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Simple Curcuit

  1. Jun 1, 2006 #1
    I'm having trouble with curcuits, my first problem is that I'm not sure if this is a combination parallel-series curcuit?

    For the simple circuit (attached diagram) determine the current through each resistor and the potential difference at the 5 ohm resistor.

    I think that the 2 ohm resistor should be considered separately from the bottom three resistors which are in parallel, but I'm not sure I can explain why even if I'm right.
     

    Attached Files:

  2. jcsd
  3. Jun 2, 2006 #2

    andrevdh

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    You know you have a parallel circuit if the current splits up into two or more branches. Starting from the positive terminal of the power supply the current continues throught the [itex]2\ ohm[/itex] resistor without splitting up. This means it is in series with the power supply. From his point onwards the current splits up throught the three branches of the bottom three resistors and joins up again after going throught the three resistors. This means you have a parallel combination with three branches in series with the [itex]2\ ohm[/itex] resistor. Try to redraw the circuit with the [itex]2\ ohm[/itex] resistor and the parallel combination in series with the power supply. This should then convince you that the circuit is still the same as the given one. Remember that circuit leads have no resistance, so you can reconnect the leads anywhere along a line of leads and the end result will still be the same.
     
    Last edited: Jun 2, 2006
  4. Jun 2, 2006 #3
    Ok so given that I do have a combination I'm not sure if I'm figuring the rest of the problem out.

    So I first figured out the resistance of the parallel resistors together.
    Rp = 1/3 + 1/4 +1/5

    And to find Req for the whole circuit 2 ohm + Rp

    Therefore the tot. current flowing from the battery is I= V/Req

    And then I'm not sure where to go to figure out I for each resistor.
    I think I have to determine V across certain sections?
     
  5. Jun 2, 2006 #4

    andrevdh

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    Correct so far.

    The current that you've got will flow undivided through the [itex]2\ ohm[/itex] resistor. This enables one to calculate the potential drop over it, let's call it [itex]V_2[/itex].

    The rest of the potential will appear over the parallel combination, [itex]V_p=12-V_2[/itex].
     
  6. Jun 2, 2006 #5

    mukundpa

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    correction

    1/Rp = 1/3 + 1/4 +1/5

    MP
     
  7. Jun 2, 2006 #6
    Sorry I had it right in my notes I just typed it incorrectly.

    Ok so I previously figured out the current flowing from the battery = I
    Then V2 over the the 2ohm resistor would be I x 2 ohms for which I got 7.2V

    And therefore Vp = 12- 7.2 = 4.8V

    And to find I for each of the parallel resistors I = 4.8/#ohms (3, 4, or 5)

    For the first resistor would I = 12/2 = 6V??

    I hope I'm doing this right!
     
    Last edited: Jun 2, 2006
  8. Jun 2, 2006 #7

    andrevdh

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    Fine up till this point, except I got the current a bit higher than yours [itex]I=3.7\ A[/itex], resulting in a bit lower voltage over the parallel combination [itex]V_p=4.7\ V[/itex]

    Then everything goes horribly wrong!

    The current flows straight through the [itex]2\ ohm[/itex] resistor and then divides up through the various branches of the parallel combination (the other three resistors). So the total current flows through the [itex]2\ ohm[/itex] resistor. The circuit in the attachment is identical to the one that you are investigating in this problem.

    OK, the weekend starts for me now. It is 18:00 here and I am going home. So I dont know when I will look at this again, but I think you will be fine. Just look at the attachment once it is approved.
     
    Last edited: Nov 29, 2006
  9. Jun 2, 2006 #8

    Andrew Mason

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    Work out the total resistance to find the total current. You haven't shown how you worked out the total resistance or current yet. I get 3.7 A. for the current.

    Then find the voltage drop over the 2 ohm resistor (v2) from that. As you point out, the voltage drop over the 5 ohm resistor is the difference (12 - v2).

    AM
     
    Last edited: Jun 2, 2006
  10. Jun 2, 2006 #9
    Based on the last two replies I'm getting a little confused.

    For the total resistance I first found Req = R1 + Rp = 2ohms + 1.28 ohms = 3.28

    So the tot current was found by I = V/Req = 12V/3.28ohms = 3.7A

    Here's were I'm a litte unsure, for the 2ohm resistor I = 3.7A ??

    And the potential difference for the 5 ohm resistor will be 12-v2 = 4.8V ??
     
  11. Jun 3, 2006 #10

    Andrew Mason

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    The circuit is equivalent to the battery in series with two resistors: R1 = 2 ohms and Rp = 60/47 ohms. The voltage across Rp (the three parallel resistors) is the same for all resistors in that group.

    AM
     
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