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Recently I started with concept of power provided by battery in moving charge from one point to other in a wire. NCERT says that

"if charge were to move in conductor free of collisions then the work done by field equals to the change in kinetic energy of the charges and hence they accelerate towards low potential".

Now what I feel in this case is, if we assume zero resistance and if resistance were to be present (by collisions with heavy fixed positive ions) then thermodynamically the velocity of each charge has equal probability to get changed in either direction after collision, hence we say due to Potential difference (constant) the electrons as a system drift with "constant" velocity and so the the work done equals the energy loss in collisions. But, why then we take P.D=0 in a wire assuming $R=0 $. Will electrons not accelerate here?

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  • $\begingroup$ Let the title show express the question. Replace P.D=o with TeX. $\endgroup$
    – David Jonsson
    Sep 2, 2021 at 7:21

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Resistance has a very specific meaning for an electrical device. It's defined by $$R=\frac VI$$ You must take account of $I$ as well as $V$. Consider a device with a very low resistance. We are free to put any pd, V, across it; the current will be in accordance with $I=V/R$. This argument will apply however low we make $R$. The equation imposes no restriction on how large we make $V$ and consequently how large an $I$ we get (if we assume $V$ to be applied for such a short time that we don't destroy, damage or change the device).

The claim that $R=0$ implies that $V=0$ is based on the unspoken assumption that $I$ remains finite. This assumption is based on the behaviour of any real power supply to which the device is connected.

So $V=IR\ \ \ $ becomes $\ \ \ V=\text{something finite} \times R$.

So as $R$ approaches zero, $V$ approaches zero.

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  • $\begingroup$ My book says that if R=0 the work done by battery equals change in kinetic energy of the charges that move through the zero resistance wire or conductor. At the same time we also take work done by battery equal to zero when R=0. The two statements contradict each other. Also if work done by battery were to be equal to zero that means there should be no electric field inside the zero resistance wire. How is this possible since battery maintains an electric field by accumulating charge on its terminals?? $\endgroup$
    – Physics is Love
    Mar 20, 2021 at 19:52
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    $\begingroup$ I'm arguing that when $R$ approaches zero, $V$ approaches zero, that is the work done by the battery per unit charge, approaches zero. I don't see how that contradicts the work done by the battery equalling the change in KE of the charges. $\endgroup$
    – Philip Wood
    Mar 20, 2021 at 23:57
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But, why then we take P.D=0 in a wire assuming $R=0 $. Will electrons not accelerate here?

They would accelerate accelerate in the wire if there was no other resistance in series with the wire. If there is resistance in series with the wire that resistance controls the current and prevents the electrons from accelerating. Since the no resistance wire is in series with the resistor, the current is constant in the wire as well.

The potential difference, $V$, between two points is defined as the work per unit charge required to move the charge between the points. Since no work is required to move the charge between any two points of a wire with zero resistance, the potential difference between any two points of the wire is zero.

A mechanical analogy involving friction might help.

Imagine you are pushing a box at constant velocity with a constant force a distance $d$ on a floor having friction.

  1. The box is analogous to the electric charge, $Q$ moving at constant velocity (analogous to constant current).

  2. The floor with friction is analogous to the resistance

  3. The force you apply to the box is analogous to the force applied by the electric field, $F=QE$

  4. The work you do moving the box the distance $d$ is analogous to the work done by the electric field moving the charge through the resistor, $W=QEd$.

  5. The work you do per unit mass the distance $d$is analogous to the work done per unit charge by the field through the resistor , namely the voltage $V=\frac{W}{Q}=Ed$.

  6. The equal negative work done by kinetic friction that dissipates heat in the floor is analogous to the negative work done by the resistor due to collisions. The collisions dissipate the energy provided by the field as resistance heating.

Now imagine after pushing the box the distance $d$ the floor becomes frictionless. This is analogous to encountering the zero resistance wire in series the resistor. You (the electric field) no longer needs to do work to keep the box (charge) moving at constant velocity (constant current) since there is no longer friction (resistance) to overcome. Since no work is required to move the charge through the zero resistance wire, the potential difference between any two points of the wire is zero.

Hope this helps.

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  • $\begingroup$ My book says that if R=0 the work done by battery equals change in kinetic energy of the charges that move through the zero resistance wire or conductor. At the same time we also take work done by battery equal to zero when R=0. The two statements contradict each other. Also if work done by battery were to be equal to zero that means there should be no electric field inside the zero resistance wire. How is this possible since battery maintains an electric field by accumulating charge on its terminals?? $\endgroup$
    – Physics is Love
    Mar 20, 2021 at 19:49
  • $\begingroup$ @PhysicsisLove "My book says that if R=0 the work done by battery equals change in kinetic energy of the charges that move through the zero resistance wire or conductor." Are you referring to the statement from NCERT you quoted? $\endgroup$
    – Bob D
    Mar 20, 2021 at 21:47
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In solving a circuit problem, taking the potential difference across a wire to be zero is a convenient approximation which assumes that the resistance of the wire is much lower than other resistances in the circuit. If the wire is long, its resistance needs to considered.

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