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Explaination of voltage drop across resistors

  1. Jun 16, 2010 #1
    1. The problem statement, all variables and given/known data


    I want to know that if there is a voltage drop across a resistor, why is it that voltage increases as resistance increases (Ohm's law)?

    2. Relevant equations


    None.

    3. The attempt at a solution

    N/A
     
  2. jcsd
  3. Jun 16, 2010 #2
    What do you mean by saying

    can you explain.
     
  4. Jun 17, 2010 #3
    In Ohm's law, resistance is directly proportional to voltage, but I want to actually know why voltage increases, as resistance increases.
     
  5. Jun 18, 2010 #4
    ohh are you asking similar to this :
    In a circuit with two resistors R1 and R2 connected in series to a voltage source V, if R2 increases why voltage across R2 increases?

    Well, the formula for voltage across R2 (say v2) is
    [tex] V_2=\frac{R_2}{R_1+R_2} \times V [/tex]

    From this relation, you can deduce that if R2 is increased, V2 is definitely going to increase.

    BTW Ohm's law says that (at constant temperature) current flowing in a conductor is proportional to voltage applied across it and the proportionality constant is 1/Resistance.
     
    Last edited: Jun 18, 2010
  6. Jun 18, 2010 #5
    No I ment like I understand why current increases, as voltage increases (cause you supplying more energy in a way which makes a stronger current). But I dont see how an increase in voltage leads to an increase in resistance.
     
  7. Jun 18, 2010 #6
    You don't get and increase in resistance by increasing voltage.

    Larger resistors have larger voltage drops. Why? Before we can answer that we must better understand what exactly a resistor does.

    A resistor resists the flow of current. This resistance means that some work must be done to "push" current through the resistor. Whenever work is done on charge, we have voltage. Thus, when current flows through a resistor, there is some voltage across the resistor.

    The larger the resistance, the more work required to "push" the current through the resistor, the more work done the larger the voltage drop across said resistor.
     
  8. Jun 19, 2010 #7
    Thanks for the explanation. And something which has been bugging me is say you had a circuit with no resistance and you wanted to calculate the current, how would this be possible since: I = V / 0 would be undefined?
     
  9. Jun 19, 2010 #8

    vela

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    The only hope to obtain a finite answer for the current would be if V=0. That's why in an ideal conductor, the voltage difference between any two points is zero.

    There are also effects like inductance and capacitance, which will keep the current from diverging due to a change in potential.
     
  10. Jun 19, 2010 #9
    How would this be possible since if V=0 then I = 0/0 would still be undefined?
     
  11. Jun 19, 2010 #10

    vela

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    Better to say indeterminate than undefined. You can avoid the division by zero if you write Ohm's law in its traditional form, V=IR. If R=0, the voltage has to be 0, but the current could be any value.
     
  12. Jun 20, 2010 #11
    You wouldn't be able to determine I using Ohm's Law right? Since a circuit must contain a resistor to obey Ohm's Law?
     
  13. Jun 20, 2010 #12

    vela

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    Right. Ohm's law doesn't tell you anything when the resistance is zero other than the voltage has to be zero as well.
     
  14. Jun 21, 2010 #13
    Realistically you would never have a case where the resistance is totally 0 would you? I mean, there would always be some infinitesimal ammount of resistance, no?
     
  15. Jun 21, 2010 #14

    vela

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    For your normal, everyday conductors, like copper wire, yes, there is always some resistance. Superconductors, however, do have 0 resistance.
     
  16. Jun 22, 2010 #15
    Voltage and current are like kinetic and potential energy, there must be conservation of energy. Therefore, as voltage increases current decreases and vice versa.
     
  17. Jun 25, 2010 #16
    Oh ok also say you had a circuit with only a +20V battery and one resistor connected in series with it to earth. Why is it that all 20V would be dropped across the resistor, even if R = 10 ohms or if R = 1k ohm?
     
  18. Jun 25, 2010 #17

    Zryn

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    We could analyse how the Voltage is slowly reduced over very small increments of wire resistance as if we are doing a voltage divider, but relatively, the resistance of wire is most likely negligable compared to the resistance of the resistor.

    For ease of calculations we assume the resistance of the wire to be zero.

    In real life this is not how it always works, for example in mines where machines operate underground on 415V sources, alot of the time generators produce ~433V because the length & resistance of the cables used drops a significant amount of voltage.
     
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