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Why do resistors always cause a voltage drop equal to the voltage applied?

  1. Mar 17, 2015 #1
    Consider a wire connected to a battery. Now,potential is analogous to the energy of the particles.And potential in a resistor drops because of the friction inside the resistor(considering there is no friction along the wire and outside the resistor).So the friction determines how much energy is lost.So why does the voltage drop to zero when the current passes from the last resistor it encounters along the wire? I mean,if there are two resistors and you have a battery of 6V,why do the resistors ALWAYS drop the voltage from 6V to 0V?Each resistor should cause a voltage drop analogous to the loss of energy due to the friction.I mean that,sometimes I(the current) should become zero(friction in resistors cause more friction than 6V-considering the convention from voltage to work W=q*ΔV) or the voltage will not reduce to zero(friction in resistors cause energy loss less than 6V).
     
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  3. Mar 17, 2015 #2

    NascentOxygen

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    The voltage at the points where you connect the resistor is set by the battery, one end being +6v more than the other. When a resistor is connected between those points the battery maintains the points at +6v and 0v, resp. It is this potential difference which determines the current, controlling it at exactly V/R amperes. No more, and no less.

    One volt = 1 Joule per coulomb, so the battery gives a fixed energy to each moving charge and the circuit resistance extracts just whatever energy each charge has, turning it into heat. The electrons that get to the +6v end of the resistor are not the same ones that entered the 0v end. As a result of energetic pushing and shoving that goes on, a different electron emerges into the external circuit.
     
  4. Mar 17, 2015 #3
    So i thought of it as I creates the drop of voltage is the resostor,whereas it is the opposite.It is the voltage that determines the current,and with its turn the current determines the voltage drop in the resistor.right?
     
  5. Mar 17, 2015 #4

    NascentOxygen

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    That's the better way of thinking of it.
     
  6. Mar 18, 2015 #5
    thanks @NascentOxygen .But,this all happens when the system reaches a steady-state.Before the steady-state,let's say the FIRST time that the current tries to perform a loop,the battery does not now how much energy to give to each charge(because the battery does not KNOW from the beginning how much energy to give to each charge).So when the charges go on their first loop,how do we know that the energy that they have is enough so that they get through the wire?
     
    Last edited: Mar 18, 2015
  7. Mar 18, 2015 #6

    NascentOxygen

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    The battery does know. It gives each charge the same energy---that's determined by the battery voltage. Each coulomb of charge gets one joule per battery volt, because the definition of a volt is one joule per coulomb.
     
  8. Mar 18, 2015 #7
    thanks a lot!
     
  9. Mar 18, 2015 #8
    I think you're giving the battery too much credit. A 6V battery is essentially just a storage device for a 6V potential difference. If you connect the two ends it will equalize through whatever you're connecting them with (the ultra-simplified definition of a DC circuit lol). To illustrate, just take any battery and measure it with a DC voltmeter, then flip the leads. I've had folks call me in a panic because their battery showed "negative voltage" when it's just the difference between electron and conventional current. The important fact is that it's simply the potential of electrical work to be done.

    If your 6V battery is wired with a single resistor, assuming negligible resistance of the circuit conductor itself, a DCV measurement between the negative terminal of the battery and the circuit after the resistor will indeed be 0V since there's no difference between them, whereas going from that point to the positive will give you your 6V measurement. The battery doesn't necessarily know or sense anything; when a means for the potential difference from the "charged" side to the "empty" side to equalize is presented electrons flow along that path until they are equalized, ie no difference in electric force potential. The resistor (or whatever else is in your circuit) is what it has to pass through on the way, with the relationship V=IR if the circuit is purely DC. So if we connect the 6V battery through let's say a 10 ohm resistor, it will create a current of .6 amps. If you make it a 20 ohm resistor, the current will drop to .3, whereas if you make it a 5 ohm resistor the current becomes 1.2.

    And if it's an issue of work, wattage=volts*amps, thus also W=(I^2)(R)
     
  10. Mar 18, 2015 #9
    Why does a roller coaster going downhill back to ground lose the same potential energy it gains when it goes uphill to the maximum height?
    This may help.
     
  11. Mar 19, 2015 #10
    yes,i already know what you are saying,but i am asking for a precise microscopical analysis.Your analysis is the easy macroscopic analysis that everyone knows.
     
  12. Mar 19, 2015 #11
    Yes,i understand this,but again,i am looking for a microscopic analysis that tells me WHY is the energy that the resistors turn into heat the same as the voltage difference of the battery(provided you multiply it by the charge)?
     
  13. Mar 19, 2015 #12

    nsaspook

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    I'll give it a try.

