If I supply 24v to a 50v rated 22000uf electrolytic capacitor, does that mean it will store 44000uf at...

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If I supply 24v to a 50v rated 22000uf electrolytic capacitor, does that mean it will store 44000uf at 24v?



If I supply 24v to a 50v rated 22000uf electrolytic capacitor, does that mean it will store 44000uf at 24v?


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$begingroup$


Is the higher voltage just a safety thing, or can the capacitor actually store more charge?
Also, if I supply it with 24v, will it output 24v or 50v?



I read the wikipedia on capacitors, watched a few youtube videos, and read a few other articles, and they were all very vague about this.










share|improve this question







New contributor




Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







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  • 6




    $begingroup$
    If you charge a capacitor to 24 volts, it will remain at 24 volts until discharged or further charged. I think this question shows a fundamental misunderstanding about what exactly capacitors do.
    $endgroup$
    – Hearth
    4 hours ago










  • $begingroup$
    Thanks, that's helpful. This question only shows how poorly these things are explained on the internet by the ones who have the knowledge. I am a programmer and am capable of understanding relatively complex concepts. Capacitors aren't a complex concept, just poorly explained.
    $endgroup$
    – Mladen
    4 hours ago








  • 2




    $begingroup$
    the voltage rating is just a limit. "halfway full" it can hold more charge, but only with higher voltages.
    $endgroup$
    – dandavis
    4 hours ago












  • $begingroup$
    @dandavis That simple sentence cleared up a lot! It should be voted the BEST ANSWER as far as I am concerned.
    $endgroup$
    – Mladen
    4 hours ago












  • $begingroup$
    From my profile, here is a link to Capacitors ibiblio.org/kuphaldt/socratic/model/mod_capacitor.pdf followed by C++ both with relatively logical approaches and simplified
    $endgroup$
    – Sunnyskyguy EE75
    3 hours ago




















1












$begingroup$


Is the higher voltage just a safety thing, or can the capacitor actually store more charge?
Also, if I supply it with 24v, will it output 24v or 50v?



I read the wikipedia on capacitors, watched a few youtube videos, and read a few other articles, and they were all very vague about this.










share|improve this question







New contributor




Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$








  • 6




    $begingroup$
    If you charge a capacitor to 24 volts, it will remain at 24 volts until discharged or further charged. I think this question shows a fundamental misunderstanding about what exactly capacitors do.
    $endgroup$
    – Hearth
    4 hours ago










  • $begingroup$
    Thanks, that's helpful. This question only shows how poorly these things are explained on the internet by the ones who have the knowledge. I am a programmer and am capable of understanding relatively complex concepts. Capacitors aren't a complex concept, just poorly explained.
    $endgroup$
    – Mladen
    4 hours ago








  • 2




    $begingroup$
    the voltage rating is just a limit. "halfway full" it can hold more charge, but only with higher voltages.
    $endgroup$
    – dandavis
    4 hours ago












  • $begingroup$
    @dandavis That simple sentence cleared up a lot! It should be voted the BEST ANSWER as far as I am concerned.
    $endgroup$
    – Mladen
    4 hours ago












  • $begingroup$
    From my profile, here is a link to Capacitors ibiblio.org/kuphaldt/socratic/model/mod_capacitor.pdf followed by C++ both with relatively logical approaches and simplified
    $endgroup$
    – Sunnyskyguy EE75
    3 hours ago
















1












1








1





$begingroup$


Is the higher voltage just a safety thing, or can the capacitor actually store more charge?
Also, if I supply it with 24v, will it output 24v or 50v?



I read the wikipedia on capacitors, watched a few youtube videos, and read a few other articles, and they were all very vague about this.










share|improve this question







New contributor




Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Is the higher voltage just a safety thing, or can the capacitor actually store more charge?
Also, if I supply it with 24v, will it output 24v or 50v?



I read the wikipedia on capacitors, watched a few youtube videos, and read a few other articles, and they were all very vague about this.







capacitor






share|improve this question







New contributor




Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






New contributor




Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 4 hours ago









MladenMladen

112




112




New contributor




Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor





Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








  • 6




    $begingroup$
    If you charge a capacitor to 24 volts, it will remain at 24 volts until discharged or further charged. I think this question shows a fundamental misunderstanding about what exactly capacitors do.
    $endgroup$
    – Hearth
    4 hours ago










  • $begingroup$
    Thanks, that's helpful. This question only shows how poorly these things are explained on the internet by the ones who have the knowledge. I am a programmer and am capable of understanding relatively complex concepts. Capacitors aren't a complex concept, just poorly explained.
    $endgroup$
    – Mladen
    4 hours ago








  • 2




    $begingroup$
    the voltage rating is just a limit. "halfway full" it can hold more charge, but only with higher voltages.
    $endgroup$
    – dandavis
    4 hours ago












  • $begingroup$
    @dandavis That simple sentence cleared up a lot! It should be voted the BEST ANSWER as far as I am concerned.
    $endgroup$
    – Mladen
    4 hours ago












