Transmitted power remaining the same, if supply voltage of a D.C. 2-wire feeder is increased 100%, saving in copper is-

Two-wire DC system with one conductor earthed

Case 1: If the supply voltage of a D.C. 2-wire feeder is V_m

Max. voltage between conductors = Vm

Power to be transmitted = P

Load current, \(I_1={P\over V_m}\)

If R1 is the resistance of each line conductor, then \(R_1={\rho l\over a_1}\)

Line losses, \(W=2I_1^2R_1=2({P\over V_m})^2({\rho l\over a_1})\)

∴ Area of X-section, \(a_1={2P^2\rho l\over WV_m^2}\)

The volume of conductor material required = 2a1l

\((Volume)_1=2({2P^2\rho l^2\over WV_m})l={4P^2\rho l^2\over WV_m^2}=4K\)

Case 2: If the supply voltage of a D.C. 2-wire feeder is increased 100%.

V = 2V_m

Load current, \(I_1={P\over 2V_m}\)

Line losses, \(W=2I_2^2R_2=2({P\over 2V_m})^2({\rho l\over a_2})\)

∴ Area of X-section, \(a_2={P^2\rho l\over 2WV_m^2}\)

The volume of conductor material required = 2a₂l

\((Volume)_2=2({P^2\rho l\over 2WV_m^2})l={P^2\rho l^2\over WV_m^2}=K\)

% change = \({final-initial \over initial}\times 100\)

% change = \({K-4K \over 4K}\times 100\)

The saving in copper is 75%