Let R be the resistance cable. Here, `P = 10 MW = 10 xx 10^(6)W = 10^(7)W`.
(i) `V_(1) = 20,000V`,
Current `I_(1) = P/V_(1) = 10^(7)/(20,000) = 500 A`
Rate of heat dissipated,
`P_(1) = I_(1)^(2) R = (500)^(2)R = 25 xx 10^(4)R W`
(ii) `V_(2) = 200 V`,
Current `I_(2) = P/V_(2) = 10^(7)/200 = 5 xx 10^(4)A`
Rate of heat dissipated,
`P_(2) = I_(2)^(2) R = (5 xx 10^(4))^(2) R= 25 xx 10^(8) R W`
Clearly `P_(1) lt P_(2)`, so there will be lesser power wastage during transmission at `20,000 V` than at `200 V`.