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Ohm's Law and series circuit

  1. Jan 28, 2004 #1
    In my physics lab we were asked a question for advanced studying. I read the chapter that it was supposed to be in, but I couldn't figure it out.

    Q: For a series circuit, what is the terminal voltage of a batter or power supply equal to in terms of the potential differences or voltage drops across circuit components?
  2. jcsd
  3. Jan 28, 2004 #2
    It's equal to the sum of the voltage drops.

    To rephrase the question: a battery of unknown voltage is connected to a series circuit containing three resistors. If the voltage drops across the resistors are 5 V, 2.5 V, and 1.5 V, what must the voltage of the battery be?

    The answer is 9 V.

    You can *kinda* create an anology between voltage for electric circuits and elevation for a water cycle: imagine the electrons in the circuit running 'downhill' through the various components. Resistors are like waterwheels mounted on vertical drops; they convert some of the 'height' (voltage) into heat energy.

    Batteries are like water towers; they come with one end full of electrons, and when all of those electrons have run 'down' to the other end, the battery is no good anymore. A DC transformer that you plug into the wall doesn't drain like a battery does cuz it's more of a vertical water pump than a tower. Of course, you can use a pump to refill a tower; that's a rechargable battery.

    Long explanation for a short answer. It's the sum of the voltages in the circuit.


    Attached Files:

  4. Jan 29, 2004 #3
    THanks, that makes a lot of sense. I guess I was just making the question harder than it actually was.
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