A pump station uses an induction motor that requires a complex power of:

$$S_1=20/_36.87^circ$$ kVA

and a synchronous motor that requires a complex power of:

$$S_2=10/_-53.13^circ$$ kVA

The total reactive power (kvar) required by the pump station is most nearly:

A.  4
B.  20
C.  22
D.  30

Let's first understand what complex power 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.

Therefore, to solve the problem, we will first find our total complex power. The imaginary portion of this number will represent our total reactive power.

To find the total complex power needed to power both the pump station as well as the synchronous motor, we must add together their individual power requirements:

$$\begin{gathered} S_T=S_1+S_2 \\ {S}_T=(20<36.87°)+(10<-53.13°) \end{gathered}$$

It is much easier to add together complex numbers when they are written in rectangular form,. To convert:

$$\begin{gathered} r<\theta =r\cos \left(\theta \right)+rj\sin \left(\theta \right) \\ 20<36.87° = \left(20\right)\left(\cos \right(36.87\left)\right)+\left(20\right)\left(j\right)\left(\sin \right(36.87\left)\right) \\ 20<36.87° = 16 + 12j \\ 10<-53.13° = \left(10\right)\left(\cos \right(-53.13)+(10\left)\right(j\left)\right(\sin (-53.13)) \\ 10<-53.13° = 6 -8j \end{gathered}$$

Now adding these together to find the total complex power:

$$\begin{gathered} S_T=(16+12j)+(6-8j) \\ {S}_T=22+4j \end{gathered}$$

Therefore, of our complex power, 22 is our real part and 4j is our imaginary part. 4 thus represents our total reactive power needed.

Reactive power is measured in Volt-Ampere-Reactive (VAR). Our answer is A, 4 kVAR.