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String efficiency: 4 units, 20% and 5%

Short answer

Solving junction equations with m=0.2 and g=0.05 gives disc voltages of 19.82%, 19.77%, 24.67%, 35.74% of V from tower to line end. String efficiency = V/(4*Vmax) = 69.96%.

Calculate the string efficiency for a 4-unit suspension insulator if the capacitance of each unit to earth and line be 20% and 5% of the self-capacitance of the unit respectively.

Each link pin has capacitance to earth mC=0.2CmC = 0.2C and to the line gC=0.05CgC = 0.05C, where CC is the self-capacitance of a unit. Number the discs 1 (tower end) to 4 (line end). Let VV be the phase voltage and Sk=V1+⋯+VkS_k = V_1+\dots+V_k.

 Tower─[C]─A─[C]─B─[C]─D─[C]─Line(V)
           |     |     |
     earth 0.2C each; line 0.05C each

Junction equation

Current from the line side = current to the tower side + current to earth − current received from the line:

Vk+1=Vk+0.2Sk−0.05(V−Sk)V_{k+1} = V_k + 0.2S_k - 0.05(V - S_k) V2=1.25V1−0.05VV3=V2+0.25(V1+V2)−0.05V=1.8125V1−0.1125VV4=V3+0.25(V1+V2+V3)−0.05V=2.8281V1−0.2031V\begin{aligned} V_2 &= 1.25V_1 - 0.05V\\ V_3 &= V_2 + 0.25(V_1+V_2) - 0.05V = 1.8125V_1 - 0.1125V\\ V_4 &= V_3 + 0.25(V_1+V_2+V_3) - 0.05V = 2.8281V_1 - 0.2031V \end{aligned}

Sum =V= V:

6.8906V1−0.3656V=V  ⇒  V1=0.1982V6.8906V_1 - 0.3656V = V \;\Rightarrow\; V_1 = 0.1982V
Disc% of VV
119.82 %
219.77 %
324.67 %
4 (line end)35.74 %

String efficiency

η=V4V4=14×0.35737=69.96 %\eta = \frac{V}{4V_4} = \frac{1}{4\times 0.35737} = 69.96\ \%

Answer: The string efficiency is 69.96 % (about 70 %).

✓ Answer checked by Claude Opus 5.5 (high effort)

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