Correct Answer - C
Power, `P = I^(2)R`
`P_(1) = I^(2)R`
and `P_(2) = (2I)^(2) R = 4I^(2)R`
`[therefore 100% " increase in current means that current becomes " 2I]`
`therefore` Increase in power dissipated `= P_(2) - P_(1)`
`= 4I^(2)R - I^(2)R`
`= 3I^(2)R` `" "(because I^(2)R = P_(1))`
Percentage increase in power dissipated `= (3P_(1))/(P_(1)) xx 100 = 300%`