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What is voltage drop?

  1. Nov 22, 2009 #1
    Is voltage the potential energy difference between source A and B in a circuit? If so in absense of resistance and the presence of a 3V battery does point A have 3V to begin with and at point B there is 0 volts. Is all potential energy converted to kinetic energy when moving from A to B? A is origin and B is the terminal end of the circuit by the way. Is that voltage drop?

    If what I have said is not correct and voltage doesn't drop like this does a circuit with no resistor and negligble resistance from the wires have 3V at point A and also 3V at point B?

    Let's say that I have a circuit like this (made it linear for ease)

    3V battery--point A----------------1ohm resistor-------------pointB

    What is the voltage of the resistor? It is 3 right. Does that when current passes through the resistor it loses 3V. If it loses all the voltage how does it move to point B?

    Thanks :smile:

    Edit: I think I'm assuming when they say voltage drops across the 1 ohm resistor is 3, the 1 ohm resistor uses all 3V. When they say voltage drop in this case are they actually referring to voltage drop from A to B. The actual amount of energy lost due to the resistor is not really 3V. It is something less. The voltage drop across the whole thing is 3 V. Is that right?
     
    Last edited: Nov 22, 2009
  2. jcsd
  3. Nov 22, 2009 #2

    Dale

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    Let's make this a little clearer:
    Code (Text):

    ┌─────┐
    a     c
    █     █
    █     █
    █     █
    b     d
    └─────┘
     
    Points a and b are the terminals of the battery and c and d are the terminals of the resistor. Between a and c and between b and d is a very conductive wire. If the voltage of the battery is exactly 3 V then a-b = 3 V. Now, there will be some voltage drop across each of the wires, but since they are very good conductors lets say that it is 0.0001 V. Then a-c = d-b = 0.0001 V. That means that the drop across the resistor is just less than 3 V: c-d = 2.9998 V. By far, the majority of the voltage drop is across the resistor and, if your measuring device is only able to measure millivolts then your device will read no drop across the wires and 3 V across the resisitor. With a sufficiently good conductor for your wire it is sufficient to ignore a-c and b-d and treat all points in the wires as being at the same voltage.
     
  4. Nov 22, 2009 #3
    Hey thanks :smile: but I still have some questions. Why should resitor voltage+wire resistor add up to 3. Is it because a-b = 3 V and b=0. If there is no resistor does all the voltage drop across the wire to make it 0. So in this case does any resistor (with any amount of resistance) have the same voltage drop?

    If volatage is P.E is it converted into K.E when going from A to B. Shouldn't this mean a voltage drop as well. Thanks!!
     
    Last edited: Nov 22, 2009
  5. Nov 22, 2009 #4

    fluidistic

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    Voltage is not electric potential energy, but electric potential difference.
    More precisely, [tex]\vec E = - \nabla V[/tex], hence [tex]V=-\int_a^b \vec E d \vec l[/tex]. The units of V are volts, while energy is always measured in joules (in the SI).
     
  6. Nov 22, 2009 #5

    jtbell

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    When you go completely around a circuit and return to your starting point, the total potential difference must be zero. In DaleSpam's example, let point b be the negative terminal of the battery. Start there and go clockwise. Then the potential differences are +3.0000 V (battery) - 0.0001 V (first wire) - 2.9998 V (resistor) - 0.0001 V (second wire) = 0.0000 V (total).

    The total potential difference can also be considered as (potential at final point) - (potential at starting point) = (potential at point b) - (potential at point b) = 0.
     
  7. Nov 22, 2009 #6

    fluidistic

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    I agree with this, and looking at [tex]V=-\int_a^b \vec E d \vec l[/tex], it's clear that if [tex]a=b[/tex], the [tex]V[/tex] is worth [tex]0[/tex].
     
  8. Nov 22, 2009 #7
    Ok thanks for the answers guys. While I'm thinking about what you guys are saying I have another question. What is power dissipiated?

    If the circuit has no resistor why is power disspiated so high and through what does the volatage drop(there is no resistor). If all energy is dissipiated as heat how can there be a current flow. What I mean is if dissipiation is turning in to heat how can there be enough energy for current flow? Thanks!!
     
    Last edited: Nov 22, 2009
  9. Nov 22, 2009 #8

    jtbell

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    Batteries have their own resistance (look in your textbook or Google for "internal resistance"). This is usually small enough that we can often ignore it when we analyze a circuit, but when it's the largest resistance in a circuit...

