Here, `V = 50` volt, `I = 12` ampere.,
`eta = 30%, R = ?`
Input electric power ` = VI = 50 xx 12 = 600` watt
As efficiency is 30%, therefore, power dissipated as heat is 70% of this power
`P = (70)/(100) (VI) = (70)/(100) xx 600 = 420 W`
As `P = I^(2) :. R = (P)/(I^(2)) = (420)/(12 xx 12) = 2.9 Omega`