  • $begingroup$
    From my profile, here is a link to Capacitors ibiblio.org/kuphaldt/socratic/model/mod_capacitor.pdf followed by C++ both with relatively logical approaches and simplified
    $endgroup$
    – Sunnyskyguy EE75
    3 hours ago
















  • 6




    $begingroup$
    If you charge a capacitor to 24 volts, it will remain at 24 volts until discharged or further charged. I think this question shows a fundamental misunderstanding about what exactly capacitors do.
    $endgroup$
    – Hearth
    4 hours ago










  • $begingroup$
    Thanks, that's helpful. This question only shows how poorly these things are explained on the internet by the ones who have the knowledge. I am a programmer and am capable of understanding relatively complex concepts. Capacitors aren't a complex concept, just poorly explained.
    $endgroup$
    – Mladen
    4 hours ago








  • 2




    $begingroup$
    the voltage rating is just a limit. "halfway full" it can hold more charge, but only with higher voltages.
    $endgroup$
    – dandavis
    4 hours ago












  • $begingroup$
    @dandavis That simple sentence cleared up a lot! It should be voted the BEST ANSWER as far as I am concerned.
    $endgroup$
    – Mladen
    4 hours ago












  • $begingroup$
    From my profile, here is a link to Capacitors ibiblio.org/kuphaldt/socratic/model/mod_capacitor.pdf followed by C++ both with relatively logical approaches and simplified
    $endgroup$
    – Sunnyskyguy EE75
    3 hours ago










6




6




$begingroup$
If you charge a capacitor to 24 volts, it will remain at 24 volts until discharged or further charged. I think this question shows a fundamental misunderstanding about what exactly capacitors do.
$endgroup$
– Hearth
4 hours ago




$begingroup$
If you charge a capacitor to 24 volts, it will remain at 24 volts until discharged or further charged. I think this question shows a fundamental misunderstanding about what exactly capacitors do.
$endgroup$
– Hearth
4 hours ago












$begingroup$
Thanks, that's helpful. This question only shows how poorly these things are explained on the internet by the ones who have the knowledge. I am a programmer and am capable of understanding relatively complex concepts. Capacitors aren't a complex concept, just poorly explained.
$endgroup$
– Mladen
4 hours ago






$begingroup$
Thanks, that's helpful. This question only shows how poorly these things are explained on the internet by the ones who have the knowledge. I am a programmer and am capable of understanding relatively complex concepts. Capacitors aren't a complex concept, just poorly explained.
$endgroup$
– Mladen
4 hours ago






2




2




$begingroup$
the voltage rating is just a limit. "halfway full" it can hold more charge, but only with higher voltages.
$endgroup$
– dandavis
4 hours ago






$begingroup$
the voltage rating is just a limit. "halfway full" it can hold more charge, but only with higher voltages.
$endgroup$
– dandavis
4 hours ago














$begingroup$
@dandavis That simple sentence cleared up a lot! It should be voted the BEST ANSWER as far as I am concerned.
$endgroup$
– Mladen
4 hours ago






$begingroup$
@dandavis That simple sentence cleared up a lot! It should be voted the BEST ANSWER as far as I am concerned.
$endgroup$
– Mladen
4 hours ago














$begingroup$
From my profile, here is a link to Capacitors ibiblio.org/kuphaldt/socratic/model/mod_capacitor.pdf followed by C++ both with relatively logical approaches and simplified
$endgroup$
– Sunnyskyguy EE75
3 hours ago






$begingroup$
From my profile, here is a link to Capacitors ibiblio.org/kuphaldt/socratic/model/mod_capacitor.pdf followed by C++ both with relatively logical approaches and simplified
$endgroup$
– Sunnyskyguy EE75
3 hours ago












4 Answers
4






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3












$begingroup$

If you exceed the voltage rating of the capacitor, the capacitor may fail.



If you put 24V across a cap, it will charge to 24V.



If it's spec'ed at 22000uF it's 22000uF for ANY voltage.






share|improve this answer









$endgroup$









  • 1




    $begingroup$
    Note that "fail" in this case probably means "explode" or at least "vent extremely hot vaporized electrolyte". Electrolytic capacitors do not like being mistreated, and I doubt you'll get 22mF out of anything other than aluminum electrolytics.
    $endgroup$
    – Hearth
    4 hours ago










  • $begingroup$
    Your final sentence needs a big asterisk for ceramics. Not that you're likely to use ceramics for 22000 uF.
    $endgroup$
    – The Photon
    3 hours ago



















3












$begingroup$


If I supply 24v to a 50v rated 22000uf electrolytic capacitor, does that mean it will store 44000uf at 24v?




No. The capacitor's capacity is a property of its dimensions.



$$ C propto frac {A}{d} $$ where A is the area of the capacitor and d is the distance between the plates. Note that voltage doesn't appear in the equation so it's still 22,000 μF even at zero volts.




Is the higher voltage just a safety thing, or can the capacitor actually store more charge?




