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Current through each individual resistor of a circuit

  1. Mar 12, 2010 #1
    1. 1082045.jpg

    What is the current through each resistor




    2. All the obvious stuff...



    3. Ok, without showing all my work, here's what I have solved and know is right:

    The equivalent resistance of the whole circuit is 56.3 ohm. From this, I used Ohm's law to get the equivalent current for the circuit, which is 0.213A.

    Where I am stuck is now using this to get the individual currents for each resistor. For example, I calculated I_2 = V/R_2 = 12V/45ohm = 0.27A

    Similarly, I found found the others, so here's what I got (which is incorrect):

    I_1 = 0.8A
    I_2 = 0.27
    I_3 = 0.48
    I_4 = 0.6

    Also, I notice that these values sum to 2.12, which approximately euqals 2.13, which is the equivalent current with the decimal point moved over.... where am I going wrong?
     
  2. jcsd
  3. Mar 12, 2010 #2
    It would help to show the values of each resistor and also showing the details for your current calculations.
     
  4. Mar 12, 2010 #3
    Ahh, sorry for that:

    V_bat = 12V
    R1 = 15ohm
    R2 = 45
    R3 = 25
    R4 = 20

    My calculated values for the equivalent resistance and current of the circuit are correct. Briefly, R3 and 4 are in series, equiv resistor is in parallel with R1, and that equiv resistor is in series with R2.
     
  5. Mar 12, 2010 #4

    vela

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    I've noticed in several of your circuit threads you keep making the same mistake. You can't just plug in 12 volts for V for every resistor. You have to use the actual voltage across the resistor.

    In the equivalent circuit, you found that the current through the battery is 0.213 A, right? Because R2 and the battery are connected in series, the current through them must be equal. Now that you know that the current through R2 is 0.213 A, you can apply Ohm's law to calculate the voltage drop across the resistor, V=(0.213 A)(45 Ω)=9.59 V. The direction of the voltage drop is from the end where the current enters the resistor to the end where it leaves the resistor.

    Now apply Kirchoff's voltage law to the lefthand loop to find the voltage across R1. Once you know that, use Ohm's law to find the current through R1.

    Then you can apply Kirchoff's current law to the top node. Since you know how much current entered via R2 and how much current left through the left branch, you can determine the amount that must have gone through the right branch. R3 and R4 are in series, so the same current flows through each.

    As a check, you should calculate the voltage drop across R3 and R4 and make sure Kirchoff's voltage law holds for the righthand loop. If it doesn't, you made a mistake somewhere.
     
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