# The Power Triangle   Whatsapp  The three quantities average power, apparent power, and reactive power can be related in the vector domain by $$\bbox[10px,border:1px solid grey]{S = P + Q } \tag{1}$$ with $$P = p \angle 0^\circ \,\, Q_L = Q_l \angle 90^\circ \,\,Q_C = Q_c \angle -90^\circ$$ For an inductive load, the phasor power S, as it is often called, is defined by $$S = P + j Q_L$$ as shown in Fig. 1. Fig. 1: Power diagram for inductive loads.
The $90^\circ$ shift in $Q_L$ from $P$ is the source of another term for reactive power: quadrature power.
For a capacitive load, the phasor power $S$ is defined by $$S = P - j Q_C$$ as shown in Fig. 2. Fig. 2: Power diagram for capacitive loads.
If a network has both capacitive and inductive elements, the reactive component of the power triangle will be determined by the difference between the reactive power delivered to each. If $Q_L > Q_C$, the resultant power triangle will be similar to Fig. 1. If $Q_C > Q_L$, the resultant power triangle will be similar to Fig. 2.
That the total reactive power is the difference between the reactive powers of the inductive and capacitive elements can be demonstrated by considering Equations of previous topics.
An additional verification can be derived by first considering the impedance diagram of a series R-L-C circuit (Fig. 3). Fig. 3: Impedance diagram for a series R-L-C circuit. Fig. 4: The result of multiplying each vector of Fig. 3 by $I^2$ for a series R-L-C circuit.
If we multiply each radius vector by the current squared ($I^2$), we obtain the results shown in Fig. 4, which is the power triangle for a predominantly inductive circuit.
Since the reactive power and average power are always angled $90^\circ$ to each other, the three powers are related by the Pythagorean theorem; that is, $$\bbox[10px,border:1px solid grey]{S^2 = P^2 + Q^2} \tag{1}$$ Therefore, the third power can always be found if the other two are known.
It is particularly interesting that the equation $$\bbox[10px,border:1px solid grey]{ S = VI^*} \tag{2}$$ will provide the vector form of the apparent power of a system. Here, V is the voltage across the system, and $I^*$ is the complex conjugate of the current. Fig. 5: Demonstrating the validity of Eq. (2).
Consider, for example, the simple R-L circuit of Fig. 5, where $$I = { V \over Z_T} = { 10 \angle 0^\circ \over 3Ω + j 4Ω}\\ = { 10 \angle 0^\circ \over 5 Ω \angle 53.13^\circ} = 2A \angle -53.13^\circ$$ The real power (the term real being derived from the positive real axis of the complex plane) is $$P = I^2 R = (2 A)^2 (3 Ω) = 12W$$ and the reactive power is $$Q_L = I^2 X_L = (2 A)^2 (4 Ω) = 16 (\text{VAR})$$ with $$S = P + j Q_L = 12 W + j 16 VAR (L) = 20 VA \angle 53.13^\circ$$ as shown in Fig. 6. Applying Eq. (1) yields $$S = V I^* = (10 V \angle 0^\circ)(2 A \angle 53.13^\circ)\\ = 20 VA \angle 53.13^\circ$$ as obtained above. Fig. 6: The power triangle for the circuit of Fig. 5.
The angle $\theta$ associated with S and appearing in Figs. 1, 2, and 6 is the power-factor angle of the network. Since $$P = VI \cos \theta$$ or $$P = S \cos \theta$$ then $$F_p = \cos \theta = { P \over S}$$

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