can a capacitor voltage be higher than the source's voltage? how? What about Kirchhoff's loop rule?
2007-07-26 15:56:04 · 8 answers · asked by Anonymous in Science & Mathematics ➔ Physics
could that be done in an RLC circuit?
2007-07-26 16:13:14 · update #1
If there's a lot of inductance in the circuit, it would be entirely possible when a switch is turned off (di/dt is extremely large), in which case it's called inductive "kick".
2007-07-26 16:07:10 · answer #1 · answered by supastremph 6 · 1⤊ 0⤋
The voltage stored in a capacitor cannot be higher than the source voltage. The reason is that in a DC circuit the capacitor acts as a wire at first, but as it starts to store up charge it will become more of an open in the circuit. The time it takes to charge is equal to 5 tau's. Tau is calculated by Rtotal * Ctotal in the circuit.
If the capacitor acts as an open than the charge it has stored in it is equal to the sum of the source charges that run through it in the circuit. In an open the charge is sitting on 1 end of the broken wire, and is equal to the source voltage going into it. The capacitor will have the same charge in it when fully charged because it acts as an open and follows the same rules.
With an RCL circuit it'll remain the same way if it's a DC source, however the inductors will act as opens initially, until they have their magnetic field fully formed, then they'll act as shorts and can be used to bypass the capacitor in a DC circuit because votage follows the path of least resistance to ground.
2007-07-26 18:08:08 · answer #2 · answered by dkillinx 3 · 0⤊ 0⤋
If the capacitor is in parallel across teh voltage source it cannot exceed the voltage source. If it is in series it still can't exceed the voltage of the source. In an ac circuit, there is a phase difference between current and voltage through the capacitor but the voltage follows the source.
2007-07-26 16:01:08 · answer #3 · answered by nyphdinmd 7 · 0⤊ 0⤋
The only way that I know of is to physically separate the plates after they have been charged. The voltage will then increase.
Edit:
Otherwise I do not believe it can be increased without active components. With diodes and AC there are circuits that will double, triple etc. the charge in a cap.
http://en.wikipedia.org/wiki/Voltage_doubler
Some semantics:
There are two different voltages associated with a capacitor; charge voltage and breakdown voltage. I (and everyone else) have been assuming you mean charge voltage.
2007-07-26 20:38:29 · answer #4 · answered by J C 5 · 0⤊ 0⤋
No.
Capacitor circuits can be designed with multiple capacitors which double or triple the voltage - but each individual capacitor can only charge up to the potential placed across its leads.
2007-07-26 16:51:10 · answer #5 · answered by LeAnne 7 · 0⤊ 0⤋
That can't be answered outside the context of a circuit. A photographer's electronic flash uses a 6 volt battery to charge the Xenon flash tube's capacitor, often to several hundred volts. KVL and KCL apply to circuits, so we can't tell you how they apply to a particular circuit without knowing the circuit.
2007-07-26 19:28:14 · answer #6 · answered by Frank N 7 · 0⤊ 0⤋
No, a capacitor is a storage device, it can only hold voltage equal to the source.
2007-07-29 10:47:54 · answer #7 · answered by johnandeileen2000 7 · 0⤊ 0⤋
if you're using a simple (passive) circuit - there's no way I know of (ok it's been 30 years since I took circuits) that the cap can have a voltage greater than the source
BUT if you put an opamp between the source and the cap it can
like i said it's been so long since i took circuits - i have forgotten kirchoffs rule
2007-07-26 16:00:33 · answer #8 · answered by lancej0hns0n 4 · 0⤊ 0⤋