The capacitor has a maximum voltage that the dialectric between its plates can withstand if you exceed this the dielectric will break down, current will flow and the capacitor will overheat and possibly explode.



The amount of charge stored is given by $$ Q = {C}{V} $$ so the higher the voltage on a given capacitor then the more charge is stored. Again, check for nonsense by testing for 0 V: at 0 V the charge, Q, will be zero. This makes sense.




Also, if I supply it with 24v, will it output 24v or 50v?




If you charge it to 24 V, disconnect the supply and measure the voltage you will get a reading of 24 V. Once you start to draw current from it the voltage will decrease.



enter image description here



Figure 1. When discharging through a resistor the time constant, τ can be calculated as RC. After RC seconds the capacitor will have discharged by 63%. After 3RC, 95% and after 5RC by about 99%. Image source: I am Technical.





Finally, a few points: SI symbols for the volt is 'V' and for the farad is 'F'. The symbols are capitalised but units named after a person are lowercase when spelled out.






share|improve this answer









$endgroup$





















    0












    $begingroup$

    I'll try to explain with a gradual approach:



    A capacitor consists of two plates or conductive surfaces separated by a gap or an insulator (called the dielectric). Voltage is potential difference between any two points. Another name for voltage is electromotive force, which starts to convey the idea that it describes how "motivated" an electric charge is to do work.



    I won't try to explain the how, but suffice to say that the positive and negative charges are attracted to each other, and will "stay" on the plates opposite the dielectric, until provided a connecting path (circuit).



    A capacitor's voltage rating is based on its ability to function correctly up to the specified voltage. It encompasses things such as properties of the dielectric and physical dimensions of the capacitor. With enough voltage, insulators will eventually conduct (usually temporarily) and break down. Thus, if a capacitor is rated for 50 volts, it means that the dielectric composition, placement, thickness, etc. are only designed to function up to 50V.



    The Farad is a unit of capacitance or amount of electric charge. (Technically it's equal to one coulomb (one ampere of current in one second) stored with the potential difference (voltage) of one volt.)



    Analogies always fail at some point, but think of a capacitor like a sphere in a water plumbing system which has a flexible diaphragm separating it into halves1. If you pump water into half of it, you're storing some quantity of water. If the pressure of the water is too high, you'll rupture the diaphragm. You can store some quantity of water up to the maximum capacity of the vessel, and it will be stored at whatever pressure was applied. (Obviously this analogy fails when you think of some small quantity of water not even causing the diaphragm to stretch at all, so you should dispense with hydraulic analogies as soon as possible...)



    You can therefore use a 50V 22000µF capacitor to store charge at 1 volt, 10 volts, or 24 volts. In all cases, it will store a maximum of 22000µF, at the same voltage it was applied. Exceeding 50V will cause it to fail. (Applying 51V will probably result in a slow failure over time whereas 100V will cause a much more abrupt and potentially dangerous failure.)



    You might be thinking "Well, if it can store 22000µF at 50V, why can't it store twice as much at half the voltage?"



    It's tempting to think this if, for example, you're familiar with pneumatic systems: You have a two liter tank and store compressed gas in it. You can store a given quantity of gas at some pressure, and a larger quantity of gas at a higher pressure. (Within the tolerances and specifications of the gas and vessel, of course.) But this is not an analogy to electric charge. In order to understand a capacitor, you have to avoid thinking about it as either a pneumatic or hydraulic system — things which many people have everyday experience with prior to learning about electronics.



    A capacitor won't store more than its rated capacitance at any voltage, and won't function properly (read: survive) exceeding its voltage rating. Also be careful to observe polarity. Generally capacitors with such capacity are polarized. Connecting positive and negative incorrectly will also result in failure, often dramatically.





    1Credit to Warren J. Beaty for this particular analogy.






    share|improve this answer











    $endgroup$













    • $begingroup$
      Or think of the capacitor as a barrel or a bucket. A 5 gallon bucket can hold 5 gallons of water, no matter the water pressure used to fill it. If you fill it at a rate of 2.4 gal/minute, it doesn't hold twice as much water as if it were filled at 5.0 gal/minute.
      $endgroup$
      – spuck
      2 hours ago



















    0












    $begingroup$

    'the voltage rating is just a limit. "halfway full" means it can hold more charge, but only with higher voltages.'



    Thank you dandavis, and the rest of you for answering it broadly






    share|improve this answer








    New contributor




    Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






    $endgroup$













    • $begingroup$
      Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review
      $endgroup$
      – JRE
      1 hour ago












    Your Answer






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    4 Answers
    4






    active

    oldest

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    4 Answers
    4






    active

    oldest

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    active

    oldest

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    3












    $begingroup$

    If you exceed the voltage rating of the capacitor, the capacitor may fail.



    If you put 24V across a cap, it will charge to 24V.



