Voltage drop and current for each resistor

  1. 1. The problem statement, all variables and given/known data
    heres a picture of the problem. Please dont give me the answer just tell me what are the steps to complete it. thanks.

    http://i307.photobucket.com/albums/nn296/alan1592/download-1.jpg
    2. Relevant equations

    V=IR

    3. The attempt at a solution

    1/6+1/3=2 Ohms
     
  2. jcsd
  3. Ok, so you now have two resistors in the circuit- you can calculate the voltage drop across each resistor and, using Ohm's law, the current. You need to be a little more specific with what you're looking for so you can get more help.
     
  4. Im looking to find those items. The voltage drop and the current. I dont know how.
     
  5. Since there is one indepedent voltage source of 20 Volts that means the voltage drop across all the resistors should equal 20V.

    I would have made this look nice but the LaTex Isn't working.

    First Combine the Resistors in Parrallel:
    1/R =(1/R_1)+(1/R_2)+...(1/R_N)
    1/R = (1/6 Ohm) +(1/3 Ohm)
    1/R = (1/6 Ohm) + (2/6 Ohm)
    1/R = (3/6 Ohm) = (1/2 Ohm)

    1/(1/2 Ohm) = R = 2 Ohm

    So now you have
    ---||----^^^^---^^^^-|
    |___________________|

    Sorry for the bad drawing
    ( --||-- = voltage source, ^^^^ = resistor)


    Now combine the resistors in series
    R= R_1 +R_2+...R_N
    R= 3 Ohm + 2 Ohm
    R = 5 Ohm

    So now you have

    ---||---^^^^--|
    |____________|

    Using V = IR you can solve for the current:
    V=IR
    (20 Volts) = I * (5 Ohm)
    I = 4 Amperes

    Now split the circuit back up so that it is 2 resistor in series.
    Resistors in series have the same current
    ---||----^^^^---^^^^-|
    |___________________|

    So for the first resistor (3 Ohm resistor)
    V = IR
    I = 4 Ampere
    R = 3 Ohms

    V = (4 Ampere)(3 Ohm)
    V= 12 Volts
    Voltage on the 3 Ohm resistor directly right of the independent voltage source is 12 volts


    Now for the second resistor (2 Ohm Resistor)
    V = IR
    I = 4 Ampere
    R = 2 Ohms

    V = (4 Ampere)(2 Ohms)
    V = 8 Volts
    Voltage on the 2 Ohm resistor directly right of the 3 Ohm resistor is 8 volts.

    Split it up so you are back with the circuit you were first with.
    ---||---^^^^-.----^^^^--|
    |___________|----^^^^--|

    Resistors in parrallel have the same voltage so:
    6 Ohm Resistor in Parrallel:

    V=IR
    (8 volts) = I * (6 Ohm)
    I = (4/3) Amperes

    3 Ohm Resistor In parrallel:

    V= IR
    (8 Volts) = I * ( 3 Ohm)

    I = (8/3) Ampere

    So,
    Resistor 1:
    Resistor(R)= 3 Ohm
    Voltage Drop(V) = 12 Volts
    Current(i) = 4 Amperes

    Resistor 2:
    Resistor(R) = 6 Ohm
    Voltage Drop(V) = 8 Volts
    Current(i) = (4/3) Ampere

    Resistor 3:
    Resistor(R) = 3 Ohm
    Voltage Drop(V) = 8 Volts
    Current(i) = (8/3) Ampere
     
  6. Err, fallen186, you may want to read the rules over real quick. Specifically this line:
     
  7. Whats the point of that rule?
    Anyway back to fallen186´s answer.
    Thank you.
    That helped me as well.
     
  8. So the members here don't do your homework
     
  9. Hahahaha that made me smile
    Oh well every forum has its own rule.
    But still its not like he is posting his whole paper or HW here, he is only posting one question so wouldnt that be something else?
    I mean mostly you can only find how something works by knowing the answer and the exact way of getting there.
    Hence you need the full explanation?
     
  10. Yeah he was very helpful to me too. I deff learned how to do the problem and thats what i wanted. Thanks man!
     
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