4-unit string with guard ring (10% and 5%)

A guard ring is a metal ring at line potential placed around the bottom of the string. It gives each link pin a capacitance to the line that partly offsets the pin-to-tower (earth) capacitance. Here the pin-to-tower capacitance is $mC = 0.1C$ and the pin-to-guard-ring capacitance is $gC = 0.05C$.

 Tower ─[C]─ A ─[C]─ B ─[C]─ D ─[C]─ Line (V)
             |       |       |
  to tower  0.1C    0.1C    0.1C
  to ring   0.05C   0.05C   0.05C

Junction equation

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

$$\omega C V_{k+1} + \omega gC(V - S_k) = \omega C V_k + \omega mC S_k$$ $$\Rightarrow V_{k+1} = V_k + 0.1S_k - 0.05(V - S_k)$$

Solution

$$\begin{aligned} V_2 &= 1.15V_1 - 0.05V\\ V_3 &= V_2 + 0.15(V_1+V_2) - 0.05V = 1.4725V_1 - 0.1075V\\ V_4 &= V_3 + 0.15(V_1+V_2+V_3) - 0.05V = 2.0159V_1 - 0.1811V \end{aligned}$$

Adding: $5.6384V_1 - 0.3386V = V$, so $V_1 = 0.2374V$.

String efficiency

$$\eta = \frac{V}{4V_4} = \frac{1}{4\times 0.29747} = 84.04\ \%$$

Answer: The voltage distribution is 23.74 %, 22.30 %, 24.21 % and 29.75 % of the phase voltage, and the string efficiency is 84.04 %. (Without the guard ring, with $m = 0.1$, the line-end disc would take 32.6 % and the efficiency would be about 76.6 %.)