0
$\begingroup$

Charges loose their energy within the circuit if they move through a resistor(like a bulb). Now, if I make a circuit with just wires and no resistor, will the charge loose its energy?

One reason of loss of energy of the charge I can think of is that as the charge reaches the positive terminal, it it getting closer to the positive charge which is attracting it, and getting closer to the source of the electric field, and hence, it is loosing potential energy(Similar to the case of the gravity of earth, as we move closer to ground, we loose potential energy). And, this potential energy is converted to the kinetic energy of the charge.

Now, using all of the things I wrote above, we can ask the question that as the charge reaches the positive terminal, where is this Kinetic energy lost? If it isn't lost, then will the battery work properly? Because on reaching the positive terminal, charge already has energy, so what will the battery do?

Of course due to the very less resistance of the wire, the charge will loose some energy, but it will be extremely small. So, is this Kinetic energy lost? If not, what is the case with this circuit?

$\endgroup$
  • $\begingroup$ The potentials on the ends of an ideal wire are the same. If you connect the wire first to ground (negative battery terminal) and then you start moving it closer to the positive terminal, the electric field between the positive terminal and the end of the wire will increase. As the distance between the two becomes ever smaller, it will, eventually, diverge. At some point there has to be field emission from the wire and the charges will be accelerated in the strong field until they hit the terminal. At that point they will lose all of their kinetic energy and heat the metal. $\endgroup$ – CuriousOne Feb 24 '16 at 10:44
  • $\begingroup$ The electrons in the wire are already moving with great velocities prior to connecting the battery; however, this motion is in essentially random directions. Look up the Drude model for conduction for an elementary analysis. When a voltage is supplied it results in a drift velocity being added to the mix. This drift velocity is quite slow in comparison to the underlying random motions. With AC there is barely any net motion of the electrons. $\endgroup$ – Peter Diehr Feb 24 '16 at 10:54
  • $\begingroup$ Assuming the battery is a physical battery (produces finite short-circuit current) and the wire is effectively ideal, the voltage across the battery terminals is effectively zero (assuming the voltage across the battery terminals with a wire connected across is equivalent to assuming the wire is a resistor). Thus, the battery itself dissipates the electrical energy by getting (possibly very) hot and, perhaps, igniting or exploding. $\endgroup$ – Alfred Centauri Feb 24 '16 at 11:46
2
$\begingroup$

Conduction is not about batteries magically delivering charges and passing them through a circuit towards the other end of the battery like a water flow. What happens in conduction is atoms of the conductive material successively losing and catching electrons in microscopic scale, resulting in a global trend of moving charges on the macroscopic scale. They do so because they are constrained by the source's potential difference, which in case of battery is the need for it to even charges inside a chemical reaction.

What will happen if you short a battery with a wire will be a huge current drawn from the battery taking place in the wire, damaging the battery until it can't function properly or until the wire melts.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.