# The Total P, Q and S   Whatsapp  The total number of watts, volt-amperes reactive, and volt-amperes, and the power factor of any system can be found using the following procedure:
• Find the real power and reactive power for each branch of the circuit.
• The total real power of the system (PT) is then the sum of the average power delivered to each branch.
• The total reactive power (QT) is the difference between the reactive power of the inductive loads and that of the capacitive loads.
• The total apparent power is $S_T = \sqrt{P^2 + Q^2_T}$.
• The total power factor is $P_T/S_T$.
There are two important points in the above tabulation. First, the total apparent power must be determined from the total average and reactive powers and cannot be determined from the apparent powers of each branch. Second, and more important, it is not necessary to consider the series-parallel arrangement of branches. In other words, the total real, reactive, or apparent power is independent of whether the loads are in series, parallel, or series-parallel. The following examples will demonstrate the relative ease with which all of the quantities of interest can be found.
Example 1: Find the total number of watts, volt-amperes reactive, and volt-amperes, and the power factor Fp of the network in Fig. 1. Draw the power triangle and find the current in phasor form. Fig. 1: Example 1.
Solution: Construct a table such as shown in Table 1.
\begin{align*} Load & W & VAR & VA \\ 1 & 100 & 0 & 100 \\ 2 & 200 & 700(L) & \sqrt{(200)^2 + (700)^2} = 728\\ 3 & 300 & 1500(L) & \sqrt{(300)^2 + (1500)^2} = 1529\\ & P_T = 600 & Q_T = 800(C) & S_T = \sqrt{(600)^2 + (800)^2} = 1000\\ \end{align*}
(Note that $S_T$ = sum of each branch:
$1000 \neq 100 + 728 + 1529.71$)
Thus $$F_p = { P_T \over S_T} = {600 W \over 1000VA} = 0.6 \text{leading (C)}$$ Fig. 2: Power triangle for Example 1.
The power triangle is shown in Fig. 2. Since $S_T = VI = 1000 VA$, $I = 1000 VA/100 V = 10 A$; and since $\theta$ of $\cos \ theta = F_p$ is the angle between the input voltage and current: $$I = 10 A \angle +53.13^\circ$$ The plus sign is associated with the phase angle since the circuit is predominantly capacitive.

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