# Complex Power

The V and I phasors in Figure A.18 are shown again in Figure A.19(b) in the form of a triangle. With the current as reference, the voltage V applied across the circuit is equal to the (a) Impedance triangle  Figure A.19 The complex power triangle phasorial addition of the voltage VR across the resistor (in phase with the current) and voltage VX across the reactor (in quadrature and leading the current). If the magnitude of each voltage phasor is divided by I a similar triangle, the impedance triangle, is obtained, shown in Figure A.19(a). If each voltage magnitude is multiplied by I, another similar triangle, the power triangle, is obtained, shown in Figure A.19(c). The hypotenuse of this triangle is of course the apparent power seen earlier. The apparent power S in VA can be visualized as a complex quantity with a real component P = VI cos 8 in watts and an imaginary component Q = VI sin 8 in VAR. The complex power S is therefore defined as

Note that the power factor angle 8 is the same angle in all three triangles.

Now the complex power is ready to be derived directly from the applied voltage and the resulting current in a circuit element. In the element of Figure A.20 the phasors of voltage and current can be expressed in general by the exponential form

Figure A.20 Reference directions for V, I,S,P and Q

where a and ¡3 are the angles by which VAB and IAB lead a reference phasor. Assume that a > ¡3. The angle by which VAB leads IAB (the power factor angle) is(a - p).

It would seem sensible that the multiplication of these two phasors gives the complex power. This task will now be carried out:

Vkbejalabejb = VabIabej(a+w = VabIab cos(a+P) + JVabIab sin(a+P)

Unfortunately, this product gives the wrong result! In fact, there is no good reason why an algebraic manipulation would invariably produce a result that complies with the adopted conventions. An alternative product will be tried using the conjugate of the current phasor I*B in which the sign of the angle is negated:

S = VabI*b = vabejalabe-Jb = VabiabeJ(a-P) = VabIab cos(a-P) + JVabIab sin(a-P) This gives the correct result; hence it can be confirmed that in general

The accepted reference directions of voltage, current, P and Q are shown in Figure A.20, which depicts any power network element. If the complex product of VAB and I*B results in positive P and Q (i.e. the active and reactive power flow into the element terminals), the element is a consumer of P and Q and operates in the first quadrant of Figure A.18. In Chapters 4, 5 and 6 it is shown that, depending on its nature, an element could operate in any of the four quadrants. 