shespuzzling,



I do too. I think you need to understand the relationship between voltage, charge, capacitance, and energy. Let's start with the basics. First, like charges repel one another. It takes energy to bring them close together. It takes more energy to concentrate like charges even closer together. This energy is stored in an electrostatic field and is potential energy. The total amount of energy accumulated divided by the amount of charge is voltage. So voltage is the energy density of the charge. You can have the same voltage (energy density) with some charge close together or less charge closer together. A capacitor is a charge separation container. When one plate accumulates a charge, the other plate depletes an equal amount of charge. There is a net gain/loss of zero charge, but the separated charge on the plates stores energy in the form of a electrostatic field. So it can be said that the capacitor becomes "charged" with energy. There will be a higher energy density on one plate and a lower energy density on the opposite plate, resulting in a voltage difference between the plates. A smaller capacitor (charge container) concentrates the charge closer together, so for the same charge, a smaller capacitor will show a higher voltage difference between the plates.

Now let's see what happens to a capacitor when a voltage is placed across it. The voltage forces the charge to accumulate at a rate dependent on the resistance in the circuit. This charge will cause a back voltage which is opposite to the charging voltage. So the displacement current becomes less and less until it becomes zero when the back voltage reaches the charging voltage. Is that what they mean when they say a capacitor resists a voltage change? A capacitor cannot have a back voltage higher than the charging voltage no matter how large its plates are. A larger capacitor (from the word capacity) can store more charge at the same voltage than a smaller one. A capacitor does not dissipate energy unless there are imperfections like leakage or dielectric absorption. A capacitor stores and releases energy to/from the circuit thereby raising or lowering its voltage.

If you have any questions about this, just ask away.

Ratch

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

[Moderator's note: @Ratch This thread created for this new post that was originally posted to a very old thread. http://forum.allaboutcircuits.com/threads/why-how-do-capacitors-resist-change-in-voltage.27121/#post-1022460]

 

What is "r" in kQ/r?

 

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).
.

 

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).
.

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.

 

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

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.