Question regarding Ferranti effect

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SUMMARY

The Ferranti effect describes the phenomenon where the receiving end voltage of a transmission line increases, particularly in open-circuit conditions, due to the line's capacitance. The discussion clarifies that the capacitive current leads the voltage by 90 degrees, which is counterintuitive since inductors typically cause a lag. The power in this scenario is derived from the voltage multiplied by the current, which remains zero in an open circuit, indicating that the increase in voltage does not require a corresponding increase in current. The references provided include detailed explanations of these concepts, emphasizing the relationship between voltage, current, and phase angles in electrical circuits.

PREREQUISITES
  • Understanding of the Ferranti effect in electrical engineering
  • Knowledge of phasor representation and phase relationships
  • Familiarity with inductive and capacitive reactance
  • Basic principles of AC circuit analysis
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  • Study the mathematical derivation of the Ferranti effect in transmission lines
  • Learn about phasor diagrams and their application in AC circuit analysis
  • Explore the concepts of inductive and capacitive reactance in detail
  • Investigate the implications of open-circuit conditions in power systems
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Electrical engineers, power system analysts, and students studying transmission line theory will benefit from this discussion, particularly those interested in the dynamics of voltage and current in AC circuits.

jaus tail
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the phasor for ferranti is

ferranti-effect-in-transmission-line1.gif

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?
 
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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].
 
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.
 
Icapacitive Xinductive.jpg

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].
 
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:)
 

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