A balanced 3-phase load is rated at 100 kVA and 0.65 pf lagging. A purely capacitive load is added in parallel with the inductive load to improve the power factor to 0.9 lagging. The capacitive load must supply a reactive power (kvar) that is most nearly:

A.  76
B.  65
C.  45
D.  31

In order to solve how much reactive power the capacitor supplies to the system, we want to calculate the total reactive power of the system before the additional load was added, the total reactive power after, and then take the difference.

We are given that the load initially is rated at 100kVA with a power factor of .65 lagging. This represents our complex, or apparent, power, represented as a sum of both real power and imaginary (reactive) powers:

$S = P+jQ$

To solve for reactive power:

$Q = S\sin \left(\theta \right)$

We must first calculate the angle between our voltage and current, $\theta$, which can be derived from our power factor as follows:

$$\begin{gathered} power factor = \cos \left(\theta \right) \\ .65 = \cos \left(\theta \right) \\ {\cos }^{-1}(.65)={\cos }^{-1}\left(\cos \right(\theta \left)\right) \\ 49.46° = \theta \end{gathered}$$

Plugging this angle back into our equation to solve the reactive power of the system prior to the capacitive load being added:

$$\begin{gathered} Q = S\sin \left(\theta \right) \\ Q = \left(100k\right)\left(\sin \right(49.46°\left)\right) \\ Q = \left(100k\right)(.75995) \\ Q = 75995VAR \end{gathered}$$

Therefore, our inductive load supplies 75995 VAR in reactive power. Since the power factor improves to .9 lagging with the addition of the capacitive load, we want to find the new angle and use it to find the reactive power of the updated system. 

$$\begin{gathered} power factor = \cos \left(\theta \right) \\ .9= \cos \left(\theta \right) \\ {\cos }^{-1}(.9)={\cos }^{-1}\left(\cos \right(\theta \left)\right) \\ 25.84° = \theta \end{gathered}$$

Since our apparent power, S, will change due to the higher power factor, we need to find its updated value. We know that the real power doesn't change because a capacitive load is purely imaginary. Therefore, we can calculate the real power of our system prior to the addition of the capacitive load, and use it to find the updated apparent power with our new power factor:

$$\begin{gathered} P = S\cos \left(\theta \right) \\ P = S·(power factor) \\ P = \left(100k\right)(.65) \\ P = 65k \end{gathered}$$

Now, using real power to solve for apparent power at new angle:

$$\begin{gathered} P = S\cos \left(\theta \right) \\ P = S·(power factor) \\ 65k = \left(S\right)(.9) \\ S = 72,222 \end{gathered}$$

Finally, we use our solved apparent power value and solved angle to find the reactive power of the new system:

$$\begin{gathered} Q = S\sin \left(\theta \right) \\ Q = (72.222k)\left(\sin \right(25.84°\left)\right) \\ Q = (72.222k)(.43586) \\ Q = 31479VAR \end{gathered}$$

Lastly, in order to find the power supplied by the capacitive load, we take the difference in reactive power between the new and old system:

$$\begin{gathered} ∆Q = 75995VAR - 31479VAR \\ ∆Q = 44516VAR \end{gathered}$$

Therefore, our capacitive load must be supplying approximately 45 kVAR, which corresponds to answer C.