• # question_answer ${{B}_{1}},{{B}_{2}}$ and ${{B}_{2}}$ are the three identical bulbs connected to a battery of steady emf with key K closed. What happens to the brightness of the bulbs ${{B}_{1}}$ and ${{B}_{2}}$, when the key is opened. A)  Brightness of the bulb ${{B}_{1}}$ increase and that of ${{B}_{2}}$ decreases. B)  Brightness of the bulb ${{B}_{1}}$ and ${{B}_{2}}$ increase C)  Brightness of the bulb ${{B}_{1}}$ decrease and ${{B}_{2}}$ increases. D)  Brightness of the bulb ${{B}_{1}}$ and ${{B}_{2}}$ decreases.

When key K is opened, bulb B^ will not draw any current from the source, so that terminal voltage of source increases. Hence power consumed by bulb increases, so light of the bulb becomes more. The brightness of bulb ${{B}_{1}}$ decreases because ${{i}_{before}}\,=\frac{V}{R+\left( \frac{RR}{R+R} \right)}$ put ${{i}_{after}}=\frac{V}{R+R}$ And $P={{i}^{2}}R$ while i decreases