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Electric Circuit using Kirchoff's rules

  1. Mar 9, 2016 #1
    1. The problem statement, all variables and given/known data
    We are given that R1 = 15.0 Ω, R2 = 10.0 Ω, and R3 = 8.00 Ω. What is the current and direction of the current through each resistor? Again, I have no idea how to do this, I got my answer from chegg, but I don't know if this is correct.

    4.PNG


    2. Relevant equations

    V = IR



    3. The attempt at a solution

    Using Kirchoff's rules , for the loop ADCBA, starting from D, we have:
    (-10ohm * i1) + (10V) - 15ohm(i1+i2) +(20V) = 0

    (25ohm * i1) + (15ohm * i2) = 30V
    (5 * i1) +(3 * i2) = 6 to the -(i) direction ?

    Now, applying Kirchoff to DCFED, starting from D:
    (-10ohm * i1) + (10V) + (8ohm * i2) = 0
    (10ohm * i1) - (8ohm * i2) = 10V
    (5 * i1) - (4 * i2) = 5V to the -(ii) direction ?
    After solving, we get:
    i1 = 1.14A for R2?
    i2 = 0.142A for R3?
    i3 = 1.256A for R1?

    Are the currents and directions correct? If not, can you tell me what the direction is at least?
     
  2. jcsd
  3. Mar 9, 2016 #2
    Have you tried reading your textbook or a textbook? This question should be one of the first things mentioned.
     
  4. Mar 9, 2016 #3

    CWatters

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    Science Advisor
    Homework Helper

    Your equations look correct but after solving I get very slightly different answers...

    I1 = 1.114A
    I2 = 0.143A
    I3 = 1.257A

    However...

    Before you apply KVL/KCL it's essential to mark up the circuit with the currents I1 and I2 including arrows to define what you mean by +ve current flow. Then, after applying KVL/KCL and solving, if I1 or I2 turns out to be +ve or -ve you know which direction that is by referring to your drawing.

    Note: You need BOTH the drawing and the calculated currents for a valid answer. That's because the direction you choose for +ve current is arbitrary. Your classmates may have chosen to define a different direction as +ve, in which case they would have got -ve answers for the currents. That would be an equally valid answer. The only time it's not an arbitrary choice is when the problem statement tells you to make a particular choice.
     
  5. Mar 9, 2016 #4
    Okay, thanks again CWaters!
     
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