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Voltage in series circuits

  1. Jan 22, 2017 #1
    Say you have two separate resistors in a circuit. When you close the switch on the circuit is it the electric field that flows through the circuit that effectively sets the voltage drops so a bigger voltage drop occurs across the higher resistor in proportion to its resistance such that the current flow in the circuit is constant.
    I'm aware of W=Q V and so as a coulomb then flow through the larger resistance more energy is transferred there than in the smaller resistance.
  2. jcsd
  3. Jan 22, 2017 #2


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    It is the electric field that sets the voltage drops (without electric field we cant have voltage anywhere, though if you study more advanced electromagnetism, it is the scalar potential V that is more fundamental than the electric field) but you forgot to mention Ohm's Law V=IR which is the basic reason that the voltage drop is bigger in the higher resistor (since current is the same, with higher resistance R there must be higher voltage V due to V=IR).
  4. Jan 22, 2017 #3
    Would a useful analogy to maybe think of a scree slope which is already mapped out with a guy at the top with a certain amount of gpe and that there is one steep drop that represents the larger voltage drop and then a shallower drop later on that represents the smaller voltage drop across the smaller resistance... of course you've got to get the guy back to the top of the scree slope to start over again (like energy input from a batter in the form of a winch say) so the guy has no choice as to where he loses his energy it is pre-determined if you like

    get a different set of resistors and a different battery and you get a new pre-determined scree slope set up.
  5. Jan 22, 2017 #4


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    Here is one way to think about it....

    The voltage and the combined resistance dictate the current that flows around the circuit...

    I = V/R or in this case...

    Then current and the individual resistance dictates the individual voltages across the resistors..

    VR1 = I * R1
    VR2 = I * R2


    V = VR1 + VR2
  6. Jan 22, 2017 #5


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    only if the initial voltage is higher, see delta's comment

    exactly !
  7. Jan 24, 2017 #6


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    "flows" is not a good term to use here. You could say that the change in field Propagates, from the instant the circuit is completed and a new steady state is reached. All the basic theory on resistive circuits deals with that steady state situation.
    However, the arrangement of (macroscopic) fields is not very relevant. The Field (Volts per metre) will vary a lot depending how you lay the circuit wires out and choose different sized components so it's not a lot of use to consider that. The layout of the charges flowing around the circuit will depend on the local field effects - interactions between neighbouring charges. The reason that Potential is used more, in circuit analysis is that it is more suitable. There is no fundamental hierarchy which makes Field or Potential more important so it's not necessary that a Force makes a change, any more than a Potential Difference.
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