1 MW power is to be delivered from a power station to a town 10 km away. One uses a pair of Cu wires of radius 0.5 cm for this purpose. Calculate the fraction of ohmic losses to power transimitted if (i) power is transformer is used to boost the voltage to 11000 V, power transmitted, then a step down transformer is used to bring voltages to 220 V. `(rho_(Cu) = 1.7 xx 10^(-8) SI`unit)

Explanation

(i) As the town is 10 km away, length of pair of Cu-wires used, `l = 20 km = 20000 m`.
Resistance of wire, `R = (rho l)/(a) = (rho l)/(pi r^(2)) = (1.7 xx 10^(-8) xx 20000)/(3.14 (0.5 xx 10^(-2))^(2)) = 4 Omega`
When Power `P = 10^(6)` watt is transmitted at V = 220 V.
Current drawn, `I = (P)/(V) = (10^(6))/(220) = 0.45 xx 10^(4) A`
Power loss, `I^(2) R = (0.45 xx 10^(4))^(2) xx 4 = 0.81 xx 10^(8) W`, which is so large.
Therefore, this method of transmission of power is not feasible.
(ii) When power `P = 10^(6)` wattis transmitted at 11000 V,
Current drawn, `I = (P)/(V) = (10^(6))/(11000) = (10^(3))/(11) A` , Power loss, `I^(2) R = ((10^(3))/(11))^(2) xx 4 = 3.3 xx 10^(4)` watt
Fraction of ohmic losses to power transmitted `= (3.3 xx 10^(4))/(10^(6)) xx 100% = 3.3%`