1. Transatlantic

    Thread Starter Member

    Feb 6, 2014
    I'm having trouble trying to understand voltage. From my understanding, it is the amount of energy required to move one Coulomb of charge from one point to another (J/C). So if you have a voltage of 9v across a resistor, then you are saying that it takes 9 joules of energy to move one Coulomb of charge through that resister. If the voltage is 10v, it takes 10 joules to move the one coulomb and so on


    But this raises two questions for me with the above circuit

    1) why is it that no matter what the value of the resistor is, it always takes 9 joules of energy to move one coulomb of charge through it? If the resistor value is increased, then surely it would take more energy to move the coulomb of charge through it? (I know this is ohms law, but I want to understand what the electorons are doing)

    2) why is it that all the voltage is dropped before reaching the end of a loop (KVL) ? or rather, why do the electrons always lose all their energy before coming backing around to the positive terminal? howcome they don't retain some of their energy when there is little resistance?

    I've tried to answer the first myself :

    1) As the voltage is fixed (meaning that each electron has the same potential energy no matter how complicated or how much resistance the circuit has), a higher resistance does not affect the amount of energy it takes to move an electron through the resistor, but rather, how long it takes to move through the resistor (I = V/R).

    A higher resistance means a lower current, in that fewer electrons are passing through the resistor per second, thus a lower C/S rate
    A lower resistance means a higher current, in that more electrons are passing through the resistor per second, thus a higher C/S rate

    In both cases - the same amount of work is done to move one coulomb. Just over a different time range.

    By increasing the resistance, you're reducing the current, thus decreasing the amount of energy that can be consumed from the battery per second (P = IV)
  2. crutschow


    Mar 14, 2008
    I think your answer to question 1 is basically correct.

    For question 2):
    The potential energy is determined by the voltage. After passing through the resistor, the voltage with respect to the negative terminal is zero, thus so is the energy.
    Hypatia's Protege likes this.
  3. Hypatia's Protege

    Distinguished Member

    Mar 1, 2015
    I suggest that your conceptual difficulties stem from misapprehension of 'voltage' as energy -- same is properly regarded as force...

    Sincere kudos on your initiative manifest in your admirable insistence upon comprehension (as opposed to blind acceptance):):):)

    Best regards
  4. TheButtonThief

    Active Member

    Feb 26, 2011
    I think you're simply having difficulty understanding the relationship between voltage, current and resistance. For any conductor of electricity (not just resistors), the ratio between Potential Difference (V) and current flowing (I) equals it's resistance (R), so try to think of volts simply as the unit of measurement for the change that happens in a circuit when the current tries to pass a resistance.

    Remember also that power (the energy in a circuit) is not solely proportional to voltage, power is (instead of a ratio between volts and amps) the product of volts and amps. Resistors limit the current flow in a circuit (not the voltage) so increase the supply voltage in your batter-resistor circuit and you'll increase the work done by the resistor.

    With regards to your second question, I think the nature of "the voltage drop" would make more sense to you once you start looking at the same circuit but with more than one resistor in series. You'd find that the supply voltage is divided between all the resistors but the value of each resistor will determine how much it's share of the voltage is, or how much of a drop in voltage it creates for itself. This deduction in voltage for each circuit component is subtracted from the supply voltage and once all other voltage drops are subtracted, the result must be 0. This is because we're talking in terms of potential difference (which is merely measured in volts) and the difference (in volts) between the positive end of a 9V battery and the negative end MUST be 9V.

    Hope this helps a little
  5. Transatlantic

    Thread Starter Member

    Feb 6, 2014
    When I think I understand the first issue, it conflicts with my understanding of the second issue and vice versa :(

    When I invision this circuit, I imagine a chemical reaction in the battery that causes a build up of electrons at the negative terminal of battery. This build up of electronics causes them to want to repel from the negative terminal of the battery to the positive. The amount of repulsion they experience is the force that governs how much energy they have. So the higher the repulsion, the more they want to move away, so the more energy they have.

    A 9V battery causes a higher repulsion at it's negative terminal than a 1.5V battery, which means an electron eminating from the 9v negative terminal has more energy than that of an electron from a 1.5v battery negative terminal.

    As the electrons move through the circuit, they lose energy due to resistance and other collisions. In this case, they lose energy through the resistor. So an electron has more "drive" to get to the positive terminal before the resistor, than after.

    But why do they lose all their energy? Why doesn't how much energy they lose depend on how much work they had to do to get through the resistor?

    If I had 2 circuits, one with a 10 ohm and one with a 100 ohm, why wouldn't the electron lose less energy when passing through the 10 ohm?
  6. Jony130

    AAC Fanatic!

    Feb 17, 2009
  7. crutschow


    Mar 14, 2008
    It does lose the same energy. It equals the electron charge times the voltage drop across the resistor.
    The value of the resistance does not enter into the equation.
    Last edited: Jul 28, 2015