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Resistors and Current

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Consider the group of resistors shown in the figure below, where R1 = 4.4 Ω and E = 8.7 V (that is supposed to be the greek "E" that looks like a backwards "3")

    Find the current for each resistor:


    2. Relevant equations

    V = IR?
    E = I(R1 + R2 +R3...)?

    3. The attempt at a solution

    The problem is that my teacher is very bad and students have actually started a petition against her because she messes everything up and confuses herself. This is the last class I need in order to graduate, so I am in desperate need of help. I know you guys dont do the problem for me, all I need is for you guys point me in the right direction for this.

    E = I(Req) so I = E/Req

    I dont even know how to start, please help.
  2. jcsd
  3. Feb 20, 2010 #2
    you can solve this problem using "kirchhoff's law" ..
  4. Feb 20, 2010 #3
    Its impossible to teach this to myself, you dont understand how bad this teacher is. Some fellow students and I were talking about this class after lecture and we all agreed that this lady should never have been hired.

    I cant figure this out at all.
  5. Feb 20, 2010 #4
    E = (I1)(R4.4) - (I2)(R1.2) - (I3)(R9.8) = 0

    That got me nowhere :(
  6. Feb 20, 2010 #5
    (I4)(R6.7) = (I2)(R1.2) got me nowhere either
  7. Feb 20, 2010 #6
    u can solve it using mesh analysis or by superposition law...
    lets solve it using mesh analysis:
    consider the left mesh with I1 traversing it in clockwise direction: then: 12v=I1(R1+1.2+9.8) so u can get I1
    consider the right mesh with I2 traversing it also in clock wise direction: then: -E=I2(1.2+6.7) so u can get I2.
    in a result:resistors (R1 and 9.8) have I1 traversing them
    resistor 6.7 has I2 traversing it
    and finally resistor 1.2 has (I1-I2) traversing it in downward direction
  8. Feb 20, 2010 #7
    12V = I1(4.4 + 1.2 + 9.8)

    12V = I1(15.4)

    I1 = 12V/15.4

    I1 = .779 which doesnt get me anywhere either :(
  9. Feb 20, 2010 #8
    yes I1=0.779 A....where is the problem?

    this the current traversing R1 (from left to right)
    and this current also traversing 9.8ohm (from right to left)..

    same way u calculate I2 (in the second mesh and in clock wise direction)....which is the current traversing 6.7 ohm.....then ( I1-I2 ) is the current traversing 1.2 ohm

    note: i test this circuit on the multisim 9 program..and i get the expected values using this method (mesh analysis)
  10. Feb 20, 2010 #9
    I have to find the current running through each resistor, and isnt that equation V=IR?

    I know V, I know R, and I know "I" so there is nothing for me to solve for. Or am I supposed to be using a different equation?

    Sorry for not knowing this better, but my teacher is the worst.
  11. Feb 21, 2010 #10
    yes sure V=IR is always valid (thats Ohm's law giving ralation between V and I across any resistor)........
    but the above method (mesh analysis) gives u directly all the currents you need in any branch of the circuit..and so, you can get V across each resistor (using Ohms law V=IR)...
    i think its hard to find V first..unless you use superposition law,but though superposition law can also find the currents instead of V....
  12. Feb 21, 2010 #11


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    Homework Helper

    Kirchoff's laws are not difficult to understand. The first says that the potential across any closed loop is 0. That's pretty obvious: every point has to have a certain potential, so if you add up the potential increases and potential decreases across every resistor in a closed loop, you'll get 0 when you reach exactly the same point you started on. When current flows through a resistor, potential DROPS by V=IR; when current flows through a battery, potential could increase or decrease, depending on whether it flows from negative to positive or positive to negative.

    The second of Kirchoff's laws is the law that current can never be "lost". If too wires combine into a single wire, the single wire must have a current of I1+I2; if it doesn't, that means some charge was lost, and that's not possible.

    To solve circuits using Kirchoff's laws, start by assuming that current flows in a certain direction in each of the wire segments. It doesn't matter which direction you chose; if you chose the wrong one, your answer will turn out to be negative, but its magnitude will still be correct. Then apply Kirchoff's voltage law to as many closed loops as you can find. Do the same with Kirchoff's current law; find as many junctions as possible. Then solve the resulting system of equations. Some may be redundant; you'll know this if you have more equations than unknowns.
  13. Feb 21, 2010 #12
    Since I have to find the current for all 4 resistors can someone please show me how to solve the problem for 1 of the resistors and then let me solve for the remaining 3 on my own?

    Would that be fair? I cant afford to let this bad teacher ruin my graduation.

    If the net potential in a closed loop has to be 0 then I would think the current flowing through all 4 resistors is the same...
  14. Feb 21, 2010 #13
  15. Feb 22, 2010 #14
    Ok, guess not :(
  16. Feb 22, 2010 #15
    Solved it! Thanks everybody ;)

    Took a loooong time but it eventually clicked
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