When the switch S is made ON in the figure, the bulb which glows brightest is
 

In series high resistor consume more power, more voltage
In parallel low resistor consume high power and high voltage
When switch is opened, ‘B’ will glow brighter than ‘A’ $\left({\text{R}}_{\text{B}}>{\text{R}}_{\text{A}}\right)$
${\text{P}}_{\text{A}}=\frac{{\text{V}}^{\text{2}}}{{\text{R}}_{\text{A}}}\Rightarrow {\text{R}}_{\text{A}}=\frac{{\text{V}}^{\text{2}}}{\text{100}},{\text{\hspace{0.17em}\hspace{0.17em}P}}_{\text{B}}=\frac{{\text{V}}^{\text{2}}}{{\text{R}}_{\text{B}}}\Rightarrow {\text{R}}_{\text{B}}=\frac{{\text{V}}^{\text{2}}}{\text{25}}$
After closing the switch B & C are in parallel
$$\begin{gathered} \therefore {\text{P}}_{\text{C}}=\frac{{\text{V}}^{\text{2}}}{{\text{R}}_{\text{C}}}\Rightarrow {\text{R}}_{\text{C}}=\frac{{\text{V}}^{\text{2}}}{\text{100}} \\ {\text{i}}_{\text{1}}{\text{R}}_{\text{C}}=\left(\text{i}-{\text{i}}_{\text{1}}\right){\text{R}}_{\text{B}} \\ {\text{i}}_{\text{1}}{\text{R}}_{\text{C}}={\text{\hspace{0.17em}i}}_{\text{2}}{\text{R}}_{\text{B}} \\ {\text{i}}_{\text{1}}\times \frac{{\text{V}}^{\text{2}}}{\text{100}}={\text{i}}_{\text{2}}\times \frac{{\text{V}}^{\text{2}}}{\text{25}} \\ \therefore {\text{i}}_{\text{1}}=4{\text{i}}_{\text{2}} \\ But \text{i}-{\text{i}}_{\text{1}}={\text{i}}_{\text{2}}\Rightarrow \text{i}-4{\text{i}}_{\text{2}}={\text{i}}_{\text{2}}\Rightarrow \text{i}=5{\text{i}}_{\text{2}} \\ \therefore {\text{i}}_{\text{2}}=\frac{\text{i}}{\text{5}},{\text{\hspace{0.17em}\hspace{0.17em}i}}_{\text{1}}=\frac{\text{4}}{\text{5i}} \\ \therefore \text{i}={\text{i}}_{\text{1}}+{\text{i}}_{\text{2}} \\ \therefore {\text{P}}_{\text{A}}={\text{i}}_{\text{2}}{\text{R}}_{\text{A}},{\text{\hspace{0.17em}\hspace{0.17em}P}}_{\text{B}}={{\text{i}}_{\text{2}}}^{\text{2}}{\text{R}}_{\text{B}},{\text{\hspace{0.17em}\hspace{0.17em}P}}_{\text{C}}={{\text{i}}_{\text{1}}}^{\text{2}}{\text{R}}_{\text{C}} \\ ={\text{i}}^{\text{2}}\frac{{\text{V}}^{\text{2}}}{\text{100}}=\frac{{\text{V}}^{\text{2}}}{\text{100}} ={{\text{i}}_{\text{2}}}^{\text{2}}\frac{{\text{V}}^{\text{2}}}{\text{25}}=\frac{\text{1}}{\text{25}}\frac{{\text{V}}^{\text{2}}}{\text{25}}=\frac{{\text{V}}^{\text{2}}}{\text{625}} \\ =\frac{{{\text{i}}_{\text{1}}}^{\text{2}}{\text{V}}^{\text{2}}}{\text{100}}=\frac{\text{16}}{\text{25}}\frac{{\text{V}}^{\text{2}}}{\text{100}}=\frac{{\text{16V}}^{\text{2}}}{\text{2500}} \end{gathered}$$
  bulb A will glow brightly.