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_{m}sinωt = q/C Where q = charge on the capacitor at time 't', To find the current, I = dq/dt, dq = Idt, q = ∫Idt q = ∫I_{m}cosωt dt q = I_{m}∫cosωt dt q = I_{m}*ω*sinωt So, V_{m}sinωt = 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?
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.
this is wrong.. q = (I_{m}/ω)*sinωt.. You are integrating. remember not differentiating. So this becomes V_{m} = I_{m}*1/ωC compare this to V = IR so 1/ωC plays the role of impeding the AC.