If the complex power is 7,500 VA with a power factor of 0.778 lagging, the reactive power (VAR) is most nearly:

To solve this question, let's first understand what complex power, power factor, and reactive power mean.

1. Complex power (S), as its name suggests, is represented by a complex number and therefore contains parts that are both real and imaginary. It can be expressed as: $S = P+jQ$
2. Reactive power (Q) is the imaginary part of the complex power as suggested by the above equation. This results when the current flow either leads or lags behind the AC voltage. It is given by the formula: $Q = S\sin ({\theta }_v-{\theta }_i)$
3. Power factor can be calculated with complex and real power according to the following equation, where P represents real power: $Power Factor = \raisebox{1ex}{$P$}\!\left/ \!\raisebox{-1ex}{$S$}\right. = \cos ({\theta }_v-{\theta }_i)$

We are given that the complex power is equal to 7,500 VA:

$S = 7,500$

We are also given that the power factor is equal to .778 lagging:

$$\begin{gathered} Power Factor = \cos ({\theta }_v-{\theta }_i) \\ Power Factor = .778 \end{gathered}$$

We can take the inverse cosine of this number in order to find the original angle that the current lags behind the voltage:

$$\begin{gathered} \cos ({\theta }_v-{\theta }_i) = .778 \\ {\cos }^{-1}\left(\cos \right({\theta }_v-{\theta }_i\left)\right) ={\cos }^{-1}(.778) \\ {\theta }_v-{\theta }_i = .679 radians \end{gathered}$$

Now, using our equation for the complex power, we can find our solution:

$$\begin{gathered} Q = S\sin ({\theta }_v-{\theta }_i) \\ Q = (7,500)\left(\sin \right(.679\left)\right) \\ Q = (7,500)(.628) \\ Q = 4710 \end{gathered}$$

Reactive power is measured in Volt-Amps-Reactive (VAR), so our final answer is 4710 VAR.