    If it's spec'ed at 22000uF it's 22000uF for ANY voltage.






    share|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Note that "fail" in this case probably means "explode" or at least "vent extremely hot vaporized electrolyte". Electrolytic capacitors do not like being mistreated, and I doubt you'll get 22mF out of anything other than aluminum electrolytics.
      $endgroup$
      – Hearth
      4 hours ago










    • $begingroup$
      Your final sentence needs a big asterisk for ceramics. Not that you're likely to use ceramics for 22000 uF.
      $endgroup$
      – The Photon
      3 hours ago
















    3












    $begingroup$

    If you exceed the voltage rating of the capacitor, the capacitor may fail.



    If you put 24V across a cap, it will charge to 24V.



    If it's spec'ed at 22000uF it's 22000uF for ANY voltage.






    share|improve this answer









    $endgroup$









    • 1




      $begingroup$
      Note that "fail" in this case probably means "explode" or at least "vent extremely hot vaporized electrolyte". Electrolytic capacitors do not like being mistreated, and I doubt you'll get 22mF out of anything other than aluminum electrolytics.
      $endgroup$
      – Hearth
      4 hours ago










    • $begingroup$
      Your final sentence needs a big asterisk for ceramics. Not that you're likely to use ceramics for 22000 uF.
      $endgroup$
      – The Photon
      3 hours ago














    3












    3








    3





    $begingroup$

    If you exceed the voltage rating of the capacitor, the capacitor may fail.



    If you put 24V across a cap, it will charge to 24V.



    If it's spec'ed at 22000uF it's 22000uF for ANY voltage.






    share|improve this answer









    $endgroup$



    If you exceed the voltage rating of the capacitor, the capacitor may fail.



    If you put 24V across a cap, it will charge to 24V.



    If it's spec'ed at 22000uF it's 22000uF for ANY voltage.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered 4 hours ago









    mike65535mike65535

    1,0862719




    1,0862719








    • 1




      $begingroup$
      Note that "fail" in this case probably means "explode" or at least "vent extremely hot vaporized electrolyte". Electrolytic capacitors do not like being mistreated, and I doubt you'll get 22mF out of anything other than aluminum electrolytics.
      $endgroup$
      – Hearth
      4 hours ago










    • $begingroup$
      Your final sentence needs a big asterisk for ceramics. Not that you're likely to use ceramics for 22000 uF.
      $endgroup$
      – The Photon
      3 hours ago














    • 1




      $begingroup$
      Note that "fail" in this case probably means "explode" or at least "vent extremely hot vaporized electrolyte". Electrolytic capacitors do not like being mistreated, and I doubt you'll get 22mF out of anything other than aluminum electrolytics.
      $endgroup$
      – Hearth
      4 hours ago










    • $begingroup$
      Your final sentence needs a big asterisk for ceramics. Not that you're likely to use ceramics for 22000 uF.
      $endgroup$
      – The Photon
      3 hours ago








    1




    1




    $begingroup$
    Note that "fail" in this case probably means "explode" or at least "vent extremely hot vaporized electrolyte". Electrolytic capacitors do not like being mistreated, and I doubt you'll get 22mF out of anything other than aluminum electrolytics.
    $endgroup$
    – Hearth
    4 hours ago




    $begingroup$
    Note that "fail" in this case probably means "explode" or at least "vent extremely hot vaporized electrolyte". Electrolytic capacitors do not like being mistreated, and I doubt you'll get 22mF out of anything other than aluminum electrolytics.
    $endgroup$
    – Hearth
    4 hours ago












    $begingroup$
    Your final sentence needs a big asterisk for ceramics. Not that you're likely to use ceramics for 22000 uF.
    $endgroup$
    – The Photon
    3 hours ago




    $begingroup$
    Your final sentence needs a big asterisk for ceramics. Not that you're likely to use ceramics for 22000 uF.
    $endgroup$
    – The Photon
    3 hours ago













    3












    $begingroup$


    If I supply 24v to a 50v rated 22000uf electrolytic capacitor, does that mean it will store 44000uf at 24v?




    No. The capacitor's capacity is a property of its dimensions.



    $$ C propto frac {A}{d} $$ where A is the area of the capacitor and d is the distance between the plates. Note that voltage doesn't appear in the equation so it's still 22,000 μF even at zero volts.




    Is the higher voltage just a safety thing, or can the capacitor actually store more charge?




    The capacitor has a maximum voltage that the dialectric between its plates can withstand if you exceed this the dielectric will break down, current will flow and the capacitor will overheat and possibly explode.



    The amount of charge stored is given by $$ Q = {C}{V} $$ so the higher the voltage on a given capacitor then the more charge is stored. Again, check for nonsense by testing for 0 V: at 0 V the charge, Q, will be zero. This makes sense.




    Also, if I supply it with 24v, will it output 24v or 50v?




    If you charge it to 24 V, disconnect the supply and measure the voltage you will get a reading of 24 V. Once you start to draw current from it the voltage will decrease.



    enter image description here



    Figure 1. When discharging through a resistor the time constant, τ can be calculated as RC. After RC seconds the capacitor will have discharged by 63%. After 3RC, 95% and after 5RC by about 99%. Image source: I am Technical.





