A resistor R₁ dissipates the power P when connected to a certain generator. If the resistor R₂ is put in series with R₁, the power dissipated by R₁

Power dissipated by R₁ initially is 

$\mathrm{P} = {\mathrm{i}}^2{\mathrm{R}}_1$

$\Rightarrow \mathrm{P} \propto \mathrm{i}$

When R₂ is connected in series, 

current $\mathrm{i} = \frac{\mathrm{V}}{{\mathrm{R}}_1+{\mathrm{R}}_2}$

Current across the resistance is decreased, therefore P decreases.