Determine the maximum voltage that the string of the suspension insulators in Fig. E.8.1 can withstand if the maximum voltage per unit is 17.5 kV.

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Let the voltages across the various units be [latex]E_{1}, E_{2}[/latex] and [latex]E_{3}[/latex] as shown such that [latex]E=E_{1}+E_{2}+E_{3}[/latex], where E is the desired withstand voltage of the string. Applying Kirchhoff's current law at A,

[latex]\begin{aligned}i_{2}=i_{1}+i_{a} &=E_{1} \omega C+E_{1} \omega \frac{C}{8}=E_{1} \omega C\left(1+\frac{1}{8} ight)=\frac{9}{8} E_{1} \omega C \&=E_{2} \omega C\ ext {or} \quad E_{2} &=\frac{9}{8} E_{1}\end{aligned}[/latex]

Similarly at B,

[latex]i_{3}=i_{2}+i_{b}[/latex]

[latex]\begin{aligned}E_{3} \omega C &=E_{2} \omega C=\frac{E_{1} \omega C}{8}+\frac{E_{2} \omega C}{8} \&=E_{2} \omega C\left[1+\frac{1}{8} ight]+\frac{E_{1} \omega C}{8} \&=\frac{9}{8} imes \frac{9}{8} \omega C E_{1}+\frac{E_{1} \omega C}{8} \&=\left(\frac{81}{64}+\frac{1}{8} ight) \omega C E_{1}=\frac{89}{64} \omega C E_{1} \E_{3} &=\frac{89}{64} E_{1}\end{aligned}[/latex]

It can be seen that the voltage across the line unit i.e., unit near the power conductor is maximum.

[latex] herefore \quad E_{3}=\frac{89}{64} E_{1}=17.5 \mathrm{kV}[/latex]

or [latex]E_{1}=17.5 imes \frac{64}{89}=12.58 \mathrm{kV}[/latex]

[latex]E_{2}=\frac{9}{8} E_{1}=\frac{9}{8} imes 12.58=14.15 \mathrm{kV}[/latex]

[latex] herefore \quad E=E_{1}+E_{2}+\mathrm{E}_{3}=12.58+14.15+17.5=44.23 \mathrm{kV} .[/latex]

Question 8.1: Determine the maximum voltage that the string of the suspens...

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