    Now consider [itex]P = V^2/R[/itex] and what happens when R is small while keeping V constant!
     
  10. Nov 22, 2009 #9

    Dale

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    jtbell's answer in post 5 is good. The only thing I have to add is that this rule is called Kirchoff's Voltage Law.
    Yes, with the additional caveat that jtbell mentioned about a real battery's internal resistance.

    In electric circuits there is never an appreciable amount of KE in the current. In most circuits the electron drift velocity is usually quite a bit less than 1 mm/s. With the low velocity and the low mass of the electrons the KE is negligible.

    The only exception that I know of is vacuum tubes.
     
  11. Nov 22, 2009 #10

    sophiecentaur

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    Current flow is not energy!
    The Voltage is, effectively, the energy. That is what gets 'used up' as you go from the positive terminal 'down' to the negative terminal. The same charge flows all the way round because there are immensely strong electric forces which ensure that every electron which moves form one atom to another will displace another one, further along in the circuit, so you cannot get a build up. Even in a battery or capacitor, you still get the same number of charges leaving or entering, so the current is the same all the way round. Like a bicycle chain, carrying energy from foot to wheel.
     
  12. Nov 22, 2009 #11

    fluidistic

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    Can someone confirm this? This would contradict the post #4. I'm starting to doubt.
     
  13. Nov 22, 2009 #12

    sophiecentaur

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    One Volt is One Joule per Coulomb.
    I am sure Wikkers will confirm that.
    There's yer energy. :-)
     
  14. Nov 22, 2009 #13
    If you could see what was actually happening inside the wires that would look like train carriges taking up the slack in their buffers ats they bump into each other. So you should in theory always have a flow of electricity after the power supply is turned off. Because the 'wave' can't move faster than light, it's not instantaneous. So I'd say there is always a build up, but you can't see it.
     
  15. Nov 22, 2009 #14
    electricity is just the kinetic motion of electrons isn't it ?
     
  16. Nov 22, 2009 #15
    http://ocw.mit.edu/NR/rdonlyres/Electrical-Engineering-and-Computer-Science/6-002Spring-2007/VideoLectures/6002_l1.pdf [Broken]

    See slides 6-18, it might help clear some things up.
     
    Last edited by a moderator: May 4, 2017
  17. Nov 22, 2009 #16
  18. Nov 22, 2009 #17
    Thanks guys but I'm still at square one. I don't think I expressed my question properly. I understand K.. law.

    00112.png

    Why do different resistances get different portions of the voltage. For example in 1--4 do have a voltage you need to have more electrons at 4 than 1. If the current (rate) flowing through each resistor is the same how can there be different electron differences. Basically why do different resistances have different voltage drops?
     
  19. Nov 23, 2009 #18

    Integral

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    Voltage is not about the number of electrons. It is about the energy available to move electrons. Outside of the battery electrons are not stored, gained, or lost anywhere in the circuit so the number of electrons passing any point in a time interval must be constant. In other words current is constant in a series circuit.

    Now do you know Ohms Law?

    E=IR

    This gives the voltage measured across a resistor given the current through it.

    In a series circuit the current is constant and the sum of the voltage drops is equal to the source voltage.
     
  20. Nov 23, 2009 #19

    sophiecentaur

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    There may an initial build up of charge in some parts of a circuit - as when a capacitor charges up - but this cannot continue. We were discussing 'current' electricity - in which a steady state has been reached in the flow through resistance only when there is no reactive component to store energy.
    If you want to show why there can't be a constant build up of charge then go and calculate the force involved when a charge of one Coulomb (that could be just 1/10Amp for 10 seconds)is brought 1m away from another 1 Coulomb. The analogy to a bicycle chain is still there; the chain stretches, initially, when you start pedalling and takes up the slack. This is equivalent to a small amount of polarisation in parts of the circuit.

    And NO: 'Electricity' is not the Kinetic Motion of Electrons. They only drift through wires at a few mm per second! 'Electricity' is not really a Scientifically defined term - it just covers the general subject of Electrical Energy, Forces, Current etc.
     
  21. Nov 23, 2009 #20

    sophiecentaur

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    They get different shares of the Voltage because they need different amounts of energy to shift charge through them. Once the situation has settled down (after 1ns or so) the KII situation will apply. The battery supplies just enough power to drive current through them all and each one dissipates just enough to balance the situation. If you short one of the resistors out, then the PD is shared out amongst the rest, after another ns and KII rules again.
     
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