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Homework Help: Potential difference in a circuit

  1. Feb 17, 2010 #1
    1. The problem statement, all variables and given/known data
    What exactly happens to the voltage...

    in between things such as resistors, and bulbs?

    It isn't a homework question it is just something that is bothering me.

    2. Relevant equations
    V=IR for resistors
    Sum of V in a closed loop is zero.

    3. The attempt at a solution

    If there was a circuit that had nothing but a battery, would the potential difference everywhere be equal to the V of the battery?

    If there was a battery, and a resistor in a circuit. Then we know that V-IR=0. And we know that V=IR. So the voltage of the battery and the resistor are equal. Well what is the voltage in the wire between the the battery and the resistor?
  2. jcsd
  3. Feb 17, 2010 #2


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    Homework Helper
    Gold Member

    In terms of circuit diagrams, think of the voltage drop across a wire as being 0. In other words, the voltage on one side of the wire is the same as the other side of the same wire.

    Voltage can be thought of as potential energy per unit charge. And like energy, it's assumed to be measured with respect to something else. For example, you could define the potential energy at the top of a hill in reference to the potential energy at the bottom of a hill. In this case you define the bottom of the hill to have a potential energy of 0, and a mass at the top of the hill to have potential energy of mgh. Of course you could have made your reference point the center of the earth. Although it changes the values you assign to the hill, it no effect on the physical hill itself.

    In circuits, the ground terminal (or "earth" if you wish to call it that) is traditionally chosen as the reference. But it's always understood that voltage is a measure of "difference" between two points on the circuit.

    Don't take this analogy too far, but you could think of voltage as simply being sort of like two different points on the height of a hill.

    Well, that depends on how you draw the circuit. If the circuit is open (meaning there is no current flowing), then yes, everything connected to the positive lead would have V volts with respect to the negative lead. Everything connected to the negative lead would have 0 volts (with respect to the negative lead).

    Now if you connect a wire between the two battery's terminals, things get more crazy. If you have an ideal voltage source for a battery, and an ideal wire, then the current blows up to infinity. In practice, when using a non-ideal battery and non-ideal wire, the situation can be modeled such that the battery's terminal voltage is equal to the IR drop across the internal resistance of the battery and the actual resistance of the wire, such that
    V = I(Ri + Rw).

    In idealized circuits (often called lumped parameter systems), the voltage across a wire is always zero. If you wanted to model a wire with a non-zero resistance, just use a resistor in your circuit diagram that you write on paper.

    In real-world circuits, wires have resistance (well, ignoring superconductors). But the resistance is comparatively small. So you could model this (using a very small resistor in the circuit diagram). The voltage drop across it would I(Rw), but since Rw is very small, the voltage drop would be very small compared to the drop across the large resistor.

    And finally, a bit of trivia for you to think about, the whole usefulness of modeling things with these idealized circuits (lumped parameter systems) starts to fall apart, or at least make everything far more difficult when the highest frequency in the circuit is greater than roughly c/(100 L) where c is the speed of light, and L is the approximate physical size of the circuit. That's because at higher frequencies, wires start to act like transmission lines. This has implications for things like USB cables, antennas, long distance AC power lines, and one reason why microprocessors have a maximum clock speed.
    Last edited: Feb 18, 2010
  4. Feb 17, 2010 #3
    The voltage of the battery and the current times the resistance are equal. Voltage as pressure in a water tank is the example that comes to mind. Think of it as the pressure in the water tank doesn't go anywhere (it is the h2o that moves because of the pressure) but it just kind of dissipates in order to equalize itself. The same thing happens with voltage. All the voltage in a battery that is found on one of the terminals equalizes itself because electrical potential energy acts in the same way as gravitational potential energy i.e. it wants to go to a lower potential.
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