If in a circuit there is real power positive and reactive power negative, what does mean in terms of power flow? I undestand that the real power is produced in the generator/source and goes to the charge and the reactive goes backwards from the charge to the generator.
UPDATE
I have read both answers, but this makes me wonder the meaning of equal real and reactive power. What does it mean?
If in a circuit there is real power positive and reactive power negative, what does mean in terms of power flow?
If you have positive real power it means that some component is consuming it and converting it into work, be it mechanical work or heat or both.
If you have negative reactive power it means that some component is changing the phase of the voltage and current, in this case in a capacitive manner (current leads voltage). This is because the impedance of a capacitor is \$\frac{1}{sc} => \frac{1}{jwc} => -j\frac{1}{wc}\$, notice the -j factor.
Real power flow is controlled primarily by angular differences, flowing from leading angles toward lagging. Reactive power is controlled by voltage magnitudes, flowing from higher voltage toward lower. This assumes a system where all bus voltages are reasonably near 1 p.u. and relatively small angles. This is the basis of a decoupled load flow by the way.
Referring to the circuit above (with \$X=X_G+X_S\$) the power flow equations are,
$$P = \frac{EVsin\theta}{X}$$
$$Q = \frac{E^2-EVcos\theta}{X}$$
So, looking at \$P\$ we can see that in a system where the bus voltages are near 1 p.u. varying the angle \$\theta\$ will be the primary way to control real power flow.
Looking at \$Q\$ we can see that for normal angular differences (few degrees) between buses the reactive power flow is controlled by varying the bus voltages. e.g. \$cos(5°)= 0.9996\$ is not dominant when compared to \$E=1.1\$ and \$V=1.05\$. Ignoring \$cos(5°)\$ the numerator is \$0.055\$. Including \$cos(5°)\$ the numerator becomes \$0.059\$, only a 7% change.
So, for example, if a generator is not producing enough reactive to satisfy the \$I^2X\$ var drop in it's step-up transformer, then you will see reactive power coming into the HV side of the transformer from the system.
We add capacitors at some load buses to provide \$I^2X_C\$ reactive power at the load point. So, this reduces the reactive power coming from the source and reduces the real power loss on the line (which is proportional to \$I^2R_{Line}\$). You will see a corresponding rise on the load bus voltage when the caps are online.
UPDATE: To answer OP question - "What is the meaning of equal real and reactive power?"
We look at P,Q quantities on an x-y coordinate system where the x-axis is the real (P) axis and the y-axis is the imaginary (Q) axis.
So, with \$|P|=|Q|\$ and \$P\$ positive and \$Q\$ negative we have the plot below. Note that by definition \$P\$ and \$Q\$ are always in quadrature (separated by ±90°).
From this we can draw the power triangle and determine power factor, apparent power (\$S\$) etc. The power factor angle, \$\theta\$, is the angle between the real (\$P\$) axis and the hypotenuse (\$S\$). Power factor is the cosine of this angle \$\theta\$.
So, when \$P\$ and \$Q\$ have equal magnitudes with \$P\$ positive and \$Q\$ negative we find,
$$\theta=tan^{-1}\frac{P}{-Q}={-45°}$$
and
$$power factor = cos({\theta}) = \frac{1}{\sqrt{2}}=0.707$$
