Circuit Behavior - Problem Solving

To fully understand this topic, the reader must have an understanding in linear algebra, basic knowledge of power calculations and kirchoff's laws.

One of the most useful electrical problems, is being able to calculate your every day power usage. The following example will not only show you how power can be calculated, but also how well your system is working.

Practical Household problem

A household is running the following equipment:

A single phase pump for the swimming pool is also connected and the nameplate reads:

\(230 \text{ V}_\text{rated}, 15\text{ A} , 0.83 \text{ pf} \).

What is the expected power factor for the entire house?

The solution to this problem is to break up the powers into its active and reactive components by using the power triangle, you are left with the resulting equation.

• \(S = P + jQ\)

The power consumed by your household is the vector sum of all the real components and the vector sum of all the imaginary components.

The real components of this household are the heater, geyser, active components of the motor and the fluorescent globes

• \(P_{total}\) = \(\text{Heater} + \text{Geyser} + (\text{Fluorescent Globe}\times75 ) + (230 \times 15 \times 0.83)\)
• \(P_{total}\) = \(5013.5 W\)

So to find the reactive components of the systems, you need to use power factor equations.

• \(pf = \frac{P}{S}\)
• \(S = \frac{10}{0.72} \times 75\)
• \(S = 1041,67 VA\)
• \(Q = S\sin(pf)\)
• \(Q = 722,8896 VAr\)

The motor is also running and consists of both active and reactive components (active component included in active power calculation above)

• For the motor:
• \(pf = \arccos(0.83)\)
• \(pf = 33.9\)
• \(S = 230 \times 15\)
• \(S = 3450 VA\)
• \(Q = 3450\sin(33,90)\)
• \(Q = 1924,72 VAr\)
• \(Qt = 1924,72 + 722,8896\)
• \(Qt = 2647,145 VAr\)

By adding the two vector sums of the active and reactive power, leaves

• St = (5013,5 + j2647,145) VA

From this equation, you can find the power factor by two methods, one being

• \(\arctan(\theta) = \frac{2647,145}{5013,5}\)
• \(\theta = 27,834\)
• \(pf = \cos(27,834)\)
• \(pf = 0.8834\)

The other being

• \(pf = \frac{5013,5}{5669}\)
• \(pf = 0.8834\)

This is the unseen power triangle representing the entire system.

Power is most effectively used when the power factor is at one. The example shows that adding swimming pool pumps and other non-resistive elements into your home increases the reactive component of the load within your home. Power distribution becomes difficult for providers as reactive components increase at the customer's side and the client would find that the client is billed on both active power usage and the maximum demand (in VA) for each passing month. This forces larger industries to move towards power corrective measures, to decrease their overall reactive demand from its provider.