Correct Answer - Option 2 : 50%
Concept:
Let L = length of each conductor in metre
σ = current density in A/m2
P = power supplied in watts
A = Area of Conductor
Volume of Cu required for both conductors is
V = 2 × A × L
Calculation:
(i) 100 V Supply
Current per feeder conductor I1 = P/100
Area of conductor required A1 = I1/σ = P/100σ
Volume of Cu required for both conductors is
V1 = 2 × A1 × L = \( {2 × P\times L \over 100\sigma }\)
(ii) 200 V Supply
Current per feeder conductor I2 = P/200
Area of conductor required A2 = I2/σ = P/200σ
Volume of Cu required for both conductors is
V2 = 2 × A2 × L = \( {2 × P\times L \over 200\sigma } = { P\times L \over 100\sigma } \)
%age saving in Cu = \( {V_1 - V_2 \over V_1} \times 100 = { {2 × P\times L \over 100\sigma } - { P\times L \over 100\sigma } \over {2 × P\times L \over 100\sigma }} \times 100\)
= 50 %