# Why flow of real power depends on the phase difference of voltage magnitudes?

Let's say that we have a generator and a RLC circuit. Real and reactive power are tranferred at this system.

Transfer of real power depends on the phase difference between the voltage of the generator and the voltage of the load and transfer of reactive power depends on the difference between the magnitudes of these two voltages( maximum voltage of generator minus maximum voltage of load).

Why is this difference? I thought that more intuitive is the real power to depend on the difference of magnitudes.

• Too indefinite (= no reason to wait proper answers) without the problematic equations and the schematic where the same voltages are clearly identified. Fix the question! – user287001 Mar 12 '17 at 17:21
• I see that you edited the question. But I am still confused. If the generator is connected to a load, the generator output voltage and load input voltage are the same. They have same magnitude and same phase. I mean, if you draw a schematic, the generator output and load input are actually the same node on the schematic. So the question is still unclear. – mkeith Mar 12 '17 at 18:11
• @mkeith Please see the answer and the link below – veronika Mar 12 '17 at 18:12
• So you are talking about power delivered to a load when there is a transmission line in between source and load? That was not at all clear from the question. The question sounds like it is about power delivered to an RLC circuit directly connected to a voltage source. – mkeith Mar 12 '17 at 18:16
• No mention of Slideshow, transmission line, buses, phase angles or what is the load wrt RLC. – StainlessSteelRat Mar 13 '17 at 16:30

With apparent-, real- and reactive- powers defined as: $$S^2 =P^2 +Q^2.$$
The real power $P$ flow between two buses is obtained by: $$P = {V_1 \cdot V_2 \cdot \sin (\delta) \over X}$$ and the reactive power $$Q = {V_1 \cdot (V_1 - V_2) \cdot \cos (\delta) \over X}.$$