Ferranti effect is due to voltage drop across the line inductance, due to charging current, being in phase with the applied voltage at the sending end of the line.
Thus both capacitance and inductance are necessary to cause this phenomenon. The capacitance, and therefore, charging current is negligible in short lines but significant in medium and long transmission lines. On a 300 km line operating on 50 Hz supply, the receiving-end voltage on open-circuit is usually found to be roughly 5% higher than the sending-end voltage.
Phasor diagram given in Fig. 5.16 illustrates the Ferranti effect.
Taking receiving-end phase voltage as reference phasor we have,
ADVERTISEMENTS:
VR = VR (1+ j 0) represented by phasor OA
Charging current, IC = j ω C VR represented by phasor OD.
Voltage at sending end, VS = VR + resistive drop in line + reactive drop in line-
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= VR + IC R (represented by phasor AB) + j IC X (represented by phasor BC)
= VR + IC (R + j X) = VR + j ω C VR (R + j ω L)
or VS = VR – ω2 C L VR + j ω C R VR represented by phasor OC.
The overhead transmission lines having long length to bring the effect into prominence usually have a fairly large reactance compared with resistance, and by taking advantage of this fact a simple approximate formula for determining the magnitude of the effect can be derived.
ADVERTISEMENTS:
Neglecting resistive drop IC R we get,
Rise in voltage = OA – OC = Inductive drop in line (refer to Fig. 5.16).
If C0 and L0 are the capacitance and inductance of the transmission line per km length and l is the length of line in km, the capacitive reactance, XC = 1/ωlC0.
As the capacitance of the line is uniformly distributed over the entire length of the line, therefore, the average current flowing throughout the line,
The reactance of the transmission line, X = ω L0l
Rise in voltage = IC X = 1/2 VR C0 ω l × ω L0 l = 1/2 ω2l2 VR C0 L0 volts
The quantity 
is constant for all overhead lines and is equal to the velocity of propagation of electromagnetic waves. It is nearly equal to the speed of light i.e., 3 × 105 km/s. So,
From Eq. (5.32) it is obvious that the excess of voltage at the receiving end of an open- circuited line is proportional to the square of the length of the line. It is found that quite a small load eliminates this effect, and where there are transformers in the circuit at the receiving end, the magnetizing current acts in such a way.
ADVERTISEMENTS:
For long high voltage and extra high voltage lines shunt reactors are provided to absorb part of charging current or shunt capacitive VA of the line on no-load and light load, in order to prevent over-voltages on the line.
Example:
Determine the percentage rise in voltage at the receiving end of a transmission line of length 200 km operating at 50 Hz.
Solution:
