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Guard ring capacitances for 5-disc string

Short answer

Require equal current through each disc. By KCL, pin-to-guard-ring capacitance at pin n must be C_n = (n/(5−n)) C1, yielding C_x1=0.25C1, C_x2=0.667C1, C_x3=1.5C1, C_x4=4C1.

An overhead line uses 5 nos. of insulator discs. The voltage along the different units of string is equalized by a suitable guard ring. Determine the theoretical value of the capacitance between the pin of different units and the guard ring as the function of capacitance between the pin and earth.

A guard (grading) ring is a metal ring connected to the line conductor and surrounding the lower units of the string. It adds capacitance CxC_x between each link pin and the line. This capacitance supplies, at every pin, the charging current that the pin-to-earth capacitance C1C_1 takes away. Then the same current flows through every disc, and the voltage is equal across all units.

Assumptions

  • 5 identical discs, each of self capacitance CC.
  • Pin-to-earth (tower) capacitance at each link pin = C1C_1.
  • Pin-to-guard-ring capacitance at the nn-th pin = CnC_n (unknown).
  • Pins are numbered n=1n = 1 to 4 from the cross-arm; the line is at voltage 5V5V, so each disc has voltage VV.
  cross-arm (earth)
     |  disc 1
     +- pin 1 at V    --C1--> earth   <--Cx1-- ring
     |  disc 2
     +- pin 2 at 2V   --C1--> earth   <--Cx2-- ring
     |  disc 3
     +- pin 3 at 3V   --C1--> earth   <--Cx3-- ring
     |  disc 4
     +- pin 4 at 4V   --C1--> earth   <--Cx4-- ring
     |  disc 5
  line (5V) ===== guard ring (at 5V)

Condition for equal voltage

Since all discs have the same voltage VV and capacitance CC, the current through every disc is the same (ωCV\omega C V). So at each pin, the current leaving to earth must be exactly supplied by the current coming from the guard ring.

At pin nn (voltage above earth =nV= nV, voltage below line =(5−n)V= (5 - n)V):

ωCn(5−n)V=ωC1(nV)\omega C_n (5 - n)V = \omega C_1 (nV) Cn=n5−n C1\boxed{C_n = \frac{n}{5 - n}\,C_1}

In general, for a string of NN units: Cn=nN−nC1C_n = \dfrac{n}{N - n}C_1.

Values for 5 discs

Cx1=14C1=0.25 C1Cx2=23C1=0.667 C1Cx3=32C1=1.5 C1Cx4=41C1=4 C1\begin{aligned} C_{x1} &= \frac{1}{4}C_1 = 0.25\,C_1 \\ C_{x2} &= \frac{2}{3}C_1 = 0.667\,C_1 \\ C_{x3} &= \frac{3}{2}C_1 = 1.5\,C_1 \\ C_{x4} &= \frac{4}{1}C_1 = 4\,C_1 \end{aligned}
Link pin (from cross-arm)Voltage to earthCxC_x to guard ring
1VV0.25 C10.25\,C_1
22V2V0.667 C10.667\,C_1
33V3V1.5 C11.5\,C_1
44V4V4 C14\,C_1

Remarks

  • The required capacitance increases sharply towards the line end. This is why the guard ring is placed around the lower discs, close to the line.
  • With exact values the string efficiency becomes 100%. In practice exact values cannot be obtained, so the voltage is only approximately equalised.

✓ Answer checked by Claude Opus 5.5 (high effort)

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