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Finding currents in branches of a circuit

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data
    The 5.00-V battery in the figure (Figure 1) is removed from the circuit and replaced by a 20.00-V battery, with its negative terminal next to point b. The rest of the circuit is as shown in the figure.

    YF-26-26.jpg

    Part A ) Find the magnitude of current in the upper branch.

    Part B ) Find the magnitude of current in the middle branch.

    Part C) Find the magnitude of current in the lower branch.

    Part D) Find the potential difference Vab of point a relative to point b.

    2. Relevant equations
    Resistors in parallel have equivalent voltages

    Reciprocal of total resistance of resistors in parallel is equal to sum of reciprocals of individual resistances

    V = IR

    Junction Rule : Summation of current at a junction = 0

    Loop Rule : Summation of voltage around a loop = 0

    3. The attempt at a solution

    I am not sure what exactly to do so I have been going back through notes and watching videos to better understand the step by step thinking used in this kind of problem. I have only tried Part A and I got it wrong, so that is the part I will be asking for help on first.

    What I ended up trying was saying that the "top branch" is the top half of the circuit, and then I made an equation. Also, I made sure to always put voltages in terms of IR. So remembering the loop rule I thought that
    I could make one flow of current going left from the top battery and a flow of current going left from the middle.

    For the current moving counterclockwise from the topmost battery and looping around the middle wire of the circuit: -10v + 4(I1) + 3(I1) = 0
    And for the current moving clockwise from the middle battery and looping upward around the top of the circuit then back to the middle: -20V + 3(I2) + 4(I2) = 0

    I cannot remember if I would also have to add 2(I1) and 4(I2) for the top and middle lines respectively due to the internal resistances of the batteries, so that might be a problem. I thought I2 would be changed from 1 to 4 by the battery replacement because I thought that the ratio of internal resistance to emf or voltage would remain the same.

    Having said that, Solving for I1 and I2 then subtracting them and taking the absolute value to correct for any directional errors, I came up with this: I1 = 10/7, I2= 20/7, I2-I1 =10/7 = 1.4 (For Part A)

    That is wrong and the answer is .4, I am not sure where I am going wrong. Thank you for any and all help.
     
  2. jcsd
  3. Oct 26, 2014 #2

    haruspex

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    I don't understand the reasoning behind that equation.
    As far as I am aware, it is not a good idea to try to solve this sort of problem piecemeal. There's too much interaction between the various voltages and currents. I proceed by assigning symbols to the unknown voltages and currents, then writing down all the equations.
     
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