## A string of suspension insulator consist of 3-units, the capacitance between each link pins to earth is one-sixth of the self-capacitance of the unit. If the maximum peak voltage per unit is not to exceed 20kv. Find the greatest working voltage and efficiency of the spring.

V = V1 + V2 + V3; = V1 + V2 + 20

Applying kcl(Kirchhoff’s voltage law) to NODE A

i3 + i1 + ic1

WC2V2 = WC1V1 + WCV1

WMCV3 = WMCV2 + WC(V1 + V2)

MV2 = MV1 + V1

MV2 = 7V1

M = C1/C ; C1 Ã· 1/6C1 = 6

6V2 = 7V1

V2 = 7/6V1

V3 = 20kv

AT NODE B

WC3V3 = WC2V2 + WC(V1 + V2)

Since M = C3/C, therefore C3 = MC

**Therefore we’ve**,

WMCV3 = WMCV2 + WC(V1 + V2)

Divide through by WC

MV3 = MV2 + V1 + V2

Recall, from the solving = 6

6V3 = 6V2 + V1 + V2

6V3 = 7V2 + V1

Recall V2 = 7/6V1

**Therefore**,

6V3 = 7V2 + V1

= ; 6V3 = 7/6 Ã— 7V1 + V1

6V3 = 49/6V1 + V1

6V3 = 55/6 V1

Cross multiply

Cross multiply

V3 = 55/36 Ã— V1

V3 = 20kv

**Therefore**, 20kv = 55/36 Ã— V1

V1 = (36Ã—20kv) Ã· 55

V1 = 13kv

V2 = 7V1/6

7(13)/6 kv

= 15.2kv; approximately 15kv

## Maximum voltage of the spring

V = V1 + V2 + V3

V = 13 + 15 + 20

V = 48kv

V = 13 + 15 + 20

V = 48kv

## Maximum efficiency of the spring

Efficiency = V Ã· (nÃ—V3) Ã— 100

= (48kv) Ã· (3Ã—20kv) Ã— 100

48/60 Ã— 100 = 80%