A resistor  ${\text{R}}_{\text{1}}$ dissipates the power P when connected to a certain generator. If the resistor  ${\text{R}}_2$ put in series with  ${\text{R}}_{\text{1}}$ the power dissipated by ${\text{R}}_{\text{1}}$ 

Given that when ${\text{R}}_{\text{1}}$ is connected to a generator $\Rightarrow \text{P}=\frac{{\text{V}}^{\text{2}}}{{\text{R}}_{\text{1}}}\text{and\hspace{0.17em}}i=\frac{\text{V}}{{\text{R}}_{\text{1}}}$

Now  ${\text{R}}_2$ is connected in series with ${\text{R}}_{\text{1}}$   then  $i^{|}=\frac{\text{V}}{{\text{R}}_{\text{1}}+{\text{R}}_{\text{2}}}$

So power dissipated across  ${\text{R}}_{\text{1}}$ decreases