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Finding current through a resistor

  1. Oct 10, 2013 #1
    1. Problem statement.
    23hlpft.jpg

    2. Known equations
    Kirchoffs law, logic

    3. Attempt

    16sxac.jpg

    I believe there are 4 currents, as shown above. i1 and i2 are coming from the 9 volt battery in a clockwise direction and i3 and i4 are from the 24v battery in a anti clockwise direction. I just apply kirchkoffs law for each current with the resistor in its path way.

    Ex. For i1 the equation would look like

    9v - i1(3ohms) -i1(6ohms) = 0

    Then

    A = i1 + i3
    B = i2 + i4
     
  2. jcsd
  3. Oct 10, 2013 #2

    jedishrfu

    Staff: Mentor

  4. Oct 11, 2013 #3
    Hmm, what I learned is at junctions, current in = current out. Would that mean that there will only be 2 currents? For example from the 9 volt battery, the current would travel clockwise through the 3ohm resistor and it has only one of two choices to go after that, which would be down to the 6 ohm resistor or towards the 4 ohm resistor?
     
  5. Oct 11, 2013 #4

    NascentOxygen

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    Staff: Mentor

    Yes, let's use that.

    No, it has 3 choices, there are 3 paths to get back to the - terminal of its 9V source, one path you have overlooked being through the 24V source.

    The easiest (?) way to analyze this is to let the voltage across the 6Ω resistor be Vx volts. :smile: Now, express the current through every resistor (and voltage source) in terms of that one unknown, Vx.

    See how you go.

    Eventually, you'll be able to solve for that unknown, and once you have, you'll know the current in every element.
     
  6. Oct 11, 2013 #5
    I am having trouble setting up the equation still. I am thinking there might be 2 equations max, where the current is coming from the 9v battery and the 24volt battery. I am also wondering if I should treat the currents as the same value as for each of the resistors.

    For example, this is what i'm thinking the equations would look like.

    9v - i1(3ohm) - i1(6ohm) - i1(8ohm) - i1(4ohm) - 24v - Vx = 0

    And then do the same again for the other voltage source.
     
  7. Oct 13, 2013 #6

    gneill

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    Staff: Mentor

    If you use NascentOxygen's suggestion and set the potential at the top of the middle resistors to Vx (the potential is with respect to a common reference point assumed to be the bottom node of the circuit), then you can write expressions for every current using Vx and the other known potentials. For example, the current from the Vx node to the 9V battery through the 3 Ω resistor would be

    ##I = \frac{Vx - 9}{3}##

    If you wrote similar expressions for all the currents for the Vx node and summed them (i.e. doing KCL at the Vx node) and set the result to zero, you'd have one equation in one unknown (Vx). Solve for Vx. Then knowing Vx you can go back and give values to all those individual currents.

    The above procedure constitutes "Nodal Analysis", which you will cover in your course at some point. But the technique should be pretty obvious if you already know KCL and KCL.
     
  8. Oct 13, 2013 #7
    I see, so if I put a node at the top of the middle resistors, (between the 3ohm and 4ohm resistors) and do a similar equation to the current from 9v to that node, Would I get 2 other currents, which goes in a anti-clockwise circle from the 6 ohm and 8 ohm resistor, and one other current from the 24v battery through the 4 ohm resistor, and then I sum them all up and equal to zero, would that be my equation?

    If so this is what I've got,

    9(volt) - 3I -Vx = 0 (which is the example you provided, just haven't sold for I)
    Vx - 6I - 8I = 0 (node, current going anti-clock wise through the 6 ohm and 8 ohm resistor)
    24(volt) - 4I - Vx = 0

    Then I add them all up,

    33(volt) - 21I - Vx = 0

    Vx = 33-21I

    Does this seem correct?
     
  9. Oct 14, 2013 #8

    gneill

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    Staff: Mentor

    No. Don't unravel the expressions for individual currents. Leave them in the form I showed. The "I" in the expression I wrote was not meant to be taken as a new variable, it's only meant to show that the expression yields a current value; it's the current in that particular branch. I suppose I should have taken your circuit diagram and placed new current labels on everything. So here:
    attachment.php?attachmentid=62870&stc=1&d=1381725873.gif

    The "I" in the expression I wrote corresponds to ##i_1## in the above diagram.

    This is KCL you're doing: The sum of all currents entering or leaving a single node is zero. The middle node has 4 branches leading from it so there will be four expressions of the form that I showed, one for each branch leaving the node. Each expression is of the form:
    $$\frac{\Delta V}{R}$$
    The ##\Delta V## for the example I gave is Vx - 9, corresponding to the potential drop across the 3 Ω resistor that sits between Vx and the 9 V potential at the top of the 9 V battery. Hence the current in that branch is:
    $$\frac{Vx - 9}{3}$$
    That will be one term of four terms that go to make your KCL expression for the node:
    $$\frac{Vx - 9}{3} + \frac{(~~~~)}{(~)} + \frac{(~~~~)}{(~)} + \frac{(~~~~)}{(~)} = 0 $$
    which corresponds to the KCL statement: ##i_1 + i_2 + i_3 + i_4 = 0## for the diagram above; each term represents one of the currents.

    Once you solve for Vx you can revisit the individual terms to find the individual currents.
     

