Parallel circuits school work

Discussion in 'General Electronics Chat' started by Dorrance, Sep 30, 2012.

  1. Dorrance

    Thread Starter New Member

    Feb 19, 2011
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    I am a Devry student and now am working on simple parallel circuits.
    I’m not looking for easy answers just a better understanding of what I am doing.
    My problem is if I have a 10V power supply and I hook it to one side (+)of lets say 3 resisters and the other side (-) to the opposite side of those same resisters, when I measure from power supply (+) to the (-) side of the resistor I get 10V?
     
  2. wmodavis

    Well-Known Member

    Oct 23, 2010
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    150
    You appear to have made a statement but ended it with a question mark. The way you stated it was not a question.

    Are you asking - "If I connect a 10V power supply across three parallel resistors will I measure 10 volts across those three resistors?"

    If that is what you are asking the answer is YES most likely but you can get other results depending on the power supply and the value of the resistors. If that is not what you are asking then you need to state your question more precisley.
     
  3. crutschow

    Expert

    Mar 14, 2008
    13,028
    3,237
    Below its current limit rating a power supply will always provide whatever voltage it is set at. The current output will change, depending upon the load, but the voltage will stay essentially the same.
     
  4. JMac3108

    Active Member

    Aug 16, 2010
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    66
    Dorrance,

    Yes, if you place several resistors in parallel, then the votlage across them is the same. If they are connected directly across a 10V power supply, then they will all have 10V across them. Its the current that splits between the resistors. A smaller value resistor gets more current, and a larger value resistor gets less current.

    Lets use easy numbers. You have a 10V power supply across three parallel resistors, R1=10ohms, R2=20ohms, and R3=30ohms. The current through each is:

    I1=10V/10ohm=1.0A
    I2=10V/20ohm=0.5A
    I3=10V/30ohm=0.33A.

    And the total current is I1+I2+I3 = 1A+0.5A+0.33A=1.83A.

    Now if instead we calculated the equivalent resistance of the three resistors in parallel, then found the current through that equivalent resistance, we would get...

    Req= 1 / [ (1/R1) + (1/R2) + (1/R3)] = 5.45ohms

    Then the current is: I = 10V/Req = 10V/5.45ohms = 1.83A

    Same answer of course!

    This is called a current divider. Do you have a handle onhow a voltage divider works? If you're going to work in electronics, then a good understanding of voltage and current dividers is essential.

    Hope this helped with your question.
     
  5. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    With resistance in series just add the resistance R + R + R

    With resistance in parallel just add the reciprocals 1/R + 1/R + 1/R to find the effective resistance.
     
  6. Austin Clark

    Member

    Dec 28, 2011
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    Incorrect.

    You add the reciprocals of the resistors resistances to find the total conductance, you then take the reciprocal of THAT, which gives you your equivalent resistance.
    Basically, conductance is the multiplicative inverse of resistance.
     
  7. Kermit2

    AAC Fanatic!

    Feb 5, 2010
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    I though someone smart enough to start learning electronics would understand the implications of basic mathematics adding a series of reciprocals will give an answer which is a reciprocal

    But thanks for clearing that up.
     
  8. Austin Clark

    Member

    Dec 28, 2011
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    That depends on the answer you're looking for. Who's to say you weren't just looking for total conductance? In that case the answer wouldn't be a reciprocal. Either way, It's important to be as clear as humanly possible when teaching anyone anything, regardless of how smart they "should" be.
     
  9. Dorrance

    Thread Starter New Member

    Feb 19, 2011
    18
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    To all that wrote I really appreciate all you have written. Between school, pregnant wife, and work I just haven’t had time to write. Hopefully I can get back on soon.
    I didn’t just want to leave everyone that helped me hanging so once again thank you.
     
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