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I bought a 'Juice jw2d 2.4 farad 24V surge low esr' capacitor. It's a type of capacitor used for car audio sound systems, 12v I assume.

I however need to use it for a 24v sound system.

Does '24V surge low esr' mean it will output 24V if give it 24V from the batteries?

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    \$\begingroup\$ I think first you need to do a little research on how capacitors work.... \$\endgroup\$
    – DerStrom8
    Commented Apr 29, 2019 at 1:19
  • \$\begingroup\$ VTC as unclear what you're asking; all I can understand of this question is that you don't really understand how capacitors work. I recommend at least reading the wikipedia page. \$\endgroup\$
    – Hearth
    Commented Apr 29, 2019 at 1:24
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    \$\begingroup\$ '24v surge low esr' is almost meaningless in engineering terms. It means "buy me". What do you want to use the capacitor for, and what do you want it to do for you? \$\endgroup\$
    – TimWescott
    Commented Apr 29, 2019 at 1:32
  • \$\begingroup\$ Useless for a Car stereo booster as the battery can deliver more far current than this tiny cap and with lower ESR. I would expect this cap to have an ESR of >1 Ohm \$\endgroup\$
    – D.A.S.
    Commented Apr 29, 2019 at 1:44
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    \$\begingroup\$ 24V surge sounds like an absolute maximum rating, so this would not be wise to use with a 24V nominal system. Voltage, depending on system, often varies somewhat around "nominal" voltage, which is why a 24V absolute maximum capacitor is necessary for a "12V" system in a car, where voltage will actually vary up to 14 or so volts, higher with an inductive surge. You can use capacitors in series to increase their voltage ratings, but additional measures(balancing circuit) may be necessary, and using them in series will decrease total capacitance, search "capacitors in series" for info. \$\endgroup\$
    – K H
    Commented Apr 29, 2019 at 2:00

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Welcome to the site.

Once you have charged the capacitor to 24V, it is able to output 24V for a very brief period of time. There is a phenomenon called 'discharging of capacitor'. The voltage across the capacitor decreases exponentially. Have a look at the figure.

enter image description here

There is a mathematical relation relating the voltage across the capacitor and time. Go through the 'capacitor' topic. You will be able to learn how it works.

If you wanna see what I mean, simply construct a circuit and see the effect of capacitor.

schematic

simulate this circuit – Schematic created using CircuitLab

The thing you always need to keep in mind is that the voltage rating in the capacitor is the maximum voltage it can maintain across its terminals. If you have a 24V capacitor, it can be used in the above schematic. But if you have capacitor with rating less than 5V, the capacitor is gonna blow out.

For constant current discharge, I recommend you to go through this explanation in this site.

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  • \$\begingroup\$ The discharge rate depends on the load - the graph is correct for a resistive load, but not for a constant current load or the sample schematic. Anyway, this is more information than what was asked in the question ;-). \$\endgroup\$
    – le_top
    Commented Apr 29, 2019 at 18:14
  • \$\begingroup\$ Yeah I missed that fact. Thanks for the edit \$\endgroup\$
    – G-aura-V
    Commented Apr 30, 2019 at 3:04

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