    Finally, a few points: SI symbols for the volt is 'V' and for the farad is 'F'. The symbols are capitalised but units named after a person are lowercase when spelled out.






    share|improve this answer









    $endgroup$


















      3












      $begingroup$


      If I supply 24v to a 50v rated 22000uf electrolytic capacitor, does that mean it will store 44000uf at 24v?




      No. The capacitor's capacity is a property of its dimensions.



      $$ C propto frac {A}{d} $$ where A is the area of the capacitor and d is the distance between the plates. Note that voltage doesn't appear in the equation so it's still 22,000 μF even at zero volts.




      Is the higher voltage just a safety thing, or can the capacitor actually store more charge?




      The capacitor has a maximum voltage that the dialectric between its plates can withstand if you exceed this the dielectric will break down, current will flow and the capacitor will overheat and possibly explode.



      The amount of charge stored is given by $$ Q = {C}{V} $$ so the higher the voltage on a given capacitor then the more charge is stored. Again, check for nonsense by testing for 0 V: at 0 V the charge, Q, will be zero. This makes sense.




      Also, if I supply it with 24v, will it output 24v or 50v?




      If you charge it to 24 V, disconnect the supply and measure the voltage you will get a reading of 24 V. Once you start to draw current from it the voltage will decrease.



      enter image description here



      Figure 1. When discharging through a resistor the time constant, τ can be calculated as RC. After RC seconds the capacitor will have discharged by 63%. After 3RC, 95% and after 5RC by about 99%. Image source: I am Technical.





      Finally, a few points: SI symbols for the volt is 'V' and for the farad is 'F'. The symbols are capitalised but units named after a person are lowercase when spelled out.






      share|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$


        If I supply 24v to a 50v rated 22000uf electrolytic capacitor, does that mean it will store 44000uf at 24v?




        No. The capacitor's capacity is a property of its dimensions.



        $$ C propto frac {A}{d} $$ where A is the area of the capacitor and d is the distance between the plates. Note that voltage doesn't appear in the equation so it's still 22,000 μF even at zero volts.




        Is the higher voltage just a safety thing, or can the capacitor actually store more charge?




        The capacitor has a maximum voltage that the dialectric between its plates can withstand if you exceed this the dielectric will break down, current will flow and the capacitor will overheat and possibly explode.



        The amount of charge stored is given by $$ Q = {C}{V} $$ so the higher the voltage on a given capacitor then the more charge is stored. Again, check for nonsense by testing for 0 V: at 0 V the charge, Q, will be zero. This makes sense.




        Also, if I supply it with 24v, will it output 24v or 50v?




        If you charge it to 24 V, disconnect the supply and measure the voltage you will get a reading of 24 V. Once you start to draw current from it the voltage will decrease.



        enter image description here



        Figure 1. When discharging through a resistor the time constant, τ can be calculated as RC. After RC seconds the capacitor will have discharged by 63%. After 3RC, 95% and after 5RC by about 99%. Image source: I am Technical.





        Finally, a few points: SI symbols for the volt is 'V' and for the farad is 'F'. The symbols are capitalised but units named after a person are lowercase when spelled out.






        share|improve this answer









        $endgroup$




        If I supply 24v to a 50v rated 22000uf electrolytic capacitor, does that mean it will store 44000uf at 24v?




        No. The capacitor's capacity is a property of its dimensions.



        $$ C propto frac {A}{d} $$ where A is the area of the capacitor and d is the distance between the plates. Note that voltage doesn't appear in the equation so it's still 22,000 μF even at zero volts.




        Is the higher voltage just a safety thing, or can the capacitor actually store more charge?




        The capacitor has a maximum voltage that the dialectric between its plates can withstand if you exceed this the dielectric will break down, current will flow and the capacitor will overheat and possibly explode.



        The amount of charge stored is given by $$ Q = {C}{V} $$ so the higher the voltage on a given capacitor then the more charge is stored. Again, check for nonsense by testing for 0 V: at 0 V the charge, Q, will be zero. This makes sense.




        Also, if I supply it with 24v, will it output 24v or 50v?




        If you charge it to 24 V, disconnect the supply and measure the voltage you will get a reading of 24 V. Once you start to draw current from it the voltage will decrease.



        enter image description here



        Figure 1. When discharging through a resistor the time constant, τ can be calculated as RC. After RC seconds the capacitor will have discharged by 63%. After 3RC, 95% and after 5RC by about 99%. Image source: I am Technical.





        Finally, a few points: SI symbols for the volt is 'V' and for the farad is 'F'. The symbols are capitalised but units named after a person are lowercase when spelled out.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 4 hours ago









        TransistorTransistor

        90.2k788194




        90.2k788194























            0












            $begingroup$

            I'll try to explain with a gradual approach:



            A capacitor consists of two plates or conductive surfaces separated by a gap or an insulator (called the dielectric). Voltage is potential difference between any two points. Another name for voltage is electromotive force, which starts to convey the idea that it describes how "motivated" an electric charge is to do work.



