Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Why does an element in a circuit cause a "voltage drop"

  1. Jul 14, 2015 #1
    I have seen this topic in other threads before, but I have not found an answer that eliminates my confusion. I know that electric potential is defined so that only the position of a charge with respect to a field determines the quantity of the electric potential (voltage). This allows us to quantify, generally, how spacial orientation with respect to a field affects the specific potential energy of a specific charge with some amount of coulombs. This all being said, I don't understand why there is a "voltage drop" after charged particles enter an element (such as a resistor) in a circuit. If we say that a battery creates an electric field and thus well-defined electric potentials around the field, then how come there is a drop in electric potential after a particle enters a a resistor, if that charged particle is roughly the same distance from the battery which the electric field comes from? I've heard people say that there is a voltage drop because the charged particles lose energy in the resistor, but if voltage is electric potential and not electric potential energy, then how does the fact that they lose energy affect electric potential (voltage), which is defined as the amount of potential energy per coulomb? I think my confusion lies in the concept of voltage. Basically I don't know how voltage is supposed to drop when there is a definite, constant amount of voltage for the battery.
     
  2. jcsd
  3. Jul 14, 2015 #2

    phinds

    User Avatar
    Gold Member
    2016 Award

    Suppose you have three resistors of equal value hooked in series with each other and with the battery. Would you expect the voltage across the middle resistor to be identical to the voltage of the battery? This would imply that the other two resistors are acting not like resistors but like wires. Ohm would be upset.
     
  4. Jul 14, 2015 #3

    Drakkith

    User Avatar

    Staff: Mentor

    My understanding, with some not-very-accurate terminology:

    Remember that the charges all interact with each other. Charges moving through the resistor require more energy to move than charges in the conductor. You could say that the charges 'pile up' on one side of the resistor until the current flow through the conductor and the resistor are equal, and that the current is mostly limited by how many charges per second can pass through the resistor.

    So the distance from the terminals doesn't matter since the charges just move around in response to each other's electric fields as well as the battery's.
     
  5. Jul 14, 2015 #4
    You have constant water pressure delivered to you house from the regulator provided by the water company. The pressure is all the same, throughout your house--34 PSI, until someone turns on the water.

    You're taking a shower, and someone turns on the dishwater. The water flow drops because the pressure at the nozzle decreases. (It is an unstated universal law, that the water always gets colder, but that's another story.)

    This is because there there is a restistance, from the sides of the pipes, to impede the water from flowing. The more water that is in demand, the lower the flow rate, and the less pressure.

    Were you asking about water pressure, right?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Why does an element in a circuit cause a "voltage drop"
Loading...