Understanding Capacitors and Reactance in AC Circuits

[jeff1evesque]

The more rapidly voltage across a capacitor changes, the more current will flow (the higher the frequency, the greater the current).


Questions
I was wondering, does the statement above deal specifically with AC circuits by the statement

The more rapidly voltage across a capacitor changes...

Or does this statement apply to DC circuits also? And I was also wondering what does it mean

...the higher the frequency, the greater the current

Is this frequency under consideration \omega for AC circuits [phasor form]? Could someone provide an equation of some sort to help my understanding?


Thanks a lot,


Jeff

 

[Fenn]

That statement refers to AC circuits, as a DC circuit is expected to maintain a constant voltage across a capacitor. Voltage spikes in a DC circuit cause current spikes across capacitors, which this statement can also apply to, but usually that is referred to as AC behaviour.

 

[cipher42]

The over-all context of the sentence would be needed to pin it down for certain, but since the sentence mentions a frequency, I would be fairly certain that we're talking about AC circuits here.

Think about it this way: we know that in a DC circuit, the more you charge a capacitor, the more it resists the flow of current (as time goes on, it will slow to a trickle as the capacitor becomes fully charged). Now switch to thinking about an AC circuit. The alternating current is continually charging the capacitor, then uncharging it, then charging it again...rinse and repeat for each cycle. But we said that the more charged up the capacitor is, the more it resists the flow of current. So (for some fixed maximum voltage) a higher frequency means that the capacitor will have less time to charge on each cycle, hence it will be less charged, and therefore it will do less to oppose the flow of current. Inversely, if the frequency is low, it will have a lot of time to build of charge, and therefore it will oppose the current greatly.

This is all summed up in the equation for capacitive reactance
X_c=\frac{1}{\omega C}
In the sense that a capacitor in an AC circuit will oppose the flow of current, it provides a resistance of sorts; this is called it's reactance. As you can see from the above equation, the higher the frequency, the lower the reactance...lower reactance means that (all other things equal) more current will flow.

 

[jeff1evesque]

The over-all context of the sentence would be needed to pin it down for certain, but since the sentence mentions a frequency, I would be fairly certain that we're talking about AC circuits here.

Think about it this way: we know that in a DC circuit, the more you charge a capacitor, the more it resists the flow of current (as time goes on, it will slow to a trickle as the capacitor becomes fully charged). Now switch to thinking about an AC circuit. The alternating current is continually charging the capacitor, then uncharging it, then charging it again...rinse and repeat for each cycle. But we said that the more charged up the capacitor is, the more it resists the flow of current. So (for some fixed maximum voltage) a higher frequency means that the capacitor will have less time to charge on each cycle, hence it will be less charged, and therefore it will do less to oppose the flow of current. Inversely, if the frequency is low, it will have a lot of time to build of charge, and therefore it will oppose the current greatly.

This is all summed up in the equation for capacitive reactance
X_c=\frac{1}{\omega C}
In the sense that a capacitor in an AC circuit will oppose the flow of current, it provides a resistance of sorts; this is called it's reactance. As you can see from the above equation, the higher the frequency, the lower the reactance...lower reactance means that (all other things equal) more current will flow.

That's awesome, thank you.