            I won't try to explain the how, but suffice to say that the positive and negative charges are attracted to each other, and will "stay" on the plates opposite the dielectric, until provided a connecting path (circuit).



            A capacitor's voltage rating is based on its ability to function correctly up to the specified voltage. It encompasses things such as properties of the dielectric and physical dimensions of the capacitor. With enough voltage, insulators will eventually conduct (usually temporarily) and break down. Thus, if a capacitor is rated for 50 volts, it means that the dielectric composition, placement, thickness, etc. are only designed to function up to 50V.



            The Farad is a unit of capacitance or amount of electric charge. (Technically it's equal to one coulomb (one ampere of current in one second) stored with the potential difference (voltage) of one volt.)



            Analogies always fail at some point, but think of a capacitor like a sphere in a water plumbing system which has a flexible diaphragm separating it into halves1. If you pump water into half of it, you're storing some quantity of water. If the pressure of the water is too high, you'll rupture the diaphragm. You can store some quantity of water up to the maximum capacity of the vessel, and it will be stored at whatever pressure was applied. (Obviously this analogy fails when you think of some small quantity of water not even causing the diaphragm to stretch at all, so you should dispense with hydraulic analogies as soon as possible...)



            You can therefore use a 50V 22000µF capacitor to store charge at 1 volt, 10 volts, or 24 volts. In all cases, it will store a maximum of 22000µF, at the same voltage it was applied. Exceeding 50V will cause it to fail. (Applying 51V will probably result in a slow failure over time whereas 100V will cause a much more abrupt and potentially dangerous failure.)



            You might be thinking "Well, if it can store 22000µF at 50V, why can't it store twice as much at half the voltage?"



            It's tempting to think this if, for example, you're familiar with pneumatic systems: You have a two liter tank and store compressed gas in it. You can store a given quantity of gas at some pressure, and a larger quantity of gas at a higher pressure. (Within the tolerances and specifications of the gas and vessel, of course.) But this is not an analogy to electric charge. In order to understand a capacitor, you have to avoid thinking about it as either a pneumatic or hydraulic system — things which many people have everyday experience with prior to learning about electronics.



            A capacitor won't store more than its rated capacitance at any voltage, and won't function properly (read: survive) exceeding its voltage rating. Also be careful to observe polarity. Generally capacitors with such capacity are polarized. Connecting positive and negative incorrectly will also result in failure, often dramatically.





            1Credit to Warren J. Beaty for this particular analogy.






            share|improve this answer











            $endgroup$













            • $begingroup$
              Or think of the capacitor as a barrel or a bucket. A 5 gallon bucket can hold 5 gallons of water, no matter the water pressure used to fill it. If you fill it at a rate of 2.4 gal/minute, it doesn't hold twice as much water as if it were filled at 5.0 gal/minute.
              $endgroup$
              – spuck
              2 hours ago
















            0












            $begingroup$

            I'll try to explain with a gradual approach:



            A capacitor consists of two plates or conductive surfaces separated by a gap or an insulator (called the dielectric). Voltage is potential difference between any two points. Another name for voltage is electromotive force, which starts to convey the idea that it describes how "motivated" an electric charge is to do work.



            I won't try to explain the how, but suffice to say that the positive and negative charges are attracted to each other, and will "stay" on the plates opposite the dielectric, until provided a connecting path (circuit).



            A capacitor's voltage rating is based on its ability to function correctly up to the specified voltage. It encompasses things such as properties of the dielectric and physical dimensions of the capacitor. With enough voltage, insulators will eventually conduct (usually temporarily) and break down. Thus, if a capacitor is rated for 50 volts, it means that the dielectric composition, placement, thickness, etc. are only designed to function up to 50V.



            The Farad is a unit of capacitance or amount of electric charge. (Technically it's equal to one coulomb (one ampere of current in one second) stored with the potential difference (voltage) of one volt.)



            Analogies always fail at some point, but think of a capacitor like a sphere in a water plumbing system which has a flexible diaphragm separating it into halves1. If you pump water into half of it, you're storing some quantity of water. If the pressure of the water is too high, you'll rupture the diaphragm. You can store some quantity of water up to the maximum capacity of the vessel, and it will be stored at whatever pressure was applied. (Obviously this analogy fails when you think of some small quantity of water not even causing the diaphragm to stretch at all, so you should dispense with hydraulic analogies as soon as possible...)



            You can therefore use a 50V 22000µF capacitor to store charge at 1 volt, 10 volts, or 24 volts. In all cases, it will store a maximum of 22000µF, at the same voltage it was applied. Exceeding 50V will cause it to fail. (Applying 51V will probably result in a slow failure over time whereas 100V will cause a much more abrupt and potentially dangerous failure.)



            You might be thinking "Well, if it can store 22000µF at 50V, why can't it store twice as much at half the voltage?"



