find unkown values in series circuit

Discussion in 'Homework Help' started by the_fundamental_freq, May 7, 2011.

  1. the_fundamental_freq

    Thread Starter New Member

    May 7, 2011
    8
    0
    hi everyone,

    have been given this circuit as revision for an upcoming exam, and really cant remember where too start (may just need a bit of memory jogging). its a series circuit so the '20mA' stated would be the total current? also i think that the way to work it out would be to use p=i^2/r (as 'P6) is given but is this not the total power dissipation?
    i realize im a total noob on here and im sorry if i have annoyed anyone by asking for help on my first post! (will try and help others if i can)

    many thanks

    link to image: http://postimage.org/image/1v373e2mc/
     
  2. jegues

    Well-Known Member

    Sep 13, 2010
    735
    43

    You can start with the middle branch. First you should find the voltage drop V2.

    After that, you know that R3=R4, so call it R, they are in series so we can combine them as 2R.

    Now write a KVL in the loop on the RHS and solve for the required voltage drop across the combined series resistor 2R and thus solve for R.

    This should be enough to get you started.
     
  3. the_fundamental_freq

    Thread Starter New Member

    May 7, 2011
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    0
    ah im with ya i think?
    v2= 2V
    R 3&4 are 285 Ohms
    V 3&4 5.7
    R5=330ohms
    thus (excluding R1&R6) Rt=1kOhm and and tot V drop 20v

    would i then just follow the same principle for R1 & R6 ie combine to make 2R
     
  4. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    1,097
    V2 is equal 2V
    V3 = V4 = 5.7V OKi
    R5 = 330Ω

    As for R6, you know the power P = V * I = 0.112W
    But we also know the Ohm's law
    I = V/R and V = I*R so
    P = V*V/R = V^2 *R = I*R * I = I^2 * R
    Form this equation we can solve for R
    R = ?
     
    Last edited: May 7, 2011
  5. the_fundamental_freq

    Thread Starter New Member

    May 7, 2011
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    i^2/p=280 ohms?
     
  6. the_fundamental_freq

    Thread Starter New Member

    May 7, 2011
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    then
    v1=5.6
    r6= 220
    v6= 4.4?
     
  7. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    P = I^2 * R
    I divided by I^2

    \frac{P}{I^2} =\frac{I^2}{I^2} * R

    \frac{P}{I^2} =  R

     R = \frac{P}{I^2}=\frac{112mW}{20mA^2} = 280\Omega

    V6 = I * R = 20mA * 280Ω = 5.6V

    V1 = 30V - 20V - 5.6V = 4.4V

    R1 = V/I = 4.4V/20mA = 220Ω
     
  8. the_fundamental_freq

    Thread Starter New Member

    May 7, 2011
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    cool thanks for that, so i basically got the values mixed up from R1 & R6, how do you know which to do first?
     
  9. Jony130

    AAC Fanatic!

    Feb 17, 2009
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    Simply you need experience to be able to solve this type of problems.
     
  10. the_fundamental_freq

    Thread Starter New Member

    May 7, 2011
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    if there is no way to determine which to do first, is there anyway of checking after?
     
  11. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    1,605
    Checking after is the easy part. Since you should now have all the resistor values of this simple series circuit then you sum them, divide the 30V source by the resistance, and see if you get the 20 mA back.

    If you don't, try, try again.
     
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