Question

If the length of the filament of a heater is reduced by 10% by cutting it, the power of the heater will?

Answer 1

we know that power P=V^2/R
P=V^2/ρL/A = V^2A/ρL
where ρ =density of the filament material, A= cross sectional area of filament
if P1 be the power after reducing 10% length
then P1=V^2A/ρ(L-10/100L) = V^2A/ρL(9/10)

or P1/P=10/9

therefore % of increase in power = [(P1-P)/P] *100
= (P1/P -1)*100 =(10/9 - 1)*100=11%

SO it increases by 11%