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Homework Help: Simple DC Circuit (That I Can't Solve!)

  1. Apr 9, 2010 #1
    1. The problem statement, all variables and given/known data
    For the circuit shown below, assume that the batteries are ideal and determine the following:

    a) The current through each resistor, including direction.
    b) The potential difference across the center branch of the circuit (the one including the 8 ohm resister and 20 V battery)

    100_0265.png
    2. Relevant equations
    V=IR, (V=voltage of source, I=total current, R=total resistance)


    3. The attempt at a solution
    Current through far left 2 ohm top resistor: I1
    Current through far right 2 ohm top resistor: I2
    Current through middle 8 ohm resistor: I3
    Current through bottom right 2 ohm resistor: I4
    Current through bottom left 2 ohm resistor: I5

    I represent the resisters as R and the amount of ohms. So a two ohm resistor I say R2.

    Using Kirchoff's Laws
    These are the equations I have for the circuit.

    left loop: -10V-I1R2-I3R8-20V-I3R2=0
    right loop: -20V+I3R8-I2R2-10V-I4R2=0
    outer loop: -10V-I1R2-I2R2-10V-I4R2-I5R2=0

    I'm not sure if these are right. I'm hoping I got Kirchhoff's Laws correctly.
     
  2. jcsd
  3. Apr 10, 2010 #2

    collinsmark

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    Homework Helper
    Gold Member

    Hello sweetdion,

    I think you might want to redo your equations.

    You have three equations, but you only need two. Once each component is associated with at least one current loop, you can (and should) stop forming equations. It turns out that only two of your three equations are actually independent. If you were to use linear algebra to solve the simultaneous equations, you would end up with degeneracy, which would just complicate things.

    As it happens, any two of the equations are independent of each other. So you can pick any two of them. Just don't use all three (true for this particular problem. More complicated problems might require more loops).

    A few pointers:

    //////////////////
    // Pointer 1
    /////////////////

    Define your current loops. Once each component is associated with at least one loop, stop there.

    //////////////////
    // Pointer 2
    /////////////////

    Get rid of the variable I3 through I5 in your equations. You only have 2 current loops (regardless of how you choose them), so you only need I1 and I2. (True for this particular problem.)

    If any component (resistor, particularly) is associated with both current loops, make the current going through that resistor the appropriate combination of I1 and I2.

    //////////////////////
    // Pointer 3
    //////////////////////

    If your current loop goes from the + to - direction of a given component, give the resulting voltage a + sign in your equation. If the current loop goes from the - to + direction, give the voltage a - sign.

    Here is a partial example. Suppose we define a current loop, I1, which goes around the left side of the circuit in the clockwise direction. Starting from the lower left hand side of the loop, we start forming our equation,

    -10V + I12Ω + ... + I12Ω = 0

    //////////////////////
    // Pointer 4
    /////////////////////

    When tracing out the current loop to form your equations, give the voltage drop across a resistor a positive (+) voltage drop for the particular current loop you are presently working with. In other words, for whichever current loop you happen to be working with at the moment, the current goes through the resistor in the + to - direction.

    Here is an example. Suppose, as before, you define I1 as a loop on the left side of circuit going in the clockwise direction. Also suppose you define the second current loop I2 going through the right side of the circuit, also in the clockwise direction. So when dealing with the 8Ω resistor, there are two currents to deal with.

    When tracing out the voltages for I1 (the left loop [clockwise]) you have,
    ... + I12Ω + (I1 - I2)8Ω + 20V + ... = 0

    When tracing out the voltages for I2 (the right loop [also clockwise]) you have,
    ... - 20V + (I2 - I1)8Ω + I22Ω + ... = 0

    Hope that helps.
     
    Last edited: Apr 10, 2010
  4. Apr 10, 2010 #3
    That helps a lot!
     
    Last edited: Apr 10, 2010
  5. Apr 10, 2010 #4
    Potential Difference down the center branch:
    1. At the top of the branch, we will assume there to be zero potential.
    2. By going over the 8 Ω resistor, we lose 4 V (8 Ω * .5 A), thus leaving us at - 4 V.
    3. The 20 V power supply adds 20 V, leaving us with 16 V.
    7. We now have a combined Potential Difference across the center branch to be 16 V.

    I'm assuming that the answer that I got in part a was correct. Is this the correct way to do it?
    *I used .5 A as the example of current going through it
     
    Last edited: Apr 10, 2010
  6. Apr 11, 2010 #5
    Is it correct to assume that at the top of the branch the potential is 0?
     
  7. Apr 11, 2010 #6
    Why are you assuming that value for the current? The voltage drop across the central "branch" depends on which way the current through it goes.

    Try writing out the two KVL equations. You shouldn't have to assume any values for currents.
     
  8. Apr 11, 2010 #7
    I was assuming just for the time being because I hadn't solve the equations for I yet. I actually found the current to be 1.43 Amps. So substitute that in for I instead.
     
  9. Apr 11, 2010 #8
    I see, but pay attention to the resulting polarity of your 8 ohm resistor when you're doing that. (ie, is the top end of it positive or negative?).
     
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