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Serial Resistor solution to finding individual voltage not adding up

  1. Apr 27, 2012 #1
    Ok. im adding this up time and time again and its just not making no sense. Im sitting here trying to brush up on technical math skills this week before i get my soldering iron so i dont mess nothing up. and here i am i cant understand the basics.

    I have a 20volt power supply and 4 resistors: R1, R2, R3 and R4. R1 = 200Ω R2 = 700
    R3 = 300Ω R4 = 400 Ok what i did was use ohms law and all the other crap. so anyways in order to determine how much voltage is going through each individual resistor i need to multiply current by resistance. I did that to each one and when i get done i keep coming up with 19.68 rather than 20. Can someone please help me figure this out before i go and blow a circuit board or something one day not knowing basic stuff? OH yeah and the current running through the circuit is 0.0123 milliamps
     
  2. jcsd
  3. Apr 27, 2012 #2
    IM SORRY, even though its the subject itself, i forgot to mention they are in serial, not parallel. :)
     
  4. Apr 27, 2012 #3

    Integral

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    You specified the banded resistance of your resistors, did you actually measure them? Remember they have a tolerance and can be off the banded number by up to 5%. There will always be some measurement error so do not expect to be perfect. It is good enough if you measure inside of 5%. The way I compute it you have about 1.6% error... Looks good to me.
     
  5. Apr 27, 2012 #4
    thanks alot. i just realised that i should of posted this under elec. tech what you say makes sense. does this still apply though if its just a simple problem though? I mean im just makin up problems and random to make sure i got the hang of it. I mean should i still have a difference in potential energy when doing it on paper versus actually having resistors on a circuit?
     
  6. Apr 27, 2012 #5

    NascentOxygen

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    Are these actual resistors, or are they just values you invented on paper to use in sample calculations?
    The theoretical value for the exact resistance values you listed is 0.0125A, so there you have a 1.6% departure from theory to start with.

    It is not clear to me whether there is any electronic hardware involved. Integral is apparently privy to more information than is available here:
    Those are definitely not run-of-the-mill preferred resistor values. :wink:
     
  7. Apr 27, 2012 #6

    Integral

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    It applies if you are using real resistors in a real circuit and you are using a real meter to make measurements. Have you measured the actual value of your resistors?

    To N.O.
    Keep in mind that he is not a engineer working in a multinational environment designing circuits for commercial value. To a high school student preferred resistor values are usually multiples of 100 for easier computation.
     
  8. Apr 28, 2012 #7

    vk6kro

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    The total resistance is 1600 ohms. (200 + 700 + 300 + 400 = 1600)

    The current is 0.0125 amps, not 0.0123 milliamps (20 volts / 1600 ohms = 0.0125 amps)

    The voltages are 2.5, 8.75, 3.75 and 5 volts which add up to 20 volts.

    So, you must be making a calculator error somewhere.
     
  9. Apr 28, 2012 #8
    thanks. maybe i just need to start doing all my math on a calculator rather than adding it up on paper. Someone told me that im not rounding right or something. Whew, i knew i should have never tried to refresh upon my math skills. At 30 im having just as much problems graspin basic mathematical concepts as a 70 year old.
     
  10. Apr 28, 2012 #9

    Integral

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    What did you measure for the voltage drop across each resistor? What did you measure for the actual resistance?
     
  11. Apr 28, 2012 #10

    vk6kro

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    It was just one calculation that was wrong. The current should have been 0.0125 amps and after that everything works out OK.

    So, don't worry about it.
     
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