91Ó°ÊÓ

• Input voltage:\({V_p} = 12.0\;{\rm{kV}}\)
• Old and new Output voltages are: \({V_o} = 335\;{\rm{kV and }}{V_n} = 750\;{\rm{kV}}\)

For a given number of turns in the primary and secondary coil, the ratio of voltage in the primary coil to the voltage in the secondary coil remains constant and is equal to the ratio of the number of turns in the primary coil to the number of turns in the secondary coil.

If\({V_p}\),\({N_p}\),\({I_p}\)are the input potential difference, the number of turns and the input current respectively, in the primary circuit and\({V_s}\),\({N_s}\),\({I_s}\)are the output potential difference, the number of coils and the output current respectively, in the secondary circuit. The relation between these variables is given as-

\(\frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}} = \frac{{{I_p}}}{{{I_s}}} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (1)\)

a)

For old output voltage\({V_{so}} = 335\;{\rm{kV}}\),the number of turns in the old secondary coil\(\left( {{N_{so}}} \right)\)can be calculated from equation (1) as-

\(\begin{aligned} \frac{{{V_{so}}}}{{{V_p}}} &= \frac{{{N_{so}}}}{{{N_p}}}\\\frac{{{N_{so}}}}{{{N_p}}} &= \frac{{335\;{\rm{kV}}}}{{12\;{\rm{kV}}}} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (2)\end{aligned}\)

For new output voltage\({V_{sn}} = 750\;{\rm{kV}}\),the number of turns in the new secondary coil\(\left( {{N_{sn}}} \right)\)can be calculated from equation (1) as-

\(\begin{aligned} \frac{{{V_{sn}}}}{{{V_p}}} &= \frac{{{N_{sn}}}}{{{N_p}}}\\\frac{{{N_{sn}}}}{{{N_p}}} &= \frac{{750\;{\rm{kV}}}}{{12\;{\rm{kV}}}} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 3 \right)\end{aligned}\)

the ratio of the number of turns in the new secondary coil\(\left( {{N_{sn}}} \right)\)to the number of turns in the old secondary coil\(\left( {{N_{so}}} \right)\)can be calculated by dividing equation (3) by equation (2).

\(\begin{aligned} \frac{{\frac{{{N_{sn}}}}{{{N_p}}}}}{{\frac{{{N_{so}}}}{{{N_p}}}}} &= \frac{{\frac{{750\;{\rm{kV}}}}{{12\;{\rm{kV}}}}}}{{\frac{{335\;{\rm{kV}}}}{{12\;{\rm{kV}}}}}}\\\frac{{{N_{sn}}}}{{{N_{so}}}} &= \frac{{750\;{\rm{kV}}}}{{335\;{\rm{kV}}}}\\ &= 2.24\end{aligned}\)

Therefore, ratio of number of turns in the new secondary coil as compared to the old secondary coil is \(2.24\).

b)

For old output voltage\({V_{so}} = 335\;{\rm{kV}}\),the output current in the old secondary coil\(\left( {{I_{so}}} \right)\)can be calculated from equation (1) as-

\(\begin{aligned} \frac{{{V_{so}}}}{{{V_p}}} &= \frac{{{I_p}}}{{{I_{so}}}}\\\frac{{{I_p}}}{{{I_{so}}}} &= \frac{{335\;{\rm{kV}}}}{{12\;{\rm{kV}}}} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (4)\end{aligned}\)

For new output voltage\({V_{sn}} = 750\;{\rm{kV}}\),the output current in the new secondary coil\(\left( {{I_{sn}}} \right)\)can be calculated from equation (1) as-

\(\begin{aligned} \frac{{{V_{sn}}}}{{{V_p}}} &= \frac{{{I_p}}}{{{I_{sn}}}}\\\frac{{{I_p}}}{{{I_{sn}}}} &= \frac{{750\;{\rm{kV}}}}{{12\;{\rm{kV}}}} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (5)\end{aligned}\)

the ratio of the output current in the new secondary coil \(\left( {{I_{sn}}} \right)\) to the output current in the old secondary coil\(\left( {{I_{so}}} \right)\) can be calculated by dividing equation (4) by equation (5)\(\begin{aligned} \frac{{\frac{{{I_p}}}{{{I_{so}}}}}}{{\frac{{{I_p}}}{{{I_{sn}}}}}} &= \frac{{\frac{{335\;{\rm{kV}}}}{{12\;{\rm{kV}}}}}}{{\frac{{750\;{\rm{kV}}}}{{12\;{\rm{kV}}}}}}\\\frac{{{I_{sn}}}}{{{I_{so}}}} &= \frac{{335\;{\rm{kV}}}}{{750\;{\rm{kV}}}}\\ &= 0.45 \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 6 \right)\end{aligned}\)

Therefore, ratio of output current in the new secondary coil as compared to the old secondary coil is \(0.45\).

(c.) If the resistance of both the transmission lines, old and new, is same and is represented by\(R\), then the ratio of power loss in new line\(\left( {{P_{sn}}} \right)\)to the power loss in old line\(\left( {{P_{so}}} \right)\)is given as-

\(\begin{aligned} \frac{{{P_{sn}}}}{{{P_s}_o}} &= \frac{{{I_{sn}}^2R}}{{{I_{so}}^2R}}\\ &= \frac{{{I_{sn}}^2}}{{{I_{so}}^2}}\\ &= {\left( {\frac{{{I_{sn}}}}{{{I_{so}}}}} \right)^2}\end{aligned}\)

From equation (6), substituting the value of the ratio of currents, we get

\(\begin{aligned} \frac{{{P_{sn}}}}{{{P_s}_o}} &= {0.45^2}\\ &= 0.20\end{aligned}\)

Therefore, ratio of power loss in new line to the old line is\(0.20\).