Inductive Circuit and Reactive Power

For a purely inductive circuit (such as that in Fig. 1), v leads i by $90^\circ$, as shown in Fig. 2.
Defining the power level for a purely inductive
Fig. 1: Defining the power level for a purely inductive load.
The power curve for a purely inductive load.
Fig. 2: The power curve for a purely inductive 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. (19.1) yields $$ \begin{split} p_L &= VI(1 - \cos 2wt) \cos 90^\circ + VI \sin 90^\circ (\sin 2wt)\\ &= 0 + VI(\sin 2wt)\\ \end{split} $$ or $$\bbox[10px,border:1px solid grey]{p_L = VI \sin 2wt} \tag{2} $$ where $VI \sin 2wt$ is a sine wave with twice the frequency of either input quantity (v or i) and a peak value of $VI$. Note the absence of an average or constant term in the equation. Plotting the waveform for $p_L$ (Fig. 2), we obtain $$ T_1 = \text{period of either input quantity}$$ $$ T_2 = \text{period of pL curve}$$ Note that over one full cycle of $p_L$ (T_2), the area above the horizontal axis in Fig. 2 is exactly equal to that below the axis. This indicates that over a full cycle of $p_L$, the power delivered by the source to the inductor is exactly equal to that returned to the source by the inductor.
The net flow of power to the pure (ideal) inductor is zero over a full cycle, and no energy is lost in the transaction.
The power absorbed or returned by the inductor at any instant of time t1 can be found simply by substituting t1 into Eq. (2). The peak value of the curve VI is defined as the reactive power associated with a pure inductor.
In general, the reactive power associated with any circuit is defined to be VI sin v, a factor appearing in the second term of Eq. (19.1). Note that it is the peak value of that term of the total power equation that produces no net transfer of energy.
The symbol for reactive power is Q, and its unit of measure is the volt-ampere reactive (VAR).
The Q is derived from the quadrature ($90^\circ$) relationship between the various powers, to be discussed in detail in a later section. Therefore, $$\bbox[10px,border:1px solid grey]{ Q = VI \sin \theta }\,\, \text{( VAR)} \tag{3}$$ where $\theta$ is the phase angle between V and I. For the inductor, $$Q_L = VI \,\, \text{( VAR)}$$ or, since $V = IX_L$ or $I = V/X_L$, $$Q_L = I^2X_L \,\, \text{( VAR)}$$ or $$Q_L = V^2/X_L \,\, \text{( VAR)}$$ 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 $$ F_p = \cos \theta = {P \over S} = { 0 \over VI} = 0$$ If the average power is zero, and the energy supplied is returned within one cycle, why is reactive power of any significance? The reason is not obvious but can be explained using the curve of Fig. 2. At every instant of time along the power curve that the curve is above the axis (positive), energy must be supplied to the inductor, even though it will be returned during the negative portion of the cycle. This power requirement during the positive portion of the cycle requires that the generating plant provide this energy during that interval. Therefore, the effect of reactive elements such as the inductor can be to raise the power requirement of the generating plant, even though the reactive power is not dissipated but simply "borrowed". The increased power demand during these intervals is a cost factor that must be passed on to the industrial consumer. In fact, most larger users of electrical energy pay for the apparent power demand rather than the watts dissipated since the volt-amperes used are sensitive to the reactive power requirement. In other words, the closer the power factor of an industrial outfit is to 1, the more efficient is the plant's operation since it is limiting its use of "borrowed" power.
The energy stored by the inductor during the positive portion of the cycle (Fig. 2) is equal to that returned during the negative portion and can be determined using the following equation: $$W = Pt$$ where P is the average value for the interval and t is the associated interval of time.
Recall from Chapter 13 that the average value of the positive portion of a sinusoid equals $2(peak value/\pi)$ and $t = T_2 /2$. Therefore $$ W_L = ({2 VI \over \pi}) \times ({T_2 \over 2})$$ and $$ \bbox[10px,border:1px solid grey]{W_L = {VI T_2 \over \pi}} \tag{3}$$ or, since $T_2 = 1/f_2$, where $f_2$ is the frequency of the $p_L$ curve, we have $$ \bbox[10px,border:1px solid grey]{W_L = {VI \over \pi f_2}} \tag{4}$$ 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 $$ W_L = {VI \over \pi 2f_1} = { VI \over w_1}$$ However, $V = IX_L = Iw_1L$, so that $$ W_L = {(Iw_1L)I \over w_1}$$ $$ \bbox[10px,border:1px solid grey]{W_L = L I^2} \tag{5}$$ providing an equation for the energy stored or released by the inductor in one half-cycle of the applied voltage in terms of the inductance and rms value of the current squared.