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Before proceeding onwards please note that I am talking about a simple circuit consisting of an ideal battery, a switch and an external resistance.

So I was told that $$ W_\text {ideal battery} = Q_\text { dissipated by the resistance} $$, but this seems wrong to me. My reasons are:

i) Electrons carry energy, they travel from the negative terminal to the positive terminal. If all the energy of the electrons were to be transferred to the resistance then in that case, electrons (after passing through the resistance) will start undergoing random motion since electric field at them is now 0, potential difference between any two points on that circuit is 0 (the wire itself is resistance less but we have attached a resistance to the wire).

ii) In my opinion what would happen is that all the potential energy will change to heat energy, the remaining kinetic energy of electrons will be constant (assuming that drift velocity is constant throughout the circuit, I know it will not be so but I am not well versed enough with the math required to find drift velocity, the statistical analysis required to do so is far beyond my current ability).

iii) If that is indeed the case then when we closed the switch, the battery would have certainly given the electrons a certain kinetic energy as that is why drift velocity is a thing, before the switch was closed there was no electric field and thus no drift velocity, the net velocity of electrons was 0.

iv) The work done by the battery $ q \times V $ is simply change in potential energy and there is no component of kinetic energy inside which means that initially the battery must have provided the circuit with additional energy, which is $ \frac{1}{2} mv_d^2$ and thus all of the energy provided by the battery is not dissipated into heat, only the potential energy is dissipated into heat.

Is my reasoning above correct, if so

$$ W_\text {ideal battery} = Q_\text { dissipated by the resistance} $$

would turn out to be incorrect right?

EDIT 1,

Rather than saying that potential energy is changing at every point which is incorrect as the wire itself is resistance less, I should have said that potential energy is changing across the resistor.

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    $\begingroup$ Not a full answer, but you might benefit from this video explaining how energy flow actually happens in a circuit youtu.be/C7tQJ42nGno $\endgroup$
    – RC_23
    Commented Jun 2 at 5:05

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There are a couple of misconceptions in the question. First the velocity of the electrons does not go to zero after passing through the resistor. The current is proportional to the number of electrons passing a given point in the circuit. the current is not zero in the final return wire. Current is given by the the $I= V/R$ relationship and when the potential difference is zero in the return wire, the current is I = 0/0$ which is indeterminate, not zero. The current entering the resistor is equal to the current exiting the resistor unless there are electrons accumulating in the resistor, which shouldn't happen if the resistor has no capacitance.

Secondly, the heat energy given off by a wire or resistor is equal to the power in Watts $ P = I^2 R$ and not to the velocity or kinetic energy of the electrons.. In the final wire the resistance is zero, so there is no heat generated there, because the resistance is zero and not because of the false assumption that the velocity of the electrons is zero.

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  • $\begingroup$ Actually. all conductors accumulate some amount of static charges on its surface that form a static field by which the electrons are moved the motion being the stationary current. This is what builds up during the initial turn-on transient without which there is no steady charge flow. $\endgroup$
    – hyportnex
    Commented Jun 2 at 7:30
  • $\begingroup$ @hyportnex As you say the capacitance of a resistor is transient as the circuit is switched on, but I think you will agree that under normal steady state conditions the amount of electrons leaving a resistor is the same as the number of electrons entering the resistor, per unit time. $\endgroup$
    – KDP
    Commented Jun 2 at 8:11
  • $\begingroup$ The distributed capacitance of the conductor along its surface is not transient but its charge accumulation is with a steady state current exactly as you said. My point was really not to criticize your post but by raising the issue to help Gauransh with an important conceptual problem: what makes charges move in a conductor in which the dissipation takes place? Without that it is unlikely to see the causes and results of dissipation. $\endgroup$
    – hyportnex
    Commented Jun 2 at 8:18
  • $\begingroup$ I never meant that the velocity is 0 in the return circuit, I meant that if all the energy were to be used up in heat energy, then in that case the there would be no kinetic energy left due to the electric field which would result in no current flowing through that area. $\endgroup$
    – Gauransh
    Commented Jun 2 at 15:12
  • $\begingroup$ You didn't really answer my question, my question is that why do we not consider the work done to give the electrons drift velocity as work done by the battery, since without the introduction of an electric field there would be no drift velocity, all electrons would be moving randomly and though there would be some kinetic energy (We will have to deal with Maxwell's bell curve for that), the net velocity without the introduction of an electric field would be 0. $\endgroup$
    – Gauransh
    Commented Jun 2 at 15:15

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