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Kirchoff's Laws Problem Current

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  1. Mar 23, 2016 #1
    1. The problem statement, all variables and given/known data
    Given a battery with voltage 35.7 V and 4 Resistors (Top left with resistance 2R, and the 3 others each with resistance R) where R = 507 Ohms, what current would an ideal ammeter read in its current position? (Diagram attached)
    yiOLA3B.jpg
    2. Relevant equations
    Kirchoff's Laws: KCL and KVL
    Ohm's Law: V = IR

    Resistors in series: R_eq=R1 + R2 + ...
    Resistors in parallel: 1/Req = 1/R1 + 1/R2 +...

    3. The attempt at a solution
    I first attempted to assign current variables (I1 leaving from battery, splits into I2 and I3 at the junction, and then current is redistributed again into I4 (bottom left) and I5, with I6 being the current at the ammeter's location).

    Since the resistor at the top right has less resistance than one at the top left, more current should flow through the right side, thereby making I6 flow from right to left.

    I then apply Kirchhoff's Laws and get a system of equations.

    I2 + I6 = I4
    35.7 - I2 * 2(507) - I4 * 507 =0 (Running from Battery, looping clockwise with I2 and I4)
    35.7 - I3 * (507) - I4 * 507 = 0 (Running from battery, looping clockwise with I3 and I4).
    -I2 * 2 (507) + I3 * 507 = 0 (Looping in the top right section).

    No matter what combination of loops I try, I seem to always have one free variable, which indicates that I'm missing a relationship between the currents.

    Does I3 = 2 * I2?

    Am I correct in viewing this as a pair of parallel resistors in series where I'm trying to find the current coming out of the first pair and into the second?

    Any help is appreciated.
     
  2. jcsd
  3. Mar 23, 2016 #2

    gneill

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    Staff: Mentor

    Hi WesT4N, Welcome to Physics Forums.

    You've written one node equation: I2 + I6 = I4. There are two other nodes that will also give you current relations.

    Your second KVL equation involves currents I3 and I4? I thought your second loop was meant to pass through the rightmost resistors, which according to your current descriptions would involve current I5.

    Is this your current layout?
    upload_2016-3-23_20-39-6.png
     
  4. Mar 23, 2016 #3
    Yeah. You have it exactly.
    So am I right in saying
    i1 = i2 +3, i4 = i2 +i6, and i1 = i4 + i5?
     
  5. Mar 23, 2016 #4

    gneill

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    Yes, and i3 = i5 + i6.
     
  6. Mar 23, 2016 #5
    Thanks! I just got the correct answer. I should've been able to see the other node equations.I was able to find the numerical value of I2 + I3 using Ohm's laws and the Req of the circuit and then use the KVL equations to gradually get all the values.
     
  7. Mar 24, 2016 #6

    gneill

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    Here's a trick that may help you with writing and solving KVL equations. When defining the currents on your circuit diagram, always try to add the minimum possible new currents at each junction. So when you come to, say, a junction where one already defined current enters and two leave, assign a new current to only one of exiting currents and form the other as a combination of the other two. Here's an example using the circuit layout of this problem:
    upload_2016-3-24_9-26-41.png
    If you begin by defining current i1 coming from the battery it enters junction a where it must split into two paths. Give one path a new current variable, i2, leaving i1 - 2 to flow out the other path. Do the same at junction b, where new current i3 is assigned and the other path is made up of existing currents. At junction c you have two already defined currents i2 and i3 entering, so the exiting current is their sum.

    Note that you now have only three current variables to deal with, and there are three loops that you can apply KVL to. Three equations in three unknowns, and you don't have to wade through all the junction KCL equations to eliminate extraneous current variables.
     
  8. Mar 24, 2016 #7
    Yeah your suggestion makes sense. I think my teacher taught us to assign different current variables for each resistor to make it easier to visualize. Your method simplifies the process a lot and makes the math much easier.
     
  9. Mar 24, 2016 #8

    epenguin

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    Which are what? (See my sig.)
     
    Last edited: Mar 25, 2016
  10. Mar 28, 2016 #9

    epenguin

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    Students are too easily content with just getting the right answer. If you have to gradually get it, you've probably missed a point.
    That it's possible to see in a jiffy that the current in the ammeter is 1/6 of the total current, which in turn is obtained by very ordinary quick calculation.
     
  11. Mar 28, 2016 #10
    I found the answer to be .01 A, I found I1 (total current leaving battery) to be .06 and then used substitution to get I6. Can you explain this very ordinary quick calculation? This was a Kirchhoff's Laws assignment so I utilized what methods were taught in order to solve it.
     
  12. Mar 29, 2016 #11

    epenguin

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    I see you did have the 1/6 ratio, whether at end or start of calculation I don't know. This is the trouble when work is not shown.

    I'm indicating what seems to me the easy approach on the understanding you will please complete the calculation and post it here so we can mark the problem as solved. Some of us helpers are quite fed up throwing out help and hints that are sucked into an information black hole. :oldfrown: Or just a message that the student has understood and solved, at the same time as there are clues he hasn't (as I verified several times).

    The key is simply that since the ideal ammeter is resistanceless, the potential at one end of each resistor is the same for all. In other words your H Is like an X with all resistors meeting at a point. So as the top two resistors meet in a V and the conductance of the right-hand one is twice that of the left one, twice as much current flows down the right as the left, or 2/3 of total current flows down the right one . For the bottom two equal resistors meeting in a Λ the same current similarly is split half and half. So that requires that
    (2/3 -1/2) = 1/6 of the current flow from right to left through the ammeter.

    For the total current you can forget about the resistanceless ammeter; two parallel conductors at the top, two at the bottom, these two are joined in series, simples.
     
  13. Mar 29, 2016 #12
    Yeah. This is how I viewed the arrangement of the circuit. Since I was so focused on using Kirchhoff's Laws, I didn't think to look at the ratio between the resistors and the ratio between the current that would go through each. Your explanation makes sense. And yeah. Sorry I didn't include my calculations, I'm new to this forum.
     
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