# Capacitive Catastrophe

1. Jan 5, 2009

### XPTPCREWX

Perhaps some one can explain to me what is meant by "current leads the voltage by 90 degrees"...let this inquiry be in regards to a purely capacitive circuit.

What seems to baffle me is why this terminology is used. Its not like the current just decided to occur before a potenial difference.

2. Jan 5, 2009

### Staff: Mentor

It's easiest to get comfortable with this concept based on the differential equations for the currents and voltages for inductors and capacitors.

For Capacitors, the driving function is the current, which charges up the capacitors:

$$v(t) = \frac{1}{C}\int{i(t) dt}$$

The voltage across a capacitor cannot change instantaneously, because that would require an infinite current.

Similarly for Inductors, the driving function is the voltage, which ramps up the current:

$$i(t) = \frac{1}{L}\int{v(t) dt}$$

The current in the inductor cannot change instantaneously, bacause that would require an infinite voltage. The current can change very quickly, however, like in a flyback power supply circuit that is pumping a high voltage...

So when we say that the current lags the applied voltage in inductors, or that the voltage lags the applied current in capacitors, we are just saying in words what you get from the differential equations that describe the physics of those components.

And when we say "the current leads the voltage...", we are not describing something non-causal. It's just a bit of a confusing way to state the relationship the other way around, and for a continuous waveform.

Hope that helps.

Last edited: Jan 6, 2009
3. Jan 5, 2009

### XPTPCREWX

Really helps....Thanks.

4. Jan 5, 2009

### yungman

To make it very simple. Before the cap can have any voltage across it's terminals, you need to charge it up. So current has to be pumped into the cap before the cap can charge up to have voltage. Current always have to come first so it lead the voltage!!! This is the ABC way to explain it!!!:rofl:

Berkeman has given you the detail of the rest.

5. Jan 6, 2009

### XPTPCREWX

yungman,

Actually it isn't that simple. there is obviously a potential difference between the applied voltage and the capacitors terminal voltage to create that current flow in the first place.

Berkeman has explained it to me the best so far.

6. Jan 6, 2009

### yungman

I am glad you are happy with berkeman's answer. This is a very simple question that you find in the first few lessons of AC circuits, so I want to give a very simple answer just in case. I just add this in because there are people that don't do calculus!!! I was one of those long time ago!!!!