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  10. Oct 14, 2013 #9
    Hmm, well using your examples, I've come up with the currents as following,

    i1 = Vx-9 / 3

    i2 = Vx-9 / 9 ,I am picturing this current came from the 9v battery and went through the 3 ohm resistor then down to the 6 ohm resistor

    i3 = Vx-24 / 12 , this one same like above but the opposite side, from the 24v battery through the 4 ohm resistor and then down the 8 ohm resistor.

    i4 = Vx-24 / 4 , this one I figured by using the same logic as the example you've given me

    I am also confused, the equation I was taught was

    V - IR = 0 (v for voltage source)

    but the way you have i1 set up it looks like it is

    V + IR = 0

    Is that a typo or is it a different equation we are using here?
     
  11. Oct 14, 2013 #10

    NascentOxygen

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    That wire along the bottom is 0 volts. It links up the 0V end of one battery and the 0V end of the other battery, and it's zero volts all the way along it.

    Where the top of a battery is +9V, the bottom is 0V.

    So, using Ohms Law: i2 = (Vx - 0) / 6

    (That - sign on a cell is a bit misleading. The bottom of the battery is not -9V, it's 0V.)
     
    Last edited: Oct 14, 2013
  12. Oct 14, 2013 #11
    Am I picturing the node in the middle as a 0v battery? so shouldn't the i2 current, from the node, go down the 6 ohm resistor, then back up to the 8 ohm resistor?

    so

    i2 = Vx / 14 ?
     
  13. Oct 14, 2013 #12
    hmm, so i3 would then be

    i3 = (Vx - 0 ) / 8

    I am having trouble picturing how the current flows or the possibilities that the current could flow, is there any tips or ways to make that more clear?

    from what I've read from the equations I am picturing this is what is happening, which I know is incorrect

    34g1po5.jpg
     
  14. Oct 14, 2013 #13

    NascentOxygen

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    Staff: Mentor

    Even were you right, in forming these equations we aren't interested in any current's full path.

    We are applying Ohms Law to individual branchs (resistors). The 6 Ohm resistor has Vx volts across it, and its current is I2. That's all you need to know to form the equation.

    As to your question, current goes from a higher potential to a lower potential. So if it goes from +Vx volts down to 0V, it's not going to then reverse and go from 0V back up to +Vx volts through another resistor. After reaching the ground wire (0V) current returns to the battery from whence it originated (and it's pumped back up to that battery's positive potential).

    EDIT: the diagram you have just added, with coloured paths, is close enough.
     
  15. Oct 14, 2013 #14
    I see, just to make sure, would the current between points a and b be i2 = a , and i3 = b or is it 0 since its a ground wire?
     
  16. Oct 14, 2013 #15

    NascentOxygen

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    That doesn't make any sense. :cry: Point a is a dot marked on a schematic. I2 is a current and it has units of amperes.

    Strictly speaking, points a and b are one and the same, they are that bottom node, and the voltage all along it is zero volts.

    But I can see what you mean, regarding that as a sketch of the physical layout. If you were to add an ammeter in between points a and b, what value of current would it measure? It would measure exactly the difference between the current from the battery and the current through the 6Ω, because at point a the current through the 6Ω joins with the current between nodes b and a to form the current into the base of the 6V battery.
     
  17. Oct 14, 2013 #16

    gneill

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    Staff: Mentor

    freshcoast, here's your circuit redrawn with points of interest indicated. The bottom rail (blue) is all one rail and is a convenient place to put our reference point for all potentials in the circuit. Hence the 0 V designation and the ground symbol on the bottom left.

    attachment.php?attachmentid=62889&stc=1&d=1381757695.gif

    The nodes at the top left and top right have their potentials fixed at +9 V and +24 V by the batteries (the batteries are directly between those points and the reference node).

    Now, you're looking to find the current between points a and b assuming that in a real implementation of this circuit there would exist a wire corresponding to a-b. I've designated that current as I in the diagram. Consider point a as a node. How might you determine what I is?
     

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  18. Oct 14, 2013 #17
    Sum of current in = sum current out?

    So with your diagram, I1 + I2 = I?
     
  19. Oct 14, 2013 #18

    gneill

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    Staff: Mentor

    That's the idea. Now all you need to do is solve for I1 and I2. Hence the nodal analysis...
     
  20. Oct 17, 2013 #19
    I am confused about how to check my answers, I know its correct but I don't know why, for the currents my values are

    I1 = 0.43A
    I2 = 1.71A
    I3 = 1.285A
    I4 = -3.43A

    I know that to make sure I'm correct all the voltages need to add up properly, but I do not see the reasoning behind this. For example, if I do

    V1 = I1 * R

    Which is

    V1 = (0.43A) * ( 3ohm) = 1.29V,

    I'm thinking 1.29 volts get dropped after passing the 3 ohm resistor? But why is I2 which I calculated as 10.26v, which is higher than the 9v battery? I don't know I'm just having trouble grasping the general concept of how the voltages should add up and where.

    Thanks so far for all the input I've received
     
  21. Oct 17, 2013 #20
    Oh and also for future problems,

    Is the best way to solve these is to look for a symmetrical location on the circuit, label that as a node with voltage Vx, then just write out current equations per each of the one resistor? So no current equation should have no more than one resistor value on it?
     
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