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

For a D.C. 2-wire feeder, the transmitted power is given by P = VI, where V is the supply voltage and I is the current.

Since the transmitted power remains constant, if the supply voltage is increased by 100%, the new voltage becomes 2V.

Therefore, the current becomes:

I' = P / (2V) = I / 2

The weight of copper required in a feeder is proportional to the conductor cross-sectional area, and for a given voltage drop, it is proportional to I².

Hence, when the current is reduced to half, the copper required becomes:

(I/2)² = I²/4

This means the new copper requirement is only 25% of the original amount.

Therefore, the saving in copper is:

100% − 25% = 75%

Thus, by doubling the supply voltage while transmitting the same power, the saving in copper is 75%.

Hence, Option C (75 percent) is the correct answer.