5-disc string at 66 kV: nearest and farthest disc

The ratio of shunt to self-capacitance is not given. Assume the usual value $m = C_1/C = 0.1$ (shunt capacitance 10 % of disc capacitance).

Data

• Phase voltage: $V = 66/\sqrt3 = 38.105$ kV
• Discs numbered 1 (farthest from the conductor, at the tower) to 5 (nearest the conductor)

Relations

$$V_{k+1} = V_k + m(V_1 + \dots + V_k)$$ $$\begin{aligned} V_2 &= 1.1V_1\\ V_3 &= 1.31V_1\\ V_4 &= 1.651V_1\\ V_5 &= 1.651V_1 + 0.1(5.061V_1) = 2.1571V_1 \end{aligned}$$ $$V = 7.2181V_1 \;\Rightarrow\; V_1 = \frac{38.105}{7.2181} = 5.279\ \text{kV}$$

String efficiency:

$$\eta = \frac{38.105}{5\times11.388} = 66.92\ \%$$

Answer (with $m = 0.1$): The disc nearest the conductor carries 11.39 kV (29.9 % of $V$), and the farthest disc carries 5.28 kV (13.9 % of $V$). The nearest disc is stressed more than twice as much as the farthest one.