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I'm a bit confused I know V=potential difference of a conductor is the work done by the battery in pushing one charge across a conductor.This means it is inversely proportional to the number of charges flowing through the conductor, thus inversely proportional to the current

So why is potential difference across a conductor directly proportional to the current flowing through it according to Ohm's law.

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  • $\begingroup$ Current in a conductor, $I =\frac {ne}{t} $. I don't think potential difference has the above stated relation with charge. $\endgroup$ – Wrichik Basu Apr 6 '17 at 17:12
  • $\begingroup$ I saw the sigma in $J_f=\sigma E$ defined as elementary charge * mobility of species * |valence of species| * number density of species in a book about fluid flows in biology. You had to assume a whole lot for Ohm's law to work. $\endgroup$ – Emil Apr 6 '17 at 17:15
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You said in your question that you knew that the potential energy per charge was the volatage. One way of writing this is that the potential energy lost per unit time (power dissipated, $P$) is the charge flowed per unit time (current, $I$) times the potential $V$. Therefore we get $P=IV$ or $V=P/I$. From this equation we see that the potential is indeed inversely proportional to the current assuming constant power.

Now in "real life" we usually don't have constant power. For example, let's assume I have light bulbs of two different resistances $R_1$ and $R_2$ (let's say with $R_1 < R_2$), and I plug them both into the wall. The way electricity from the wall works is that it supplies a constant voltage $V$. Now we expect that with constant $V$, the light bulb with a lower resistance will allow more current to flow. In fact, the current $I$ resulting from a voltage across a resistance is given by $I=V/R$. Therefore we will get $I_1=V/R_1$, and $I_2=V/R_2$. Since $R_1$ is smaller, its current $I_1$ will be bigger and so the power $P_1 = I_1 V$ will also be bigger. Since the power is not the same for the two light bulbs, we do not get a contradiction with your equation from the first paragraph.

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Like Hooke's Law for springs and Boyle's Law for gases, Ohm's Law is an empirical law. In many situations we find that $V \propto I$, but there is no Law of Nature which says this has to be true. The constant of proportionality is defined to be the resistance. So $V=IR$ because that is how $R$ is defined.

These empirical laws are not derived from theories about the microscopic nature of matter. The Kinetic Theory of gases, and the Electron Theory of current, came later. The theories can be used to relate microscopic properties which we cannot see (the mass and velocity of molecules, the drift velocity and number density of electrons) to the macroscopic properties we can see (volume and pressure of gases, current and voltage in an electric circuit). If these theories did not fit with the empirical laws, the theories would have to be revised, not the empirical laws.

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Unit of Voltage is Joules per Coulomb. Means how much energy there is to move charges in a conductor. More will be the energy to move charges, faster they will move and because current is rate of flow of charge so more will be the current. Ohms law basically states that for a constant resistance Voltage is proportional to current/rate of flow of charge.

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  • $\begingroup$ Fache i think it does depend...the pd accros a conductor which alows a high number of particle to flow through it for a constant electrical energy supplied by the battery would be less thatn that that allows few charges to flow through it $\endgroup$ – Lee Apr 6 '17 at 21:41
  • $\begingroup$ For the same amount of electrical energy $\endgroup$ – Lee Apr 6 '17 at 21:42
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I know V=potential difference of a conductor is the work done by the battery in pushing one charge across a conductor.This means it is inversely proportional to the number of charges flowing through the conductor, thus inversely proportional to the current.

This is not true.
If the charge is $q$ then the work done on the charge when it moves across a potential difference $V$ is $qV$.

If $N$ charges traverse that potential difference then the work done is $NqV$.
The potential difference does not depend on the number of charges.

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