$B_{1}$,$B_{2}$ and $B_{3}$ are three identical bulbs connected as shown in the figure. When all the three bulbs glow,a current of $3 \, A$ is recorded by the ammeter $A$.
$(i)$ What happens to the glow of the other two bulbs when the bulb $B_{1}$ gets fused?
$(ii)$ What happens to the reading of $A_{1}$,$A_{2}$,$A_{3}$ and $A$ when the bulb $B_{2}$ gets fused?

(N/A) $(i)$ Since the bulbs are connected in parallel,the potential difference across each bulb remains the same $(4.5 \, V)$. Therefore,if bulb $B_{1}$ fuses,the other two bulbs ($B_{2}$ and $B_{3}$) will continue to glow with the same intensity.
$(ii)$ Initially,the total current $I = 3 \, A$ is divided equally among the three identical bulbs in parallel,so each bulb draws $I_{1} = I_{2} = I_{3} = 1 \, A$.
When bulb $B_{2}$ fuses,it creates an open circuit in its branch,so the current through it becomes zero. Thus,the reading of $A_{2}$ becomes $0 \, A$.
Since $B_{1}$ and $B_{3}$ remain connected in parallel,they still experience the same potential difference and draw $1 \, A$ each. Therefore,$A_{1}$ shows $1 \, A$ and $A_{3}$ shows $1 \, A$.
The total current $A$ recorded by the main ammeter will be the sum of the currents in the remaining branches: $A = 1 \, A + 1 \, A = 2 \, A$.