    The batteries chemical reaction causes charge separation (ion movements in the electrolyte to each plate) inside the battery that creates an electric field that contains that energy. Once the voltage potential of the field from that charge separation from anode to cathode reaches equilibrium with the voltage of the battery cell reaction voltages the reactions stop as there is no place for energy to be stored and no path for charges to flow (electrons) until an external circuit is made. When a path for electron flow is connected (wire and resistor) across the battery we now also have charge separation on the terminals of the resistor and an electric field across the resistor meaning the resistor (and the wire due to very small surface charge) are no longer electrical neutral so the charges flow to neutralize the fields across them. With charge flow we now have an electric and magnetic field at points in the circuit. If we assume the 'wire' is a good conductor from the 'battery to resistor' the electric/magnetic field power product across those points will be close to zero (low voltage, current with slow speeds, large number of electrons with few collisions) but across the resistor we have the full electric field from the battery and the magnetic field from the charge flow limited (full voltage, same current with faster speeds, smaller number of electrons with many collisions) by the value of the resistor. We now have EM energy (power) moving from the battery electric field (causing chemical energy reactions to restart to maintain voltage equilibrium as charge flows around the circuit), as mainly magnetic fields across the length of both wires and then to the resistor with both fields and power (Watts) into the resistor as Joule_heating
     
    Last edited: Mar 19, 2015
  14. Mar 19, 2015 #13
    If i understood correctly,your are saying that the energy of the battery is not drained by the wire(approximately) and so the whole energy of the battery(say 6V)are drained in the resistors.
    But,there is a problem with that,which is the core of the question of this thread.Why is the exact energy of the battery drained by the resistor?The first time the electrons move through the circuit,they come across the resistor and each charge has some finite amount of energy.Now,the resistor causes collisions inside(much more than outside is the wire is a good conductor) and thus loss of energy.So how do we know that before the system comes into steady-state(current does not change),the energy that a single charge needs to get through the resistance is exactly the voltage of the battery?(assuming that the circuit has only got one resistor)
     
  15. Mar 19, 2015 #14
    I thought your answer again and something bothers me.Say,the first time that the charges try to make a loop in a circuit,you said that each charge has 1Joule/Voltage of the battery.So,how do we know that that precise energy that the charge has is just enough so that it gets it through the resistor?
     
  16. Mar 19, 2015 #15

    sophiecentaur

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    I think you need to break this down into several different things going on. If you want to know how electrons, passing through a conductor, transfer thermal energy to the molecules of the conductor then you need to look at the quantum mechanics of the process. You can choose whether or not to go into that "microscopic process"! Suffice to say that the energy is conserved, one way or another. PD to Heat out per Coulomb.
    There is also the simple logic that a battery will supply a certain amount of energy to each (+) charge that leaves it. The current that flows will depend on the so-called 'resistance' of the resistor and you can guarantee that charges flowing back into the negative terminal will have no energy left (it they were to have any more then simply more of them would flow until they didn't). Every real battery (or any electrical power source) has limited efficiency and there is always some energy dissipated inside it when energy is supplied to a load (described as its Internal Resistance - and it's never '0') this fact 'makes things right' at the end of the process. If you somehow imagine the charges emerging from the end of a wire with some energy left, this would have to be dissipated in the contact resistance / air gap.
    You dismissed the model of a roller coaster, earlier on, but a mechanical analogy is valid. If you used the potential energy of the coaster to heat up its brakes (perhaps) . If it got to the bottom of the track and was stationary then all the gpe would have been used up. If it was still moving (corresponding to charges still having some energy) then it would dissipate its energy running into the buffers. But it would still have lost all the gpe it had at the top (which it had been given by the lift engine (corresponding to the battery chemistry).
     
  17. Mar 19, 2015 #16

    sophiecentaur

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    Where would the imbalance come from or go to?
    Having just read that, would I be right in thinking that you are expecting the charges to steadily speed up or slow down if there was not equality each time they went round the loop?
     
  18. Mar 19, 2015 #17

    nsaspook

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    You can't look at one part in isolation. The total energy travels not in charge from the battery (the electrons in the wire have very little kinetic energy) but in the fields of the circuit that cause electron acceleration (kinetic energy into the resistor electrons from those fields) in the resistor causing it to heat from collisions.
     
  19. Mar 19, 2015 #18
    During steady-state,no the charges have non-changing speed.Before that though,the current is changing(read it in a book of solid state physics).
     
  20. Mar 19, 2015 #19
    Yes,i understand what you are saying.I have thought of another way of seeing things.What about saying that potential energy means coulomb forces acting on charges,which in order to get the charges past the resistor,it has to do work equal to the energy dissipated by the battery(V=IR),so the work done by the field is equal to the energy dissipated by all the resistors in the circuit.But,i come to the same paradox in this way of thinking.Why does the resistor dissipate as much energy as the voltage difference of the battery?
     
  21. Mar 19, 2015 #20

    nsaspook

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    The resistor only dissipates the amount of power given by the voltage (electric field energy) and current (magnetic field energy) product. If the voltage is steady across the resistor then the value of resistance determines the current and the power delivered into the resistor for heating. Maybe you need to change your way of thinking.
     
    Last edited: Mar 19, 2015
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