I have a 25V 100uF capacitor and a 50V 100uF one. If I charge both up to 25 volts, then is the first one going to have 100uF of actual charge, and the second one only 50uF?
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3Capacitance is not a measure of charge. It is defined as the ratio of the charge stored to the voltage across it. – jramsay42 Feb 10 '18 at 22:55
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3Nope. Both will have 100µF at 25V. The 25V one will just not last very long... – Turbo J Feb 10 '18 at 22:56
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Thanks for the comments, now i see i need to learn a lot more about capacitors. – electroexplorer Feb 10 '18 at 23:11
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2Please first work on how to ask questions in a better way – Dumbo Feb 10 '18 at 23:23
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I fixed up the mess. Maybe OP will get a clue. – Feb 11 '18 at 03:03
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100 uF means capacitance. It seldom changes with the applied voltage. If you charge both the capacitors to 25 V, both will have accumulated a charge.
Q = CV = 100 uF x 25 V = 2500 micro coulombs.
Meenie Leis
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1For consistency you should probably write either '100u x 25' (no units) or '100 uF x 25 V' (both units). – Transistor Feb 10 '18 at 23:22
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2If you think capacitance "never changes with the applied voltage" you need to avoid designing with ceramic capacitors above a few nF. – The Photon Feb 11 '18 at 00:20
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1@ThePhoton is correct. Super-capacitors and large electrolytics in general change value with voltage, just like solid carbon resistors do. Plus they change with age, which is why the tolerance is often -10%/+20%. – Feb 11 '18 at 03:01
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1So do the small, high-value multilayer ceramics so popular on boards today. – Chris Stratton Feb 11 '18 at 03:11
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1Nobody has taught us that at school. Thanks for new knowledge ... :-D – Meenie Leis Feb 11 '18 at 04:56
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1https://electronics.stackexchange.com/questions/216367/why-does-capacitance-value-changes-with-applied-voltage . found reasons here. – Meenie Leis Feb 11 '18 at 05:02