String efficiency: 3 units, 25% and 10%, 25 kV

Assumption: 25 kV is the voltage across the whole string (phase to earth). Each unit has self-capacitance $C$; each joint has capacitance $mC = 0.25C$ to earth (tower) and $nC = 0.1C$ to the line conductor.

   Tower (earth)
      |
    [C] V1
      +-- 0.25C to tower, 0.1C to line (joint A)
    [C] V2
      +-- 0.25C to tower, 0.1C to line (joint B)
    [C] V3
      |
   Line (25 kV)

Joint equations

Current flows from line towards the tower. At each joint, current entering (through lower disc + from line capacitance) = current leaving (through upper disc + to earth capacitance).

Joint A (potential $V_1$; line-to-joint voltage $V_2+V_3$):

$$V_2 + n(V_2+V_3) = V_1(1+m) \quad\Rightarrow\quad 1.1V_2 + 0.1V_3 = 1.25V_1$$

Joint B (potential $V_1+V_2$; line-to-joint voltage $V_3$):

$$V_3(1+n) = V_2 + m(V_1+V_2) \quad\Rightarrow\quad 1.1V_3 = 1.25V_2 + 0.25V_1$$

Solving (put $V_1 = 1$)

From joint B: $V_3 = (1.25V_2 + 0.25)/1.1$. Substituting in joint A:

$$\begin{aligned} 1.1V_2 + \frac{0.1(1.25V_2+0.25)}{1.1} &= 1.25\\ 1.21V_2 + 0.125V_2 + 0.025 &= 1.375\\ V_2 &= \frac{1.35}{1.335} = 1.0112,\qquad V_3 = \frac{1.25(1.0112)+0.25}{1.1} = 1.3764 \end{aligned}$$

Sum $= 1 + 1.0112 + 1.3764 = 3.3876$, so $V_1 = 25/3.3876 = 7.380$ kV.

String efficiency

$$\eta = \frac{V}{n\,V_3} = \frac{25}{3\times10.157}\times100 = 82.0\ \%$$

The line capacitance partly cancels the earth capacitance current, so the distribution is better than with earth capacitance alone (with the 0.25 earth capacitance alone, $\eta$ would be only 74.7 %). If 25 kV were the line voltage, every disc voltage would be divided by $\sqrt3$, but the efficiency stays 82.0 %.

Answer: $V_1 = 7.38$ kV, $V_2 = 7.46$ kV, $V_3 = 10.16$ kV; string efficiency ≈ 82.0 %.