## If you can screen capture the circuit (alt+printscreen, then paste into Word) I’d be happy to take a look at it, but wouldn’t be able to respond until Sunday morning. Others here might help you more quickly. I’ll be online today until about 9PM to check this thread but no access to email.

In general you shouldn’t have to understand electricity to understand simple logic circuits… or even complicated ones, really. That’s the beauty of them, in a way. Apart from excessive fan-out (too many components hanging off of a single output) there isn’t much “electricity-wise” to think about.

I think your questions have been answered well enough without trying to delve into AC theory too much. But I will anyway.

In general, however, capacitors stop DC from flowing and only pass AC. To AC they look almost like a resistor; to DC they look like an open circuit. The equivalent resistance, X, an AC wave sees at the capacitor is determined by its frequency, f, and the capacitance, C, in the formula:

X = 1/(C*2*pi*f)

(technically, then, as f-> 0, X -> infinity, which is why DC can see capacitors as open circuits; this is really not what happens, but good enough for government work)

When you have a parallel circuit it behaves as **Chronos** implied. If the resistance down each path is equal then they will receive the same current. If a single AC source is driving the circuit you describe, and if X is the same value as the resistance in the other leg, then each path here will also get the same amount of current.

For the formula to figure stuff out when the values aren’t equal, you must find what the equivalent resistence of a parallel circuit is. It will equal

1 / [(1/R[sub]1[/sub]) + (1/R[sub]2[/sub])]

If you have three paths then you can just keep adding terms in the denominator. If the frequency isn’t changing then you can substitute any “R[sub]n[/sub]” with “X” of a capacitor.

Was this too complicated? Any more questions?