In short transmission line, the effect of shunt capacitance is ignored for the sake of calculation and performance analysis. This is because, the length of short line is less and the voltage at which power is transmitted is also less. As the power is transmitted at relatively higher voltage and the length of line is quite large i.e. 50 to 150 km for Medium Transmission line, therefore the effect of shunt capacitance cannot be ignored to maintain accuracy in the calculation and performance analysis.

The shunt capacitance of line is uniformly distributed over the entire length. But if we assume the capacitance to be lumped at some particular point then calculation becomes quite easy at a cost of little error in calculation. Therefore, different line models are used for Medium Transmission line. The most commonly used models for solution and performance analysis of such lines are as follows:

- End Condenser Model.

- Nominal T Model.

- Nominal π Model.

The above models of line are based on the position of lumped shunt capacitance of line. In this post we will be discussing End Condenser Model.

### End Condenser Model

In End Condenser Model, shunt capacitance of line is lumped at the receiving end. Notice that, we are localizing the net shunt capacitance of line at the receiving end even though shunt capacitance is uniformly distributed. This means that, we are over estimating the effect of shunt capacitance in End Condenser Model.

Figure below shows End Condenser Model for single phase of three phase system as it is easier to work with single phase system.

Suppose,

I

_{R}= Load Current per phase
Vs = Sending end phase voltage

V

_{R}= Receiving end phase voltage
C = Capacitance of line per phase

X = Reactance per phase

R = Resistance per phase

Let us now draw the phasor diagram of the End Condenser Model. In the phasor, V

_{R}is taken as reference and load current I_{R}lagging by an angle Ø_{R}. Current I_{C}through shunt capacitance is leading by 90° with respect to V_{R}. Voltage drop in line resistance IsR is in phase with Is and voltage drop in line reactance IsX is leading Is by 90°.
As V

_{R}is considered reference, therefore V_{R}= V_{R}+ j0
Load Current I

_{R}= I_{R}∠-Ø_{R}(since I_{R}lagging)
= I

_{R}(CosØ_{R}- jSinØ_{R})
Shunt Current through Capacior I

_{C}= I_{C∠}90°
= I

_{C}(Cos90° + jSin90°)
= jI

_{C}_{ }

But Ic = V

_{R}/ (1/ωC) = ωCV_{R}_{ }

Ic = jωCV

_{R}= j2πfCV_{R}_{ }

Since sending end current Is is phasor sum of load current I

_{R}and I_{C}, therefore
Is = I

_{R}+ I_{C}_{ }

= I

_{R}(CosØ_{R}- jSinØ_{R}) + j2πfCV_{R}_{ }

= I

_{R}CosØ_{R}+ j(2πfCV_{R}- I_{R}SinØ_{R})
Again, sending end voltage is the phasor sum of receiving end voltage, drop in resistance and reactance, therefore

Vs = V

_{R}+ IsR + jIsX ………(please correlate with phasor for better understanding)
= V

_{R}+ Is[R + jX]
Thus we calculated the sending end voltage and sending end current. The sending end power factor can also be calculated. For calculating sending end power factor, just represent sending end voltage and current in angle form and then find the difference in angle. The cosine of the angle difference will be the sending end power factor. The voltage regulation of medium line by using end condenser method is

% Voltage Regulation = (Vs – V

_{R})x100 / V_{R}%
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