2
$\begingroup$

If a AA battery was kept at a constant temperature and was connected in a simple circuit will the battery drain faster if the wires are kept at a cold temperature or at a hot temperature?

I understand that the cooler the wire is the less resistance there will be for the electrons to flow through. My current thinking is that if the wires have less resistance the battery would drain faster as the electrons aren't being effected as much as the hot wire, however if the wire was hot, do the electrons need more energy to go through the wire so then the battery would drain faster?

$\endgroup$
2
$\begingroup$

To answer this, one needs to understand a couple of somewhat independent concepts. First is that of resistance at cool temperatures.

Assuming that we are talking of a regular conductor (metal) and a temperature variation of, say, 0C to 100C, the cooler conductor will indeed have a lower resistance. Now, the question of how much lower is a separate one, but let's assume that it is measurably lower. Lets mark $R_1(T_1) < R_2(T_2)$ where $T_1<T_2$ are the temperatures.

If the same battery is connected to these resistors, we will have $I_1(R_1) > I_2(R_2)$

Now let's turn to the second concept. Namely it is that of the battery capacity. From Wikipedia:

A battery's capacity is the amount of electric charge it can deliver at the rated voltage. The more electrode material contained in the cell the greater its capacity. A small cell has less capacity than a larger cell with the same chemistry, although they develop the same open-circuit voltage.[30] Capacity is measured in units such as amp-hour (A·h). The rated capacity of a battery is usually expressed as the product of 20 hours multiplied by the current that a new battery can consistently supply for 20 hours at 68 °F (20 °C), while remaining above a specified terminal voltage per cell. For example, a battery rated at 100 A·h can deliver 5 A over a 20-hour period at room temperature. The fraction of the stored charge that a battery can deliver depends on multiple factors, including battery chemistry, the rate at which the charge is delivered (current), the required terminal voltage, the storage period, ambient temperature and other factors.[30]

The higher the discharge rate, the lower the capacity.[31] The relationship between current, discharge time and capacity for a lead acid battery is approximated (over a typical range of current values) by Peukert's law:

${\displaystyle t={\frac {Q_{P}}{I^{k}}}}$

$Q_P$ is the capacity when discharged at a rate of 1 amp.

${\displaystyle I}$ is the current drawn from battery (A).

${\displaystyle t}$ is the amount of time (in hours) that a battery can sustain.

${\displaystyle k}$ k is a constant around 1.3.

Using this law, you can see that for lower current, the discharge (drain) time is larger. Vice versa, for higher current, the discharge time is smaller.

In other words, $t_1(T_1) < t_2(T_2)$ or for higher temperature the discharge time is longer.

$\endgroup$

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.