Series circuits with missing R value

Discussion in 'Homework Help' started by mwootan, Apr 8, 2010.

  1. mwootan

    Thread Starter Member

    Apr 16, 2009
    15
    0
    I am struggling with this series circuit. A classmate and I are confused. I think It = 15Amps like the circuit shows but my class mate thinks it should read It = 1.5 Amps.
    My math:
    IR2 = E/R = 120v / 25Ω = 4.8A
    IR3 = E/R = 120v / 25Ω = 4.8A

    Combine IR2 + IR3 = 9.6 Amps
    Now subtract this from It. 15A - 9.6A = 5.4Amps.

    R1 = E/I = 120v / 5.4A = 22.222Ω so R1 = 22.222Ω

    Rt= 25Ω + 25Ω + 22.222 = 72.222Ω

    4.8A + 4.8A + 5.4A = 15Amps Right??????? or Am I wrong????
     
  2. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
    348
    What can you say about current in a series circuit? Is it the same or different for each component? After you have resolved those questions, are you sure of the values you posted? R2 and R3 are both Twenty five ohms? Total voltage is one hundred twenty volts and total current is Fifteen amps? Something doesn't add up.
     
    Last edited: Apr 8, 2010
  3. mwootan

    Thread Starter Member

    Apr 16, 2009
    15
    0
    Current is the same.
     
  4. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
    348
    Right! Current IS the same. Now if you use ohms law to calculate the total resistance based upon the total voltage and total current you have shown, do you see any conflict?
     
  5. mwootan

    Thread Starter Member

    Apr 16, 2009
    15
    0
    It = 1.6A to be correct.
    R1= 25 ohms
    Rt = 75 ohms
    75ohms x 1.6A = 120volts!
    Hows this figure.
     
  6. BillB3857

    Senior Member

    Feb 28, 2009
    2,400
    348
    If the total current value is 1.6 amps, then RT=75 is correct. By posting misleading information, it makes it difficult to assist. Ohms law for what you are working with a this time is a very simple 3 term equation. By rearranging the formula in order to use known values, the unknown is easy to obtain.

    As you originally posted, with 120V and 15A, the resistance would have been 8 ohms TOTAL. Since your known resistors added up to 50 ohms, you had an impossible situation.
     
  7. mwootan

    Thread Starter Member

    Apr 16, 2009
    15
    0
    Yes originally, I did post 120v and 15 amps. There must have been a miss print on this practice circuit with it saying 15 amps. I could not even get it to work with 1.5 amps. The calculations were still off.

    Figuring IR2 = E/R = 120v/25 ohms = 4.8A
    R1 = E/I = 120v/4.8A = 25 ohms
    Now figuring Rt = 25ohms + 25ohms + 25ohms = 75 ohms

    But checking my work. E = R x I = 75 ohms X 4.8A = would give me 360v (thats not the given 120v)

    I took the I / number of resistors = 4.8/3 that gave me 1.6A
    Now my calculations:
    Et (source) = 120v
    If, It = 1.6A
    R= E/I = 120v / 1.6A = 75ohms - subtract 50 ohms for the other two resistors and you get 25 ohms for R1.
    Then to check: E = R x I = 75 ohms x 1.6A = 120v the given voltage.

    The problem all together was messed up from the start I think!
     
  8. The Electrician

    AAC Fanatic!

    Oct 9, 2007
    2,281
    326
    It works out just fine with 1.5A. Using your calculation with the changes in red:

    R= E/I = 120v / 1.5A = 80ohms - subtract 50 ohms for the other two resistors and you get 30 ohms for R1.

    Then to check: E = R x I = 80 ohms x 1.5A = 120v the given voltage.
     
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