            It's tempting to think this if, for example, you're familiar with pneumatic systems: You have a two liter tank and store compressed gas in it. You can store a given quantity of gas at some pressure, and a larger quantity of gas at a higher pressure. (Within the tolerances and specifications of the gas and vessel, of course.) But this is not an analogy to electric charge. In order to understand a capacitor, you have to avoid thinking about it as either a pneumatic or hydraulic system — things which many people have everyday experience with prior to learning about electronics.



            A capacitor won't store more than its rated capacitance at any voltage, and won't function properly (read: survive) exceeding its voltage rating. Also be careful to observe polarity. Generally capacitors with such capacity are polarized. Connecting positive and negative incorrectly will also result in failure, often dramatically.





            1Credit to Warren J. Beaty for this particular analogy.






            share|improve this answer











            $endgroup$













            • $begingroup$
              Or think of the capacitor as a barrel or a bucket. A 5 gallon bucket can hold 5 gallons of water, no matter the water pressure used to fill it. If you fill it at a rate of 2.4 gal/minute, it doesn't hold twice as much water as if it were filled at 5.0 gal/minute.
              $endgroup$
              – spuck
              2 hours ago














            0












            0








            0





            $begingroup$

            I'll try to explain with a gradual approach:



            A capacitor consists of two plates or conductive surfaces separated by a gap or an insulator (called the dielectric). Voltage is potential difference between any two points. Another name for voltage is electromotive force, which starts to convey the idea that it describes how "motivated" an electric charge is to do work.



            I won't try to explain the how, but suffice to say that the positive and negative charges are attracted to each other, and will "stay" on the plates opposite the dielectric, until provided a connecting path (circuit).



            A capacitor's voltage rating is based on its ability to function correctly up to the specified voltage. It encompasses things such as properties of the dielectric and physical dimensions of the capacitor. With enough voltage, insulators will eventually conduct (usually temporarily) and break down. Thus, if a capacitor is rated for 50 volts, it means that the dielectric composition, placement, thickness, etc. are only designed to function up to 50V.



            The Farad is a unit of capacitance or amount of electric charge. (Technically it's equal to one coulomb (one ampere of current in one second) stored with the potential difference (voltage) of one volt.)



            Analogies always fail at some point, but think of a capacitor like a sphere in a water plumbing system which has a flexible diaphragm separating it into halves1. If you pump water into half of it, you're storing some quantity of water. If the pressure of the water is too high, you'll rupture the diaphragm. You can store some quantity of water up to the maximum capacity of the vessel, and it will be stored at whatever pressure was applied. (Obviously this analogy fails when you think of some small quantity of water not even causing the diaphragm to stretch at all, so you should dispense with hydraulic analogies as soon as possible...)



            You can therefore use a 50V 22000µF capacitor to store charge at 1 volt, 10 volts, or 24 volts. In all cases, it will store a maximum of 22000µF, at the same voltage it was applied. Exceeding 50V will cause it to fail. (Applying 51V will probably result in a slow failure over time whereas 100V will cause a much more abrupt and potentially dangerous failure.)



            You might be thinking "Well, if it can store 22000µF at 50V, why can't it store twice as much at half the voltage?"



            It's tempting to think this if, for example, you're familiar with pneumatic systems: You have a two liter tank and store compressed gas in it. You can store a given quantity of gas at some pressure, and a larger quantity of gas at a higher pressure. (Within the tolerances and specifications of the gas and vessel, of course.) But this is not an analogy to electric charge. In order to understand a capacitor, you have to avoid thinking about it as either a pneumatic or hydraulic system — things which many people have everyday experience with prior to learning about electronics.



            A capacitor won't store more than its rated capacitance at any voltage, and won't function properly (read: survive) exceeding its voltage rating. Also be careful to observe polarity. Generally capacitors with such capacity are polarized. Connecting positive and negative incorrectly will also result in failure, often dramatically.





            1Credit to Warren J. Beaty for this particular analogy.






            share|improve this answer











            $endgroup$



            I'll try to explain with a gradual approach:



            A capacitor consists of two plates or conductive surfaces separated by a gap or an insulator (called the dielectric). Voltage is potential difference between any two points. Another name for voltage is electromotive force, which starts to convey the idea that it describes how "motivated" an electric charge is to do work.



            I won't try to explain the how, but suffice to say that the positive and negative charges are attracted to each other, and will "stay" on the plates opposite the dielectric, until provided a connecting path (circuit).



            A capacitor's voltage rating is based on its ability to function correctly up to the specified voltage. It encompasses things such as properties of the dielectric and physical dimensions of the capacitor. With enough voltage, insulators will eventually conduct (usually temporarily) and break down. Thus, if a capacitor is rated for 50 volts, it means that the dielectric composition, placement, thickness, etc. are only designed to function up to 50V.



            The Farad is a unit of capacitance or amount of electric charge. (Technically it's equal to one coulomb (one ampere of current in one second) stored with the potential difference (voltage) of one volt.)



