Question

a) Derive the expression for the current flowing in an ideal capacitor and its reactance when connected to an ac source of voltage V=V_osinωt. 

b) Draw its phasor diagram. 

c) If resistance is added in series to capacitor what changes will occur in the current flowing in the circuit and phase angle between voltage and current.

Answer 1

(a) We have: v = v_osinωt

Also, v = \(\frac{q}{c}\) ; q = charge on capacitor

v_osinωt = \(\frac{q}{c}\)

or, q = cv_osinωt

\(\therefore\) I = \(\frac{dq}{dt}\) = \(\frac{d}{dt}\)(cv_osinωt) = cv_osinωt.ω

\(\therefore\) I = \(\frac{v_o}{\frac{1}{\omega c}}\) sin(ωt + \(\frac{\pi}{2}\))

Max. current, I_o = \(\frac{v_o}{\frac{1}{\omega c}}\) x 1 when sin(ωt + \(\frac{\pi}{2}\)) = 1

∴ I = I_o sin(ωt + \(\frac{\pi}{2}\))

Comparing with ohm's law: I = \(\frac{V}{R}\) to equation I_o = \(\frac{v_o}{\frac{1}{\omega c}}\)

We have, capacitive reactance, x_c = \(\frac{1}{\omega c}\)

(b) Phasor Diagram

(c) 

Here, effective impedance, z = \(\sqrt{R^2+\mathrm{x}_c^2}\)

As z > R, current will be reduced

In Rc circuit, voltage lags behind the current by a phase angle ϕ, where

tanϕ = \(\frac{v_c}{v_R}\) = \(\frac{-I_o\mathrm{x}_c}{I_oR}\) = \(-\frac{\mathrm{x}_c}{R}\)

Negative sign is for lagging of alternating voltage behind alternating current.

Answer 2

(a) Derivation of instantaneous current i=i_osin (ωt +\(\frac{\pi}{2}\)) 

Reactance X_c=\(\frac{1}{\omega C}\) 

Phasor diagram showing v and i relation in pure C 

(b) Explanation that adding R it will behave RC series ac circuit Calcuation of current and phase angie