1. In a string of three insulator units, the capacitance of each unit is C, from each conductor to
       ground is C/3, a...
And                 v3 = 40.196% of V




With Grading Ring




The first equation does not change, i.e. 20v1 = 18v2 + 3v3...
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Grading The Insulator Problem

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Doubts regarding Grading problem

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Grading The Insulator Problem

  1. 1. 1. In a string of three insulator units, the capacitance of each unit is C, from each conductor to ground is C/3, and from each connector to the line conductor is C/5. Calculate the voltage across each unit as a % of the total voltage. To what value the capacitance between the connector of the bottom unit and the line has to be increased by a guard ring to make the voltage across it equal to that across the next higher unit? [31.5%, 28.3%, 40.196%; 0.6833C] Solution: Apply KCL at node A I1+Ia = i2 + Ix V1(jwC) + v1(jwC/3) = v2(jwC) + (v2+v3)(jwC/5) 20v1 = 18v2 + 3v3 ………(1) Apply KCL at node B i2 + Ib = I3 + Iy 20v2 +5v1 = 18v3 ………(2) Eliminating v3 from the above 2 equations V2 = 115/128 v1 And , then v3 = 1.276 v1 Hence, v1 + v2 + v3 = V or v1 = 1/3.174 V = 31.5% of V Similarly, v2 = 28.3% of V
  2. 2. And v3 = 40.196% of V With Grading Ring The first equation does not change, i.e. 20v1 = 18v2 + 3v3 But it is given that v2 = v3 Therefore v2 = (20/21)v1 Applying KCL at node B, I2 + Ib = i3 + Iy V2(jwC) + (v1+ v2)(jwC/3) = v3(jwC) + v3(jwCy) But v2 = v3 1/3(v1 + v2)C = (v2)Cy Cy = 0.6833C

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