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In the diagram above, the resistor has a constant resistance R.

We know that:

$$R=\frac{V}{I}$$ $$V\propto I$$

If the current increases, that means the rate of flow of charge (i.e. the speed at which the charge is flowing) has increased.

However, why should the potential difference across the resistor decrease to maintain the proportionality? The resistance of the resistor hasn't changed, so each coulomb of charge still needs to do the same amount of work on the resistor to flow through it.

In other words, the potential difference across a component is defined as the work each coulomb of charge needs to do on the component to flow through it. In that case, shouldn't the rate of flow of charge and the work done per unit charge be completely independent?

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marked as duplicate by Yashas, honeste_vivere, David Hammen, Jon Custer, Kyle Kanos Jul 5 '17 at 12:24

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    $\begingroup$ Possible duplicate of The Difference Between voltage and current $\endgroup$ – QuirkyTurtle98 Jul 4 '17 at 15:40
  • $\begingroup$ I think you are confusing the two expressions for potential and current as being different entities. If you look at the second form and assume a constant resistance, then how can the potential decrease when the current increases? $\endgroup$ – honeste_vivere Jul 4 '17 at 18:52
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What happens if we increase the voltage?

  • The force on each electron will increase.
  • Thus, the electrons drift faster toward the anode.
  • Therefore, the current increases.

Hence, the question is how does the current increase? Answer: Such that $U/I$ is constant.

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  • $\begingroup$ If we imagine that there is a variable resistor connected in series with the normal resistor in the diagram, then by decreasing its resistance one could increase the current through the normal resistor, which would also increase its potential difference. Could one explain this as more electrons trying to 'push' their way through the normal resistor at the same time, thereby colliding with the atoms in the resistor more frequently and doing more work overall, thus increasing the potential difference across the normal resistor? $\endgroup$ – Pancake_Senpai Jul 4 '17 at 17:11
  • $\begingroup$ @Pancake, see Drude model. $\endgroup$ – The Photon Jul 4 '17 at 17:34

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