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UPPCL JE Previous Paper 8 (Held On: 27 November 2019 Shift 2 )

Option 4 : divided into two halves

CT 1: Network Theory 1

11667

10 Questions
10 Marks
10 Mins

__Nominal - π representation of the medium transmission line:__

In this representation, the lumped series impedance is placed in the middle while the **shunt admittance is divided into two equal parts and placed at two ends**. The nominal - π representation is shown in the below figure,

Let us define three currents I1, I2, and I3 as indicated in the above figure. Applying KCL at nodes M and N we get,

IS = I1 + I2 = I1 + I3 + IR

\( = \frac{Y}{2}{V_S} + \frac{Y}{2}{V_R} + {I_R}\) ------ (1)

VS = Z I2 + VR

\( = Z\left( {{V_R}\frac{Y}{2} + {I_R}} \right) + {V_R}\)

\( = \left( {\frac{{YZ}}{2} + 1} \right){V_R} + Z{I_R}\) ------ (2)

Substitute (2) in (1) we get,

\({{\rm{I}}_{\rm{S}}} = Y\left( {\frac{{YZ}}{4} + 1} \right){V_R} + \left( {\frac{{YZ}}{2} + 1} \right){{\rm{I}}_{\rm{R}}}\) ------- (3)

Therefore from (2) and (3) the ABCD parameters of nominal - π network are

\(A = D = \left( {\frac{{YZ}}{2} + 1} \right)\)

B = Z ohm

\(C = Y\left( {\frac{{YZ}}{4} + 1} \right)\) mho

Where

Z is the series impedance of the line, Y is shunt admittance of line

__Important Points__

__Nominal - T representation of medium transmission line:__

In this representation, the **shunt admittance is placed in the middle** and the series impedance is divided into two equal parts and these parts are placed on either side of the shunt admittance. The nominal - T representation is shown in the below figure,

Let us denote the midpoint voltage as VM, then the application of KCL at the midpoint results in,

\(\frac{{{V_S} - {V_M}}}{{Z/2}} = Y{V_M} + \frac{{{V_M} - {V_R}}}{{Z/2}}\)

\({V_M} = \frac{2}{{YZ + 4}}\left( {{V_S} + {V_R}} \right)\) ------- (1)

\({I_R} = \frac{{{V_M} - {V_R}}}{{Z/2}}\) ----- (2)

Substituting the value of VM from (1) into (2), and rearranging the equating we get,

\({V_S} = \left( {\frac{{YZ}}{2} + 1} \right){V_R} + Z\left( {\frac{{YZ}}{4} + 1} \right){I_R}\) --------- (3)

\({I_S} = Y{V_M} + {I_R}\) ------ (4)

Substitute the value of VM from (1) into (4), we get

\({I_S} = Y{V_R} + \left( {\frac{{YZ}}{2} + 1} \right){I_R}\) ----- (5)

Therefore from (4) and (5) the ABCD parameters of nominal - T network are,

\(A = D = \left( {\frac{{YZ}}{2} + 1} \right)\)

\(B = Z\left( {\frac{{YZ}}{4} + 1} \right)\) ohm

C = Y mho

Where

Z is the series impedance of the line, Y is shunt admittance of line