Capacitive Circuit

For a purely capacitive circuit (such as that in Fig. 1), i leads v by $90^\circ$, as shown in Fig. 2.
Defining the power level for a purely
capacitive load.
Fig. 1: Defining the power level for a purely capacitive load.
The power curve for a purely capacitive load.
Fig. 2: The power curve for a purely capacitive load.
$$ \bbox[10px,border:1px solid grey]{p = VI(1 - \cos 2wt) \cos \theta + VI \sin \theta (\sin 2wt)} \tag{1}$$ Therefore, in Eq. (1), $\theta = -90^\circ$. Substituting $\theta = -90^\circ$ into Eq. (1), yields $$ \begin{split} p_C &= VI(1 - \cos 2wt) \cos -90^\circ + VI \sin -90^\circ (\sin 2wt)\\ &= 0 + -VI(\sin 2wt)\\ \end{split} $$ or $$\bbox[10px,border:1px solid grey]{p_C = -VI \sin 2wt} \tag{2} $$ where $-VI \sin 2wt$ is a sine wave with twice the frequency of either input quantity (v or i) and a peak value of $VI$. Again, note the absence of an average or constant term in the equation. Plotting the waveform for $p_C$ (Fig. 2), we obtain $$ T_1 = \text{period of either input quantity}$$ $$ T_2 = \text{period of pC curve}$$ Note that the same situation exists here for the $p_C$ curve as existed for the $p_L$ curve. The power delivered by the source to the capacitor is exactly equal to that returned to the source by the capacitor over one full cycle.
The net flow of power to the pure (ideal) capacitor is zero over a full cycle,
and no energy is lost in the transaction. The power absorbed or returned by the capacitor at any instant of time t1 can be found by substituting t1 into Eq. (2).
The reactive power associated with the capacitor is equal to the peak value of the $p_C$ curve, as follows: $$Q_C = VI \,\, \text{( VAR)}$$ or, since $V = IX_C$ or $I = V/X_C$, $$Q_C = I^2X_C \,\, \text{( VAR)}$$ or $$Q_L = V^2/X_C \,\, \text{( VAR)}$$ The apparent power associated with an inductor is $S = VI$, and the average power is $P = 0$, as noted in Fig. 2. The power factor is therefore $$ F_p = \cos \theta = {P \over S} = { 0 \over VI} = 0$$ The energy stored by the capacitor during the positive portion of the cycle (Fig. 2) is equal to that returned during the negative portion and can be determined using the equation $W = Pt$.
Proceeding in a manner similar to that used for the inductor, we can show that $$ W_C = ({2 VI \over \pi}) \times ({T_2 \over 2})$$ and $$ \bbox[10px,border:1px solid grey]{W_C = {VI T_2 \over \pi}} \tag{3}$$ or, since $T_2 = 1/f_2$, where $f_2$ is the frequency of the $p_C$ curve, we have $$ \bbox[10px,border:1px solid grey]{W_C = {VI \over \pi f_2}} \tag{4}$$ Since the frequency $f_2$ of the power curve is twice that of the input quantity, if we substitute the frequency f1 of the input voltage or current, Equation (4) becomes $$ W_C = {VI \over \pi 2f_1} = { VI \over w_1}$$ However, $I = V/X_C = V w_1 C$, so that $$ W_C = {(Vw_1L)V \over w_1}$$ $$ \bbox[10px,border:1px solid grey]{W_C = C V^2} \tag{5}$$ providing an equation for the energy stored or released by the capacitor in one half-cycle of the applied voltage in terms of the capacitance and rms value of the voltage squared.
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