Why/how do capacitors resist change in voltage? Part 2

Discussion in 'General Electronics Chat' started by shashank95, Jul 21, 2016.

1. shashank95 Thread Starter New Member

Mar 22, 2016
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Hi, I have a doubt. You just said that a higher capacitance could store more charge at the same voltage. It is justified by Q = CV. But if the amount of stored charge on a capacitor increases, won't the Voltage across the capacitor increase ? As V = kQ/r . How is this happening actually

Last edited by a moderator: Jul 22, 2016
2. crutschow Expert

Mar 14, 2008
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What is "r" in kQ/r?

3. profbuxton Member

Feb 21, 2014
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161
Capacitors do not "resist change in voltage" as such. When you apply a voltage(dc) to a discharged capacitor at t=0 it appears as a "short circuit" to the power supply(battery or whatever). Now a short circuit has no volts across it and takes a large current(as you know). So a large current will flow in the capacitor and charge up the plates(or whatever it is made of). Obviously the voltage will be low at this time. As the capacitor charges the current drawn from the supply is reduced and the voltage across the plates will increase to approx full supply volts.
So you see the cap did not resist change in voltage. It also means that the current "leads" the voltage by some amount(depends on circuit resistance).
.

4. Papabravo Expert

Feb 24, 2006
11,737
2,483
Ah...resisting a change in voltage means that the voltage cannot change instantaneously. A capacitor certainly satisfies this requirement. You realize this must be the case for a capacitor when you understand that it is the time integral of current which determines the voltage. The rate of increase in voltage across the capacitor decreases as the voltage rises and approaches the charging voltage asymptotically while never actually reaching the charging voltage.

5. ErnieM AAC Fanatic!

Apr 24, 2011
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If you put more charge into one capacitor then yes the voltage will increase.

But here you say you are looking at two different caps. The cap with the higher capacitance needs more charge than the smaller one to get to the same voltage, as the formula you offer implies.