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Consider a three-phase, 50 Hz, 11 kV distribution system. Each of the conductors is suspended by an insulator string having two identical porcelain insulators. The self capacitance of the insulator is 5 times the shunt capacitance between the link and the ground, as shown in the figure given below. The volt age across the two insulators are
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- e1 = 3.74kV, e2 = 2.61 kV
- e1 = 3.46 kV, e2 = 2.89 kV
- e1 = 6.0 kV, e2 = 4.23 kV
- e1 = 5.5 kV, e2 = 5.5 kV
Correct Option: B
Using KCL at point ℗, we have
⇒ I = I1 + I2
⇒ V1 (5Cω) = V2 (Cω) + (5Cω)V2
⇒ 5V1 = 6V2 ...(A)
and V1 + V2 = 11 kV ...(B)
Equation (A) and Equation (B) gives,
V1 = 3.46 kV, V2 = 2.89 kV
