# Purely Capacitive AC - voltage and current relationship

Studying electronics, capacitors in AC circuits currently. Can not understand two formula's.

I see it stated as fact: "In purely capacitive AC circuit Current leads Voltage by 90 degrees." I understand this relationship when viewing Waveform 90 degree phase difference Current leads Voltage.

I can't however find any example with circuit values to prove this is so.

I have two formula that prove this is so.

i = dq / dt

i = C * (dv / dt)

Can anybody show me how this works with actual circuit values? If I say (random example) AC 10V 200Hz with 0.0005F capacitor

Then i = dq / dt = ?? / ??

or

i = C * (dv / dt) = ? * ?? / ??

Example with number please for above circuit?

I am so confused I don't know if I just take "Current leads Voltage 90 degrees" at face value, but can it be proven with actual numbers? Maybe there is no example because there is no circuit without resistance or truly purely capacitive? Maybe it is just a theory not real world formula?

• Don't you know how to derive sin(x) or what? it's cos(x). U=Urms * 1.41 * sin(omega * t); omega=2pif; dv/dt =Urms*1.41*2pifcos(2pif*t). – Marko Buršič Apr 20 '16 at 19:37

$$I=\frac{U}{j\omega\;C}$$ $$|I|=\frac{10}{2\pi*200*0.0005}$$