<p>Heads up: I'm not good with electronics and only have a vague idea of it's inner workings.</p>

<p>I have a <a href="https://ru.aliexpress.com/item/3D-Printer-V6-Wade-Short-distance-J-head-Hotend-12V-for-1-75mm-3-0mm-Extruder/32810022530.html?spm=a2g0v.10010108.1000016.1.197a7c35uzmRpw&amp;isOrigTitle=true" rel="noreferrer">E3D V6 Extruder rated for 24&nbsp;V</a>, that i plan to use in my 3D printer. Will there be any problems with it if powered by 12&nbsp;V? Will it take longer to heat up? Will it be able to heat up enough to melt PLA? Will it work at all for that matter?
If there are any other quirks or potential problems that I overlooked, please let me know.</p>

Electrical engineering can be quite complex, but in this case you can save yourself with same simple equations/relations. Using the following formulae:
Voltage (
$\ U$
) equals current (
$I$
) multiplied by the electrical resistance (
$R$
)
$$ U=I \times R $$
and
Power (
$P$
) equals the square of the current multiplied by the electrical resistance
$$ P=I^2 \times R $$
can be rewritten using the first formula to:
$$ P= \frac{U^2}{R} $$
Applying these formulae to a
40 Watt, 24 V
heater element, the electrical resistance (in
$\Omega $
) is calculated by:
$$ \frac{{(24\ V)}^2}{40\ W}=14.4\ \Omega $$
Running this heater element with 12 V will lead to a power of
$$ \frac{{(12\ V)}^2}{14.4\ \Omega}=10\ W $$
The heat produced is proportional to the square of the current multiplied by the electrical resistance,
halving the voltage
is
quartering the heat output
. This will heat up very slowly! If it is able to reach the required temperature that is. Calculating the temperature is far more difficult, but if you are interested in doing so, please look into
this answer
from the
Electrical Engineering
Stack Exchange.