell how to use the matrix form step by step? can you give me a sample? thanx!
  10. Jul 12, 2006 #9
    here's an example (it's attached):

    Attached Files:

  11. Jul 15, 2006 #10
    how did you get the [-3]
    [ 0 ]
    in KVL how did you get that?

    should i use that gaussian elimination to get the currents? how...?

    i did'nt already know how to solve the matrix form... (im so sorry) thats why i asked u....

    can you tell me how you solve the matrix form and gausian elim.
    why (i1)(2k) + (i1)(4k) + (i2)(2k) + 3 = 0
    i know it is an equation... why + 3?

    in collect terms.. why -3

    hope you understand my questions....

    by the way.... thankz a lot to you......
  12. Jul 15, 2006 #11

    I'm tired, so I'll respond with more answers tomorrow. But real quick.

    Guassian elimination is NOT just an equation. It is typically taught in a linear algebra course. You can read about it here:

    If you are allowed to use a calculator, then you can read how to do it for your calculator. That would be the easiest way. If you are not comfortable using matrix's, and you are currently taking a class and need to understand this circuit material, I wouldn't recommend spending the time to learn the matrix forms for it. I'm guessing that your circuits will not be overly complicated, so just using algebra should be enough. That is up to you.

    As far as an equation simply goes, you can use Cramers rule to solve a smaller system of linear equations. Here's a more gentle introduction then that of Wolfram's site,

    Well... I guess I'm awake enough to keep typing. Well in KVL, you do a loop and write down each voltage term (I'm assuming you know how to do this).

    So you get,
    [tex] I_1(2k) + I_1(4k) + I_2(2k) + 3 = 0 [/tex]

    You first factor for [itex] I_1 [/itex]
    [tex] I_1(2k + 4k) + I_2(2k) +3 = 0 [/tex]

    Add what's in the parenthesis:
    [tex] I_1(6k) + I_2(2k) +3 = 0 [/tex]

    Now subtract -3 from both sides:
    [tex] I_1(2k + 4k) + I_2(2k) = -3 [/tex]

    This is so you can put it into matrix form. I could explain why you do this, but it would take awhile, and like I said above... it would probably be easier just to solve the system of equations with algebra.

    The problem you do will probably be a circuit that requires you to write the equations for KVL and KCL. It will give you maybe 3 or 4 unknowns. You then use algebra to solve form them.
  13. Jul 15, 2006 #12
    yeah yeah... were allowed to use calc....
    is that true?? there is a way to solve the circuit using calculator?? i mean the final answer....?? the current....?? eheheheh!!! i hope so... or it use to solve the simple addition, multiplication, division... ect.... am i right??
  14. Jul 15, 2006 #13
    You still have to setup the equations algebraically, convert them over to the matrix form, and then use a calculator to solve the matrix.

    If you are really good, you can skip the algebraic step, and simply right down the expression in matrix form.
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