I have some questions that have been bugging me for the past few days, and I am here in hope of some help.
My Understanding of how battery works: "The battery (say 9V DC) does not store charge, it just provides voltage for the electrons in the wire to flow, causing current."
Q1) If my understanding is correct, the amount of current (AMPS) should depend on the thickness of wire, as there are more free electrons available. Is this true?
Q2) If I short the two terminals of a battery (9V DC) using a wire (of almost zero resistance) why would the wire heat up and burn? If I had used a thicker wire would the current passing through any point in the wire be greater than if I had used a thinner wire?
Best Answer
You are absolutely right. The amps flowing through any wire has to deal with cross-sectional area or lets say thickness of the wire. The relation between them is given by: I=VenA; I is current, e is charge of an electron, n is no. of free electrons, and A is the cross sectional area of the conductor. So increasing the area increases the current through the wire.
Another way to view this is through the idea of resistance. Since all the practical conductors have certain resistance, its resistance is expressed as R=(r*l)/A ; r is the resistivity of the material, l is length of the conductor and A is the area. Here, you can see that the higher is the cross sectional area, lower is the resistance of the conductor. So more current can flow through the wire, thanks to Ohm's law.
Coming to your second question, the conductor with zero resistance doesn't heat. The heating of the practical conductors is due to their resistance. The heat energy is generated due to the loss of energy of electron in conductor. The electron can only lose energy if and only if it has some resistance in its path. So in case of ideal conductors, there is no any resistance offered, so there is no any way a electron can lose its energy in any form(heat energy, light energy, sound enery or whatever).