Hey there,I've been learning about AC circuits recently and while

[EEWannabe]

Hey there,

I've been learning about AC circuits recently and while considering and LCR circuit it occurred to me; it's possible for the voltage drop across a capacitor/inductor/both to be greater than the voltage source itself however I can't think of the reason for this, any hints or reasons would be greatly appreciated.

The only idea I can think of is that since energy is stored in both of them the potential difference across them increases, but I don't think that's right

 

[Drakkith]

If it's an AC circuit, would inductance have anything to do with it? Or maybe something about the capacitor being charged or not? I don't know enough about electrical circuits to help you I'm afraid.

 

[Petr Mugver]

Try posting the circuit and we'll see...

Yes, if you charge a capacitor in an RC circuit with a certain voltage source and then you disconnect the generator and substitute it with a smaller one, then you have the situation like the one you described, at least for some time.

 

[EEWannabe]

Well I'm just considering a simple series RLC circuit.

Because the impedances of the capacitor and inductor cancel due to their phase difference, it's actually possible for the impedance of the capacitor to be greater than the overall impedance of the circuit. And since the V through the Capacitor for example is V(c) = Vo Z(capacitor)/ Z(total).

I don't see how that could be true though...

 

[PrakashPhy]

i think the problem here is in understanding peak value of voltage drop and instantaneous value of voltage.
it is in fact true that the peak value of voltage drop across a capacitor or a inductor can be greater than the peak value of e m f because there is a phase difference between current and voltage. but the sum of instantaneous values of voltage drop across all circuit elements is equal to instanteneous emf.

 

[Bill_K]

For a series RLC circuit, Z = R + iωL + 1/iωC. At resonance ω = √LC, the impedances of the last two will be equal and opposite, and the current will be determined by R alone. If the two terms are individually both large (not small) you will get a large voltage drop across each one, in opposite directions, which may exceed the supplied voltage.

 

[pallidin]

Of course, as special circuitry(ie. transformer or other) increases voltage the current is reduced, satisfying the need for the power factor to be the same or usually always slightly less(losses).
Inversely, one can increase current with a required reduction in voltage.

The Power Factor MUST remain the same or less in these types of cases.