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2 series resistorss are in parallel with another resistor

  1. Feb 28, 2007 #1
    2 series resistors are in parallel with another resistor..

    1. The problem statement, all variables and given/known data
    Resistors R1, R2, R3 have resistances of 15ohms, 9, 8.. R1 and R2 are in series and the combination is in parallel with R3 to form the load across a 12 volt battery.
    a)diagram
    b)total current
    c)Current across R3 branch?
    d)voltage/potential drop across R2?

    2. Relevant equations
    !!!!!!Series and Parallel and Combo circuit equations..


    3. The attempt at a solution
    1. The problem statement, all variables and given/known data
    I drew it.. 1st thing I did was figure total resistance.. to do that, I combined R1 and R2 serially by adding..15+9 is 24.. Then, now that I have one and one parallel (R[1,2] and R3), I do 1/Rt=1/R1+1/R2... And got R(t) to be 6 ohms..

    With R(t) calculated, and the given V(t), I can figure B, total current
    I(t)=V(t)/R(t)... ding.. 2 A or 2 amps..
    ------
    Now, I'm kind of lost for the rest of the problem.. What is the current in R3 branch? Erm..To find individual current of a parallel resistor, I need to know the voltage and the resistance.. I would think that the voltage of R3 be 12V.. but my teacher said it was 6V..and I had to compact the whole circuit, then repackage it to get 12V..and I'm lost:confused:
    --
    Well, assuming that for some reason, V3 is 6V.. I did I3=V3/R3... 6V/8 ohms=0.75A...that would be C
    Then.. for D..I did I(t)=I(3)+I(1,2).. I(1,2)=1.25A..
    So.. V2 would be I2R2.. If I(1,2)=1.25A, then I1 and I2 also equal 1.25A?
    Then.. 1.25A*9 ohms.. 11.25V equals V2?
    ---
    meh.. ok.. Can anyone confirm whether 2A, total current, and I3 of 0.75 A is right? And I'm pretty sure the rest is wrong.. PS, any freeware circuit drawers?
     

    Attached Files:

    Last edited: Feb 28, 2007
  2. jcsd
  3. Feb 28, 2007 #2

    Doc Al

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    Staff: Mentor

    All good!
    Your initial thought was correct. R3 is connected right across the 12 V battery, so of course its voltage is 12 V.
     
  4. Feb 28, 2007 #3
    Your calculation look good until the 6V appears. Check that your teacher is saying R1 and R2 are in parallel with R3, rather than R1 being in series with R2 and R3 in parallel.

    Answer A - Yes you could do with a circuit drawing package but it matches the description.
    Answer B is correct for the circuit.
    Given the description it is as you think 12V dropped across R3 so recalculate answer C.
    You can proove that this is correct because you can use the same method to calculate the current in the R[12] branch and the sum of the two will equal answer B.
    Your calculations for D are correct but the numbers are screwed up by the 6V again.

    As to freeware circuit drawing packages not sure. You could download ICAPS4Win spice programme from www.intusoft.com, this is a full circuit modelling package but allows you to draw schematics.
     
  5. Feb 28, 2007 #4
    Direct quote from worksheet: "Restors[sic] R1, R2, R3 have resistances of 15 ohms, 9 ohms, and 8 ohms, respectively. R1 and R2 are in series and the combination is in parallel with R3 to form the load across a 12 volt battery.
    a)Draw the circuit diagram
    b)Determine the total current in the circuit
    c)What is the current in the R3 branch?
    d)What is the potential drop across R2"

    :( My teacher.. this other kid was at his desk, and showed him "3/2" for R3and he said yes, that looks like what some other kids of other classes got.. 1.5 A[12V/8 ohms] He never said 0.75A was wrong... But meh..