            Analogies always fail at some point, but think of a capacitor like a sphere in a water plumbing system which has a flexible diaphragm separating it into halves1. If you pump water into half of it, you're storing some quantity of water. If the pressure of the water is too high, you'll rupture the diaphragm. You can store some quantity of water up to the maximum capacity of the vessel, and it will be stored at whatever pressure was applied. (Obviously this analogy fails when you think of some small quantity of water not even causing the diaphragm to stretch at all, so you should dispense with hydraulic analogies as soon as possible...)



            You can therefore use a 50V 22000µF capacitor to store charge at 1 volt, 10 volts, or 24 volts. In all cases, it will store a maximum of 22000µF, at the same voltage it was applied. Exceeding 50V will cause it to fail. (Applying 51V will probably result in a slow failure over time whereas 100V will cause a much more abrupt and potentially dangerous failure.)



            You might be thinking "Well, if it can store 22000µF at 50V, why can't it store twice as much at half the voltage?"



            It's tempting to think this if, for example, you're familiar with pneumatic systems: You have a two liter tank and store compressed gas in it. You can store a given quantity of gas at some pressure, and a larger quantity of gas at a higher pressure. (Within the tolerances and specifications of the gas and vessel, of course.) But this is not an analogy to electric charge. In order to understand a capacitor, you have to avoid thinking about it as either a pneumatic or hydraulic system — things which many people have everyday experience with prior to learning about electronics.



            A capacitor won't store more than its rated capacitance at any voltage, and won't function properly (read: survive) exceeding its voltage rating. Also be careful to observe polarity. Generally capacitors with such capacity are polarized. Connecting positive and negative incorrectly will also result in failure, often dramatically.





            1Credit to Warren J. Beaty for this particular analogy.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 3 hours ago

























            answered 3 hours ago









            JYeltonJYelton

            16.5k2891194




            16.5k2891194












            • $begingroup$
              Or think of the capacitor as a barrel or a bucket. A 5 gallon bucket can hold 5 gallons of water, no matter the water pressure used to fill it. If you fill it at a rate of 2.4 gal/minute, it doesn't hold twice as much water as if it were filled at 5.0 gal/minute.
              $endgroup$
              – spuck
              2 hours ago


















            • $begingroup$
              Or think of the capacitor as a barrel or a bucket. A 5 gallon bucket can hold 5 gallons of water, no matter the water pressure used to fill it. If you fill it at a rate of 2.4 gal/minute, it doesn't hold twice as much water as if it were filled at 5.0 gal/minute.
              $endgroup$
              – spuck
              2 hours ago
















            $begingroup$
            Or think of the capacitor as a barrel or a bucket. A 5 gallon bucket can hold 5 gallons of water, no matter the water pressure used to fill it. If you fill it at a rate of 2.4 gal/minute, it doesn't hold twice as much water as if it were filled at 5.0 gal/minute.
            $endgroup$
            – spuck
            2 hours ago




            $begingroup$
            Or think of the capacitor as a barrel or a bucket. A 5 gallon bucket can hold 5 gallons of water, no matter the water pressure used to fill it. If you fill it at a rate of 2.4 gal/minute, it doesn't hold twice as much water as if it were filled at 5.0 gal/minute.
            $endgroup$
            – spuck
            2 hours ago











            0












            $begingroup$

            'the voltage rating is just a limit. "halfway full" means it can hold more charge, but only with higher voltages.'



            Thank you dandavis, and the rest of you for answering it broadly






            share|improve this answer








            New contributor




            Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$













            • $begingroup$
              Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review
              $endgroup$
              – JRE
              1 hour ago
















            0












            $begingroup$

            'the voltage rating is just a limit. "halfway full" means it can hold more charge, but only with higher voltages.'



            Thank you dandavis, and the rest of you for answering it broadly






            share|improve this answer








            New contributor




            Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$













            • $begingroup$
              Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review
              $endgroup$
              – JRE
              1 hour ago














            0












            0








            0





            $begingroup$

            'the voltage rating is just a limit. "halfway full" means it can hold more charge, but only with higher voltages.'



            Thank you dandavis, and the rest of you for answering it broadly






            share|improve this answer








            New contributor




            Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$



            'the voltage rating is just a limit. "halfway full" means it can hold more charge, but only with higher voltages.'



            Thank you dandavis, and the rest of you for answering it broadly







            share|improve this answer








            New contributor




            Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            share|improve this answer



            share|improve this answer






            New contributor




            Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.









            answered 2 hours ago









            MladenMladen

            112




            112




            New contributor




            Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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            New contributor





            Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            Mladen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.












            • $begingroup$
              Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review
              $endgroup$
              – JRE
              1 hour ago


















            • $begingroup$
              Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review
              $endgroup$
              – JRE
              1 hour ago
















            $begingroup$
            Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review
            $endgroup$
            – JRE
            1 hour ago




            $begingroup$
            Please don't add "thank you" as an answer. Instead, accept the answer that you found most helpful. - From Review
            $endgroup$
            – JRE
            1 hour ago










            Mladen is a new contributor. Be nice, and check out our Code of Conduct.










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