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Line-to-pin capacitances for uniform string voltage

Short answer

For equal voltage distribution, line-to-pin capacitances from cross-arm are C1 = 0.25C, C2 = 0.667C, C3 = 1.5C, C4 = 4C, independent of self-capacitance.

A string of five suspension insulators is to be fitted with a grading ring. If the pin to earth capacitances are equal to C, find the values of line to pin capacitances that would give a uniform voltage distribution along the string.

A grading (guard) ring is a metal ring connected to the line conductor and placed around the lower units of the string. It adds a capacitance CkC_k from the line to each cap/pin junction. If these capacitances are chosen correctly, the charging current through the line-to-pin capacitance supplies exactly the current leaking to earth through the pin-to-earth capacitance, so all units carry the same current and share the voltage equally.

Arrangement

   Cross-arm (earth)
        |
     [Unit 1]   self cap  (same for all)
        +---- J1 --- C ---> earth,  C1 <--- line
     [Unit 2]
        +---- J2 --- C ---> earth,  C2 <--- line
     [Unit 3]
        +---- J3 --- C ---> earth,  C3 <--- line
     [Unit 4]
        +---- J4 --- C ---> earth,  C4 <--- line
     [Unit 5]
        |
   Line conductor (V) ---- grading ring

Junctions J1…J4J_1 \dots J_4 are numbered from the cross-arm. Let the total string voltage be VV.

Condition for uniform distribution

For equal voltage across each of the 5 units, each unit carries V/5V/5, so the voltage of junction JkJ_k above earth is

Vk=k V5,k=1,2,3,4V_k = \frac{k\,V}{5}, \qquad k = 1,2,3,4

The voltage between the line and JkJ_k is V−Vk=(5−k)V5V - V_k = \frac{(5-k)V}{5}.

Since all units have the same self-capacitance and the same voltage, the currents through the units are equal. Applying KCL at JkJ_k, the current that enters through the line-to-pin capacitance CkC_k must equal the current leaving through the pin-to-earth capacitance CC:

ωCk(V−Vk)=ωC VkCk=C VkV−Vk=C k5−k\begin{aligned} \omega C_k (V - V_k) &= \omega C\, V_k \\ C_k &= C\,\frac{V_k}{V - V_k} = C\,\frac{k}{5-k} \end{aligned}

Values

Junction (from cross-arm)kkCk=k5−kCC_k = \dfrac{k}{5-k}C
J1J_1 (between units 1 and 2)1C/4=0.25CC/4 = 0.25C
J2J_2 (between units 2 and 3)22C/3≈0.667C2C/3 \approx 0.667C
J3J_3 (between units 3 and 4)33C/2=1.5C3C/2 = 1.5C
J4J_4 (between units 4 and 5)44C4C

Answer: C1=0.25C, C2=0.667C, C3=1.5C, C4=4CC_1 = 0.25C,\ C_2 = 0.667C,\ C_3 = 1.5C,\ C_4 = 4C (numbered from the cross-arm end). The value does not depend on the unit's self-capacitance.

The line-to-pin capacitance must rise sharply towards the line end, which is why the ring is placed close to the units nearest the conductor. In practice exact values are hard to obtain, so the grading ring only improves the distribution and string efficiency, while also reducing corona at the line-end fittings.

✓ Answer checked by Claude Opus 5.5 (high effort)

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