Voltage Drop

Discussion in 'General Electronics Chat' started by aje8127, Sep 21, 2007.

  1. aje8127

    Thread Starter New Member

    Sep 21, 2007
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    When it is said that the voltage "drops" across a resistor, does this mean that the voltage actually decreases after going through a resistor? Then shouldn't the current flowing after the resistor also decrease in proportion to the decrease in voltage? I'm kind of confused about voltage drop and Kirchoff's voltage law. I just don't get what your actually doing when you calculate the voltage drops across resistors in series. Are you calculating how much the voltage decreases after going through each resistor? If so, then after going through all resistors in series is there no more emf left to push the current? Any help would be greatly appreciated.

    Anthony
     
  2. beenthere

    Retired Moderator

    Apr 20, 2004
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    All the potential difference is still there to propel electrons through the resistor string. With series resistors, though, the total resistance in the circuit is equal to the sum of the individual resistances. Using I = E/R, we can find the total current in the circuit. Using E = IR, we see that the amount of voltage "dropped" across each resistor depends on its value in ohms. When we add the individual voltage drops together, the sum equals the total voltage in the circuit.
     
  3. steeve_wai

    Active Member

    Sep 13, 2007
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    i have some doubts regarding voltage myself(read my posts if you care).but i can tell you one thing THE battery/voltage source does work to maintain potential difference...so the potential at which charge carriers are will decrease.the potential difference across the resistor is maintained by the battery...i hope that helped.
     
  4. aje8127

    Thread Starter New Member

    Sep 21, 2007
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    This is how I understand voltage thus far. Work is required to cause a difference in the amount of electrons on two bodies. When a conductive path is given, the electrons will flow to the body with a deficiency of electrons and perform work on any resistance within the circuit. A battery is constantly doing work to maintain a potential difference. Here's what I don't get:

    If I have a battery with an electrical potential of 10 Joules per Coulomb (volts) and I have a 2 Ohm resistor (R1) and a 3 Ohm resistor (R2) in series, there will be a current of 2 Amps flowing through the circuit. In the 2 Ohm resistor there will be a drop in electrical potential energy of 4 Joules per Coulomb. Therefore, the voltage right after R1 should be 6 Joules per Coulomb. Then, the current will go through R2 and the voltage should drop to 0 Joules per Coulomb. My question is, if there is 0 Joules per Coulomb after R2, how does the current make it to the positive terminal of the battery? It just makes sense that after going through each resistors, the potential energy of each charge (the joules per coulomb; volts) should decrease.

    Anthony
     
  5. beenthere

    Retired Moderator

    Apr 20, 2004
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    Say R1 is tied to the negative terminal and R2 to the positive. There is no potential difference between the R1 lead and the negative terminal it is connected with, nor is there any between the lead of R2 tied to the positive terminal and the positive potential. The two potential drops are across the resistances of R1 and R2. Work takes place as the electrons flow through the resistors and expend kinetic energy, and in the source to maintain the potertial difference between the terminals.
     
  6. ixisuprflyixi

    Active Member

    Sep 16, 2007
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    Technically even a copper wire conductor is slightly resistive and thus if you were to measure the voltage across the wire you would (most likely) have a voltage drop from one end of the wire to the other end of the wire (or in this case from the lead of R2) the reason you do not see this "voltage drop" is because the wire is such a very small resistance the picovolt that you would measure is outside of the range of most DMM's however if you check with a good Oscope you will probably see that small voltage. So to answer your question there are small (very small indeed) voltage drops from one end of the circuit to the other but the voltage drops across the resistors are so much more pronounced than the voltage drop on the wire we take more notice of the resistors than the resistive effects in the wire. Hope that helps.
     
  7. aje8127

    Thread Starter New Member

    Sep 21, 2007
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    Thank you everyone for all of your answers. I now understand that the voltage does indeed drop throughout every part of the circuit including the wire. The wire resistance is negligible so we don't account for it. The only question I have now is, if the voltage decreases after going through a resistance, then shouldn't the current it can handle also decrease? Then wouldn't this contradict the fact that current is the same at every point in the circuit (the current would be less after going through the resistance)? Thanks again.

    Anthony
     
  8. thingmaker3

    Retired Moderator

    May 16, 2005
    5,072
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    Current is measured in coulombs per second (Amperes).

    The Electromotive Force (Volts) is "dropped" across each resistance, but Ohm's Law shows that current does not change. (Ohm's Law agrees with Kirchhoff's Current Law.)

    It takes a certain amount of force (electromotive force) to overcome each impedance. The series current is what gets pushed through ALL of the series impedances by the electromotive force.

    I don't actually think of a "drop" as a "decrease." I think of it as the portion of the total voltage which is required to push the charge carriers through a given part of the circuit. We "drop" so many volts here and "drop" so many volts there so that they may push charge carriers for us. We don't "lose" them, we "use" them.

    I truly hope I've not added confusion here.
     
  9. niftydog

    Active Member

    Jun 13, 2007
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    The current flow is the reason for the voltage drop, the two are inextricably linked. So, yes, given the same resistance, less voltage means less current. But the conditions in the circuit dictate how much current flows and how much voltage is therefore present.

    It's important to realise that voltage is not a substance, it's a state, or a condition that is caused by current flow. Current is a substance - electrons moving through wire, and the flow of electrons can be measured by the voltage they create as they go. The voltage decreases because of the current flow - the two are inextricably linked.

    Sorry if that seemed obvious to you, I just got a feeling that perhaps you're not visualising things quite right.

    As for current being the same in every part of the circuit, well, that is not entirely accurate and it's ignoring a lot of things that go on in the meantime. It's more accurate to say that the current leaving a battery to travel through a circuit is the same as the current returning to the battery. Current can be branched off many times through that circuit, but the sum of all those branch currents always equals the total.
     
  10. steeve_wai

    Active Member

    Sep 13, 2007
    47
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    the force due to electric field accelerates the electrons.this causes a gain in kinetic energy which is lost to ions by collision with the metal "ions".the average kinetic energy remains at a value (1/2)mv^2...where v is the drift velocity of the electron/charge carrier.this enables them to move in a circuit
     
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