Does the Impedance of a Capacitor Affect AC Circuits?

[Prashasti]

In a purely capacitive ac circuit, we get,
I_m = V_m*ω*C, ...(1)
Where, I_m = Amplitude of the current
V_m = Amplitude of the voltage

Now, what I think is,
We know that in a purely capacitive circuit, voltage lags behind current by a phase difference of ∏/2 rad. So, at any time 't',
I = I_m sin(ωt+∏/2)
V = V_m sinωt

Using Kirchhoff's Loop Rule,

V = V_msinωt = q/C

Where q = charge on the capacitor at time 't',

To find the current, I = dq/dt,
dq = Idt,

q = ∫Idt

q = ∫I_mcosωt dt
q = I_m∫cosωt dt
q = I_m*ω*sinωt

So, V_msinωt = I_m*ω*sinωt /C
V_m = I_m*ω/C

I_m = V_m*C/ω, which is apparently, not equal to equation (1).

Am I wrong in my approach?

 

[Simon Bridge]

Have you applied a result to a situation where it does not occur.
If there is no resistance in the circuit, then the voltage across the capacitor is going to be the same as the applied voltage - instantly. Draw the circuit diagram and see.

 

[maheshshenoy]

q = Im∫cosωt dt
q = Im*ω*sinωt

this is wrong..

q = (I_m/ω)*sinωt.. You are integrating. remember not differentiating.

V_m = I_m*ω/C

So this becomes

V_m = I_m*1/ωC

compare this to V = IR

so 1/ωC plays the role of impeding the AC.