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Voltage distribution of 3-unit string with guard ring

Short answer

Applying KCL with C1=0.15C and C2=0.05C gives voltage distribution: V1=0.302V, V2=0.3125V, V3=0.3854V. String efficiency = 1/(3×0.3854) ≈ 86.5%.

A 3 unit insulator string is fitted with guard ring. The capacitances of the link pins to metal work and guard ring can be assumed to be 15% and 5% of the capacitance of each unit. Determine the voltage distribution and string efficiency.

With a guard ring, each link pin has two shunt capacitances: one to the earthed metalwork (tower), which takes current away, and one to the guard ring (at line potential), which supplies current. Kirchhoff's current law at each pin gives the voltage distribution.

Data

  • 3 identical units, each of capacitance CC
  • Pin-to-metalwork capacitance C1=0.15CC_1 = 0.15C
  • Pin-to-guard-ring capacitance C2=0.05CC_2 = 0.05C
  • Units numbered 1 (top, at cross-arm) to 3 (bottom, at line), with voltages V1,V2,V3V_1, V_2, V_3; line voltage V=V1+V2+V3V = V_1 + V_2 + V_3
 cross-arm (earth)
    | unit 1  (V1)
    A ---0.15C--> tower ; <--0.05C-- ring
    | unit 2  (V2)
    B ---0.15C--> tower ; <--0.05C-- ring
    | unit 3  (V3)
 line (V) ======= guard ring (V)

KCL at pin A (potential V1V_1 above earth, V−V1V - V_1 below ring)

Current in from unit 2 + current from ring = current into unit 1 + current to tower:

CV2+0.05C(V−V1)=CV1+0.15CV1V2=1.2V1−0.05V(1)\begin{aligned} CV_2 + 0.05C(V - V_1) &= CV_1 + 0.15CV_1 \\ V_2 &= 1.2V_1 - 0.05V \qquad (1) \end{aligned}

KCL at pin B (potential V1+V2V_1 + V_2)

CV3+0.05C(V−V1−V2)=CV2+0.15C(V1+V2)V3=0.2V1+1.2V2−0.05V(2)\begin{aligned} CV_3 + 0.05C(V - V_1 - V_2) &= CV_2 + 0.15C(V_1 + V_2) \\ V_3 &= 0.2V_1 + 1.2V_2 - 0.05V \qquad (2) \end{aligned}

Solving

Substitute (1) into (2):

V3=0.2V1+1.2(1.2V1−0.05V)−0.05V=1.64V1−0.11V\begin{aligned} V_3 &= 0.2V_1 + 1.2(1.2V_1 - 0.05V) - 0.05V \\ &= 1.64V_1 - 0.11V \end{aligned}

Then V=V1+V2+V3V = V_1 + V_2 + V_3:

V=V1+(1.2V1−0.05V)+(1.64V1−0.11V)1.16V=3.84V1V1=0.3021V\begin{aligned} V &= V_1 + (1.2V_1 - 0.05V) + (1.64V_1 - 0.11V) \\ 1.16V &= 3.84V_1 \\ V_1 &= 0.3021V \end{aligned}

Hence

V2=1.2(0.3021V)−0.05V=0.3125VV3=1.64(0.3021V)−0.11V=0.3854V\begin{aligned} V_2 &= 1.2(0.3021V) - 0.05V = 0.3125V \\ V_3 &= 1.64(0.3021V) - 0.11V = 0.3854V \end{aligned}

Check: 0.3021+0.3125+0.3854=1.0000.3021 + 0.3125 + 0.3854 = 1.000.

Voltage distribution

UnitVoltage (fraction of VV)Percent
1 (top)0.3021 (= 29/96)30.2%
2 (middle)0.3125 (= 30/96)31.25%
3 (line end)0.3854 (= 37/96)38.5%

String efficiency

η=V3×V3=13×0.3854=96111=0.8649=86.5%\begin{aligned} \eta &= \frac{V}{3 \times V_3} = \frac{1}{3 \times 0.3854} \\ &= \frac{96}{111} = 0.8649 = 86.5\% \end{aligned}

For comparison, without the guard ring (C2=0C_2 = 0, m=0.15m = 0.15) the efficiency would be about 82.0%. The guard ring improves the voltage distribution.

Answer: V1=0.302VV_1 = 0.302V, V2=0.3125VV_2 = 0.3125V, V3=0.385VV_3 = 0.385V; string efficiency ≈ 86.5%.

✓ Answer checked by Claude Opus 5.5 (high effort)

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