5-disc string at 66 kV, ratios 5 and 6

With no guard ring, the voltage across the discs increases towards the line, because each link pin leaks charging current to the tower through its shunt capacitance $C_1 = mC$.

Data

Phase voltage:

$$V = \frac{66}{\sqrt 3} = 38.105\ \text{kV}$$

Number the discs 1 (tower end) to 5 (line end). At each junction, the current through the lower disc equals the current through the upper disc plus the current to earth:

$$V_{k+1} = V_k + m(V_1+\dots+V_k)$$

Case 1: $C = 5C_1$, so $m = 0.2$

$$\begin{aligned} V_2 &= 1.2V_1\\ V_3 &= V_2 + 0.2(2.2V_1) = 1.64V_1\\ V_4 &= V_3 + 0.2(3.84V_1) = 2.408V_1\\ V_5 &= V_4 + 0.2(6.248V_1) = 3.6576V_1 \end{aligned}$$

$V = 9.9056V_1$, so $V_1 = 38.105/9.9056 = 3.847$ kV.

$$\eta_1 = \frac{V}{5V_5} = \frac{38.105}{5\times 14.070} = 54.16\ \%$$

Case 2: $C = 6C_1$, so $m = 1/6$

$$\begin{aligned} V_2 &= 1.1667V_1,\quad V_3 = 1.5278V_1\\ V_4 &= 2.1435V_1,\quad V_5 = 3.1165V_1 \end{aligned}$$

$V = 8.9545V_1$, so $V_1 = 4.255$ kV and $V_5 = 13.262$ kV.

$$\eta_2 = \frac{38.105}{5\times 13.262} = 57.47\ \%$$

Answer: The string efficiency is 54.16 % when $C = 5C_1$. Moving the insulators further from the tower lowers the shunt capacitance so that $C = 6C_1$. The efficiency then rises to 57.47 %, and the line-end disc voltage falls from 14.07 kV to 13.26 kV. A lower ratio $m = C_1/C$ always gives a more uniform distribution.