1

We know that $V$ is directly proportional to $I$ for the same resistance. That means if we double the voltage, the current gets doubled. We also know that if a bulb or any other resistor works on a battery of say 2V, the resistor “takes away” 2 Joules of energy per Coulomb charge that passes through it.

Now that we have established what we already know, My question is: Say there is a circuit, with a battery of 2V, and the current flowing through it is 3A. The resistor present in the circuit is a bulb which by definition of potential difference “takes away” 2 Joules of energy per Coulomb charge that passes through it. If we double the P.D., the current should become 6A, but the bulb is still same!!! It should still only take 2J energy for every coulomb. It’s not like we change the bulb too every time we change Voltage. Therefore the energy for every coulomb charge should increase 2J for every rotation, increasing its energy indefinitely.

I may have the wrong understanding of Ohm’s law so please correct me if I’m wrong.

Qmechanic
  • 220,844
Arsh Hussain Naqvi
  • 51

2 Answers2

2

A close analogy is slides in a children's park: the speed with which a child moves on a slide is the current, the height of the slide is voltage. Now imagine several slides, all of them having the same length, but different angle and hence different height above the ground. For completeness we could imagine that the slides have strong friction, so that a child slides slowly (image source).

Very similar viscous friction models are really used to explain the resistance (they are closely related to Drude model, which has a few additional complications.)

Sliding along the slides also brings us to another important point - not all of the potential energy that electrons have is spent in the resistor - just like not all the energy of the sliding child is wasted on friction - they arrive at the bottom of slides with different speed. The energy that is actually produced in the resistor is given by Joule's law (see here for Drude-like derivation).

enter image description here

Roger V.
  • 68,984
-1

Say there is a circuit, with a battery of 2V, and the current flowing through it is 3A.

Sure, we can take this as true.

The resistor present in the circuit is a bulb which by definition of potential difference “takes away” 2 Joules of energy per Coulomb charge that passes through it.

There is no need to quote "takes away" because it is basically exactly correct, at least by the definitions and premises you started out with.

If we double the P.D., the current should become 6A, but the bulb is still same!!!

By this assertion, you immediately imply that the bulb is ohmic, which is a tolerably good approximation / assumption.

It should still only take 2J energy for every coulomb.

This is wrong. It will now take 4J of energy for every Coulomb worth of charge. You are just mistaken in your understanding here.

Worse, originally, it was taking 6W of power, and now it has jumped to 24W of power.

Therefore the energy for every coulomb charge should increase 2J for every rotation, increasing its energy indefinitely.

What rotation? And no, you are wrong.

naturallyInconsistent
  • 14,919