**Ferranti Effect**

**What is a short, medium, and long transmission line?**

**Why Does Ferranti Effect occur?**

**The formula of the Ferranti Effect**

_{R }= Receiving end voltage = OA

V_{S }=
Sending end voltage = AC

I_{C }=
Charging current = OD

I_{c}X
= Reactive drop = CB

I_{c}R
= Resistive drop = AB

V_{R} is a receiving end voltage and V_{R }is
taken as a reference phaser and represented by OA.

V_{R }= V_{R}(1+j0)

Charging
current is given by

I_{C}
= jωCV_{R}

Where C is the line capacitance.

The sending
end voltage V_{S} is the phaser sum of receiving end voltage V_{R},
resistive drop I_{c}R, and reactive drop I_{c}X.

V_{S}
= V_{R} + resistive drop + reactive drop

= V_{R}
+ I_{c}R + jI_{c}X

= V_{R
} + I_{C}(R + jX)

Put the
value of I_{C} = jωCV_{R} and X = jωL where L is the inductance
of the transmission line in the above equations

= V_{R} + jωCV_{R}(R + jωL)

= V_{R} + jωCV_{R}R - ω^{2}CLV_{R}

In a long
transmission line, reactance is very high compared to the resistance of the
line. So resistive drop I_{C}R can be neglected.

Consider C_{0}
and L_{0} are the capacitance and inductance of the transmission line
and L is the length of the line

Now Capacitive
reactance of the transmission line X_{C} is given by

X_{C} = 1 / ωlC_{0}

In a long
transmission line, the capacitance of the line is uniformly distributed throughout
the line. The charging current flowing through the line is given by the

I_{C} = 1 V_{R } / 2X_{C}

Now put the
value of the X_{C} in above equations

I_{C} = V_{R}C_{0}ωl
/ 2

The Inductive
reactance of the transmission line is given by

X = ωL_{0}l

The rise in
voltage is given by

= I_{C}X

Put the
value of I_{c} and X in the above equation

= V_{R}C_{0}ωl / 2 X ωL_{0}l
= ω^{2}l^{2}V_{R}C_{0}L_{0} / 2 Volts

Where 1 / √L_{0}C_{0}
is the velocity of propagation of light and it is constant for all transmission
lines and its value is given by 3 X 10^{5 }km/s.

Now we put the
above value in the rising voltage equation

Rising
Voltage = ω^{2}l^{2}V_{R} / 2 X (3 X 10^{5})^{2}

= ω^{2}l^{2}V_{R} / 2 X 9 X 10^{10 }

= ω^{2}l^{2}V_{R}
X 10^{-10} / 18 Volts

The above equation
is for the rising voltage which is occurred at receiving end voltage due to the
Ferranti effect.

**Important
Points:**

From the above
equation, we can conclude that the rise in voltage at the receiving end of the
line is proportional to the square of the length of the transmission line which
means if the length of the transmission line increased than the voltage at the receiving
end increases.

The Ferranti effect is more occurs in short line cables due to very high capacitance.

**If the length
of the transmission line is 300km and the frequency of the supply voltage is 50
Hz then the percentage rise in voltage is calculated by **

Rise in
voltage = ω^{2}l^{2}V_{R} X 10^{-10} / 18

Percentage rise
in Voltage = ω^{2}l^{2}V_{R} X 10^{-10} X 100 /
18 X V_{R}

= ω^{2}l^{2} X 10^{-8} / 18

= (2πf)^{2} (300)^{2} 10^{-8} / 18 = (2π X 50)^{2}
X (300)^{2} X 10^{-8} / 18

= 4.93 % is equal to 5 %.