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If current leads the voltage in a load, it means the load is mostly capacitive and reactive power is negative. What I don't understand is that reactive power depends on sin(phase of voltage-phase of current) so if current is leading it means I have smaller value for phase of current and sin has a positive value. So how come we alway talk about negative reactive power for current leading loads?

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3 Answers 3

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A load that has a lagging power factor is, by convention, said to be receiving reactive power from the source. A load that has a leading power factor is, by convention, said to be delivering reactive power to the source. In a vector representation of AC circuits, inductors are given the value +jX and capacitors are given the value -jX, so that leads to the the use of +jI for lagging reactive currents and -jI for leading reactive currents. Therefore, lagging reactive power is positive and leading reactive power is negative.

Sometimes a leading power factor is given a negative sign, but that is not a good practice, because power = V X I X pf and negative real power is power that is returned to the source by the load as a motor does during regenerative braking or a utility customer does by generating more power than they use with a renewable energy system.

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Good question.

Wikipedia's article on Power Factor states:

Lagging and Leading Power Factors:

In addition, there is also a difference between a lagging and leading power factor. A lagging power factor signifies that the load is inductive, as the load will “consume” reactive power, and therefore the reactive component Q is positive as reactive power travels through the circuit and is “consumed” by the inductive load. A leading power factor signifies that the load is capacitive, as the load “supplies” reactive power, and therefore the reactive component Q is negative as reactive power is being supplied to the circuit.

I can't quite make sense of the logic here. I suspect that it might be just convention in that most industrial load is inductive and that power-factor measurements were always 0 to 1. For the occasions when it goes capacitive the sign is changed.

I would be happy to be educated further on this.

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    \$\begingroup\$ The reactive component is neither consumed or supplied power. Instantaneously there will be power transfer. With a purely reactive load for half a cycle power will be transferred to the load and for half a cycle power will be recovered from the load back to the source. Reactive loads store power. This is true for capacitive and inductive loads. The only difference is which half cycle. The negative sign is purely a mathematical convention, it all works if you do the maths in the complex plain and use the conventional signs. \$\endgroup\$ Feb 10, 2016 at 17:00
  • \$\begingroup\$ What is really supplied is a complementary storage element. If the load is inductive, the supply must have a capacitive storage element that has an equivalent storage capacity. \$\endgroup\$
    – user80875
    Feb 10, 2016 at 17:54
  • \$\begingroup\$ I don't like that explanation from Wikipedia. I don't see a direct correlation between "A lagging power factor signifies that the load is inductive" and "as the load will 'consume' reactive power, and therefore the reactive component Q is positive [...]". Instead, a leading power factor is called like that because current leads voltage. A lagging power factor is called like that because current lags voltage. The convention is to compare current with voltage. \$\endgroup\$
    – alejnavab
    Nov 5, 2020 at 10:35
  • \$\begingroup\$ @AlejandroNava, I remember it from the acronym 'CIVIL': In a C I leads V which leads I in an L. \$\endgroup\$
    – Transistor
    Nov 5, 2020 at 11:39
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There're two ways to realize why reactive power is negative in a mostly capacitive load. The first one is by using the equation you used, \$Q = V I \sin{(\theta_v - \theta _i)}\$, the second one is by using \$Q = I^2 X\$. As usual, here, \$V\$ and \$I\$ stand for the RMS value of voltage and of current.

Using the equation \$Q = V I \sin{(\theta_v - \theta _i)}\$, the factors \$V\$ and \$I\$ are always non-negative (i.e. they're zero or positive) because they're the magnitude of complex numbers. Thus the sign of \$Q\$ depends only on the sign of \$\sin{(\theta_v - \theta _i)}\$; if the latter is positive, so is the former; if the latter is negative, so is the former. Recall that \$\sin{(\theta)}\$ is positive for \$0° < \theta < 180°\$, and that it's negative for \$-180° < \theta < 0°\$, where \$\theta = \theta_v - \theta_i\$. Also recall that in an inductive load, \$0° < \theta < 90°\$, and in a capacitive load, \$-90° < \theta < 0°\$. Therefore \$Q > 0\$ for a RL load, and \$Q < 0\$ for a RC load.

Using the equation \$Q = I^2 X\$, the factor \$I^2\$ is always non-negative. Thus the sign of \$Q\$ depends only on the sign of \$X\$, and in fact is equal to it (because a positive [\$I^2\$] times a positive [if \$X > 0\$] equals a positive, and, a positive times a negative [if \$X < 0\$] equals a negative). In the case of inductive loads, we know that \$X > 0\$ so \$Q > 0\$. In capacitive loads, \$X < 0\$ so \$Q < 0\$.

[...] so if current is leading it means I have smaller value for phase of current and sin has a positive value.

No. If current is leading voltage, it means it is somewhere between 0° and 90° ahead of the voltage phasor or the voltage waveform, but not that its phase angle is smaller than the voltage's.

Why \$-90° < \theta < 0°\$ in a capacitive load? A quick way to realize why, is to first remember that, in a passive load, the angle \$\theta = \theta_v - \theta_i\$ is also equal to the angle of the complex impedance. Then, also remember that the complex impedance of a capacitive load lays in the fourth quadrant of the complex plane, so its phase angle is between 270° and 360°, or equivalently between -90° and 0°, since its real part (resistance) is positive and its imaginary part (capacitive reactance) is negative.

And why is the reactance of a capacitive load negative? Well, remember that a derivative in the time domain is equivalent to multiplying by \$j \omega\$ in the phasor domain (we're assuming sinusoidal steady-state). Also remember that the \$i\$-\$v\$ relationship of an ideal capacitor is \$i(t) = C \, dv(t)/dt\$. So, in the phasor domain, we have that \$\tilde I = C (j \omega \tilde V) = (j \omega C) \tilde V\$. Since the complex impedance is defined as the ratio of voltage phasor to current phasor, the impedance of an ideal capacitor can be obtained from the previous equation as \$ \tilde V / \tilde I = Z = 1/(j \omega C) = -j/\omega C\$. Notice the imaginary part (reactance) is negative since \$\omega = 2 \pi f\$ and \$C\$ are always positive in phasor analysis.

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