    K, so question.. Is V1=12V? Is V2=12V?...I don't think they are but I think that V(1,2) is 12V. Why? Well, if I(t) [yes I realize that's I] is 2A, and suppose that 1.5A is right for I3.. If
    V1=12V.. I(1)=12V/15 ohms is 0.8 A.. but I3+I(1) then would be more than I(t)..
    For V(1,2)=12V though.. 12V/24 ohms=0.5A, fits perfectly with the R3's I3 of 1.5A, totaling up to 2A.

    Also, if I'm going to say that V1 and V2 aren't 12V.. then V3 also can't be 12V..
    -------------
    So..maybe I have just discovered that the "parallel law" of V(t)=V1=V2 only applies to when it is one resistor parallel with another resistor, not one parallel with two.. Great.. So how am I suppose to figure voltage drop of the individual V2 then..
    ---
    http://www.physics.uoguelph.ca/tutorials/ohm/Q.ohm.intro.combo.html
    Found this link, not that it helps much..
     
    Last edited: Feb 28, 2007
  6. Feb 28, 2007 #5

    Doc Al

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    Staff: Mentor

    Nonsense. The total voltage drop across R1 & R2 is 12V, so V1 + V2 = 12V. R3 is parallel to R1+R2, so the voltages across both branches are equal.
    I think it's beginning to click now. :wink:

    Find the current through each branch. Then use ohm's law.
     
  7. Feb 28, 2007 #6
  8. Feb 28, 2007 #7

    Doc Al

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    Staff: Mentor

    Sure. I = V/R = 12/8 = 1.5 A.
     
  9. Feb 28, 2007 #8
    Looks like your teacher made an error. If that's how the problem was worded, your diagram is correct. V3 = 12V as you thought, and V(1,2) is also 12V. I(3) = 1.5A as you calculated, and the total current, as you calculated is 2A.
    And you're further currect that since V(1,2) = 12V and R(1,2) = 24ohms, then I(1,2) = .5A

    If you need to know the potential difference across either R1 or R2, simply use the current. You'll find that the two potential differences do indeed add up to 12V.

    Your teacher had one of those brain farts?
     
  10. Feb 28, 2007 #9

    ranger

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    Gold Member

    Well the voltage, V3 is 12V becuase it connected in parallel with the voltage source and you know the current in that branch. The current flowing in the branch with R1 and R2 is the difference of It and I3. Since the same current flows through both R1 and R2, just apply ohms law to find the individual voltage drops V1 and V2. You will find that V1+V2 is 12V.

    wow.....was my reply late or what :)
     
  11. Feb 28, 2007 #10
    So..
    I(1,2)=I1=I2..

    d)voltage/potential drop across R2?
    V2=I1R2..
    Since 0.5A=I(1,2)..I1=0.5A..?
    0.5A*9 ohms=4.5V=V2..
    Hrm. Let's figure V1 for fun..0.5A*15 ohms = 7.5V..
    (V1+V2)=V(t)=V3..

    Everything looks great:D I think.. My teacher confused me with his "at first, V3 is 6V and blah"..
    And um.. Man. I wish I had some interactive flash app to verify that..:)
     
    Last edited: Feb 28, 2007
  12. Feb 28, 2007 #11

    ranger

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    Gold Member

  13. Feb 28, 2007 #12
    Hey, ranger, that's a very cool link!
    "$ 1 5.0E-6 10.812258501325767 50 5.0 43
    w 96 112 176 112 0
    v 176 112 256 112 0 0 40.0 12.0 0.0
    w 256 112 448 112 0
    w 448 112 448 144 0
    w 448 144 368 160 0
    w 448 144 528 160 0
    r 368 160 368 192 0 15.0
    w 368 192 368 224 0
    r 368 224 368 256 0 9.0
    r 528 160 528 256 0 8.0
    w 368 256 448 320 0
    w 528 256 448 320 0
    w 448 320 96 320 0
    w 96 320 96 112 0"
    Import into the applet..
    Does that seem accurate, relative to my problem? The "answers" are consistent..
     
  14. Feb 28, 2007 #13

    ranger

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    Gold Member

    Yup, your netlist is exactly the circuit you've described.
     
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