Mumbai University > FE > Sem 1 > Basic Electrical and Electronics Engineering

Marks: 4 M

Year: Dec 2013

Circuit diagram:

Current:

The alternating current “i” is given by,

i $=C \dfrac{dv}{dt} \\ =C \dfrac{d}{dt}(V_m \sin⁡ωt) \\ =ωCV_m \cos⁡ωt \\ =ωCV_m \sin⁡(ωt+90^0) \\ =I_m \sin⁡(ωt+90^0 )………….(I_m= ωCV_m )$

Here the current leads voltage by 90°

Waveforms:

Phasor diagram:

$Z= \dfrac VI= \dfrac{V_m}{I_m} = \dfrac{V_m}{ωCV_m}= \dfrac{1}{ωC} \\ \text{ωC is capacitive reactance.} \\ \text{The capacitor acts as open circuit for a DC supply,} f = 0, X_C= ∞$

*Power: *

Average power for one cycle = 0

$\text{Instantaneous power}P$ $= V_i= V_m \sin⁡ωt I_m \sin⁡(ωt+90°) \\ = V_m I_m \sin⁡ωt \cos⁡ωt \\ \dfrac{V_m I_m}{2} \sin⁡2ωt$