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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?

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  • \$\begingroup\$ the current depends on the weakest link between + and -, which is not always the wire, usually it's the load, and on small batteries, the terminals themselves have considerable current-limiting resistance. \$\endgroup\$ – dandavis Aug 24 '18 at 19:35
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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).

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  • \$\begingroup\$ Thanks for the reply, could you please explain what happens if i short a 9V DC battery with a small wire? \$\endgroup\$ – Ateeb Ahmed Aug 24 '18 at 4:04
  • \$\begingroup\$ You can see the relation I have mentioned. Under constant voltage across the terminals, the lower current flows due to lower cross sectional area. \$\endgroup\$ – JuneStar_2918 Aug 24 '18 at 4:08
  • \$\begingroup\$ no i mean, can i damage the WIRE or BATTERY by shorting(battery) it? and what is the reason behind this DAMAGE if any? \$\endgroup\$ – Ateeb Ahmed Aug 24 '18 at 4:13
  • \$\begingroup\$ depends upon the material of the wire...the damage is usually caused when large amount of current flows through it. So try doing some calculations and see how much current can flow and make changes on your circuit. The damage is in your wire and battery. The battery finishes very fast and the sparks can be seen. \$\endgroup\$ – JuneStar_2918 Aug 24 '18 at 4:20

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