Question regarding Ferranti effect

[jaus tail]

the phasor for ferranti is

Why is BC drawn facing left? I get it's voltage of capactive current across the line inductance. But why is it leading capacitive current by 90 degrees.

Also physically how does receiving end voltage increase without any drop of receiving current? Where does the power come from?

 

[Babadag]

In my opinion your sketch is taken from:

http://wangyj.ee.yuntech.edu.tw/PQLAB2_Web/elec_machine/Ferranti%20Effect.pdf

you said:

"But why is it leading capacitive current by 90 degrees."

See [for instance]:

http://www.colorado.edu/physics/phys1120/phys1120_fa09/LectureNotes/Voltage.pdf

you asked:

“Where does the power come from?”

As you see from above articles the voltage it is not "a power" but the work of the force exercised on an unit charge by the electric field on the distance between 2 points of a circuit.

The power is the voltage multiplied by current which is zero [in open circuit case].

 

[jaus tail]

I went through the second link but still didnt understand why the voltage drop across inductor leads capacitive current by 90 degrees. if anything the inductor must cause a lag in phase.

 

[Babadag]

The line current- if the end is open and no other load current will flow through- it is only the leakage capacitive

current [neglecting the current through air insulation].

If I [line current] is inductive or capacitive [it does not matter] I*jXL will be 90o leading the current any way.

But, since the current leads the voltage Vr you’ll need to supply less Vs in order to get the same Vr. That means

for the same Vs you'll get more Vr[see the sketch].

 

[jaus tail]

Thanks. That's helpful. in inductor current lags voltage by 90 degrees, so voltage leads the current across it by 90 degrees. gave me a smile at how basic the answer was:)