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Simple Circuit Problem

  1. Nov 1, 2015 #1
    1. The problem statement, all variables and given/known data
    Assume the electric potential is zero at the negative terminal of the battery. Calculate:

    a). the equivalent resistance of the circuit
    b). the electric potential at position a
    c). the magnitude and direction of the current through each resistor

    Circuit.png

    2. Relevant equations


    3. The attempt at a solution
    a). I said that the whole equivalent resistance of the circuit is 3000[itex]\Omega[/itex]. I combined the two on the right in series, then the 1200 with that in parallel. Then I added 1500 and 900 to that and got 3000.

    b). Since the source voltage is 30V and the equivalent resistance is 3000[itex]\Omega[/itex], then the current is 30/3000=0.01 A. So then I said that the voltage at point a is 0.01A*900[itex]\Omega[/itex]= 9V

    c). I wasn't sure how to go about this one.. Does this use Kirchoff's law?
     
  2. jcsd
  3. Nov 1, 2015 #2

    andrewkirk

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    That can't be correct, because there is a path between the battery terminals with resistance of only 1500, so the equivalent resistance must be less than that.

    The 1500 is in parallel with the other branches not containing the cell, not in series.

    It may help to redraw the diagram, rearranging it to put parallel wires next to one another and series wires above and below one another.
     
  4. Nov 1, 2015 #3
    Would it just be 1500 then? I'm a bit confused then about how to calculate it.. The two on the far right are in series so you add them and then next is the two that are on the right in parallel, so you would add the reciprocals. Then..? =/
     
  5. Nov 1, 2015 #4
    You add those two together and that combination is in parallel with the 1200 ##\Omega## resistor. So figure out the value and redraw the circuit with that equivalent resistor. You will then see that it is not in parallel with the 900 ##\Omega## resistor.

    Even if 0.01 A were the current through the equivalent resistance, the current through the 900 ##\Omega## resistor would not be 0.01 A.
     
  6. Nov 1, 2015 #5
    Circuit2.jpg
    Is this correct so far?
     
  7. Nov 1, 2015 #6

    SammyS

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    Yes.
     
  8. Nov 2, 2015 #7
    Okay, so now aren't these in series then? So you would just add them normally?
     
  9. Nov 2, 2015 #8
    Two resistors are in series with eachother, and in parallel with another...
     
  10. Nov 2, 2015 #9
    Okay, so the two that are in series on the two on the right, correct? And then in parallel would be the two ends?
     
  11. Nov 2, 2015 #10
    Yep, exactly. Use the same formulas you used to condense the right side to condense the entire circuit. ##\frac{1}{R} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_n...}## is one of them
     
  12. Nov 2, 2015 #11
    Ok! So I got that the equivalence resistance is 750 ohms. So now would I take 30V/750ohms to find the current?
     
  13. Nov 2, 2015 #12
    Correct. Then you need Kirchhoff. Start with your 40 mA total, and reconstruct your circuit piece by piece to calculate the current and voltage through each resistor.
     
  14. Nov 2, 2015 #13
    Okay, so if I look at the joint above the voltage source.. there is a current entering from the bottom and current leaving on both the left and right sides. So
    I1-I2-I3=0

    At join 2 (to the right of the resistor of the top), there is current entering from the left, and leaving out the right and bottom,so
    I2-I3-I4=0

    Is this correct?
     
    Last edited: Nov 2, 2015
  15. Nov 2, 2015 #14
    Yes.
     
  16. Nov 2, 2015 #15
    Will I have to each of the 4 joins since so far I only have 2 equations and 4 unknowns? Won't I end up with some more unknowns?
     
  17. Nov 2, 2015 #16
    Yes, there are 4 junctions. You will end up seeing that several of the unknowns are equal to each other. Also the ratio of ##\frac{l_2}{l_1} = \frac{\text{Resistance of the left side of the circuit}}{\text{Total resistance of the circuit}}##
     
  18. Nov 2, 2015 #17
    If you look at the circuit you drew in Post #5 there is no "Joint 2".

    There is no need to use Kirchhoff's Laws to analyze this circuit. Just look at the equivalent circuits you constructed on your way towards the one you drew in Post #5 and use the facts that the current through resistors in series is the same, and the voltage across resistors in parallel is the same.

    For example, if you look at the 1500 ##\Omega## resistor in the original circuit you can see that the voltage across it is 30 V (because it's in parallel with the battery). You can therefore immediately find the current through it.

    Alternatively, you could use Kirchhoff's Laws to analyze the original circuit. You would have four unknowns and four equations.

    Or use Kirchhoff's Laws to analyze any of the equivalent circuits you drew, such as the one in Post #5.
     
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