# AC applied to a capacitor.

1. Aug 31, 2014

### Prashasti

In a purely capacitive ac circuit, we get,
Im = Vm*ω*C, ....(1)
Where, Im = Amplitude of the current
Vm = 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 = Im sin(ωt+∏/2)
V = Vm sinωt

Using Kirchhoff's Loop Rule,

V = Vmsinωt = q/C

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

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

q = ∫Idt

q = ∫Imcosωt dt
q = Im∫cosωt dt
q = Im*ω*sinωt

So, Vmsinωt = Im*ω*sinωt /C
Vm = Im*ω/C

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

Am I wrong in my approach?

2. Aug 31, 2014

### 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.

3. Aug 31, 2014

### maheshshenoy

this is wrong..

q = (Im/ω)*sinωt.. You are integrating. remember not differentiating.
So this becomes

Vm = Im*1/ωC

compare this to V = IR

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