Voltage distribution on 4-disc string at 33 kV

No capacitance ratio is given. Assume the usual value: the shunt (pin-to-tower) capacitance $C_1$ is 10 % of the self-capacitance $C$ of a disc, so $m = C_1/C = 0.1$.

Data

• Phase voltage across the string: $V = 33/\sqrt3 = 19.053$ kV
• Discs numbered 1 (tower end) to 4 (line end)

 Tower─[C]─A─[C]─B─[C]─D─[C]─Line
           |     |     |
          mC    mC    mC  (to tower)

Junction equations

Current through the lower disc = current through the upper disc + current to earth:

$$\begin{aligned} V_2 &= V_1(1+m) = 1.1V_1\\ V_3 &= V_2 + m(V_1+V_2) = 1.31V_1\\ V_4 &= V_3 + m(V_1+V_2+V_3) = 1.651V_1 \end{aligned}$$ $$V = (1+1.1+1.31+1.651)V_1 = 5.061V_1 \;\Rightarrow\; V_1 = \frac{19.053}{5.061} = 3.765\ \text{kV}$$

Voltage distribution

String efficiency:

$$\eta = \frac{V}{4V_4} = \frac{19.053}{4\times6.215} = 76.64\ \%$$

Answer (with $m = 0.1$): The disc voltages are 3.77, 4.14, 4.93 and 6.22 kV (tower to line end), and the string efficiency is 76.64 %. The disc nearest the line is the most stressed. For any other $m$, use the same relations $V_2 = (1+m)V_1$, $V_3 = (1+3m+m^2)V_1$, $V_4 = (1+6m+5m^2+m^3)V_1$.