4-unit string with guard ring (10% and 6%)

With a guard ring, each link pin has capacitance to the tower (earth), $mC$, and to the guard ring (which is at line potential), $gC$. Here $m = 0.1$ and $g = 0.06$.

Junction equation

Number the discs 1 (tower end) to 4 (line end). Let $V$ be the line-to-earth voltage and $S_k = V_1+\dots+V_k$. Current balance at the junction below disc $k$:

$$V_{k+1} = V_k + mS_k - g(V - S_k)$$

Solving in terms of $V_1$

$$\begin{aligned} V_2 &= V_1 + 0.1V_1 - 0.06(V - V_1) = 1.16V_1 - 0.06V\\ V_3 &= V_2 + 0.16(V_1+V_2) - 0.06V = 1.5056V_1 - 0.1296V\\ V_4 &= V_3 + 0.16(V_1+V_2+V_3) - 0.06V = 2.0921V_1 - 0.2199V \end{aligned}$$

Sum $= V$:

$$5.7577V_1 - 0.4095V = V \;\Rightarrow\; V_1 = 0.2448V$$

Voltage distribution

The guard ring makes the distribution much flatter than a plain string. Disc 2 has the smallest share, because the guard ring current is largest near the top.

String efficiency

$$\eta = \frac{V}{4V_4} = \frac{1}{4\times 0.29223} = 85.55\ \%$$

Answer: The discs carry 24.48 %, 22.40 %, 23.90 % and 29.22 % of the phase voltage (tower end to line end), and the string efficiency is 85.55 %.