What is the percentage saving in feeder copper if the line voltage in a 2-wire DC systems is raised from 100 V to 200 V for the same power transmitted over the same power distance and having the same power loss?

Consider 200 V (Fig i) and 400 V (Fig ii) system as shown below"

We have,

P₁ = V₁I₁ = 200I₁  and P₂ = V₂I₂ = 400I₂ 

As the same power is delivered in both cases,

Therefore,

P₁ = P₂

200I₁ = 400I₂ 

I₂ = \(\frac{200}{400} = 0.5\)I₁

Now, power loss in 200V system (W₁) = 2I₁²R₁

And, power loss in 400V system ( W₂) = 2I₂²R₂  =  2(0.5I₁)²R₂ = 0.5I12R2

As power loss in the two cases is the same, W₁ = W₂

2I₁²R₁ = 0.5I₁²R₂;

⇒ R₂/R₁ = 4

We know resistance is inversely proportional to area.

Therefore:  A₁/A₂ = 4,  Also, ratio of volume ; v₁/v₂ = 4

⇒ v₁ = 4v₂ 

(Note: 'v' abbreviated for volume)

Hence,

Percentage Saving in copper = \(\frac{V_{1}-V_{2}}{V_{1}}\) × 100 = \(\frac{4v_2-v_2}{4v_2}\times 100 =75\%\)