$B_1, B_2$ and $B_3$ are three identical bulbs connected as shown in the figure. When all 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?
$(iii)$ How much power is dissipated in the circuit when all three bulbs glow together?

(N/A) $(i)$ Since the bulbs are connected in parallel,the potential difference across each bulb remains the same $(4.5\,V)$. Therefore,the glow of the bulbs $B_2$ and $B_3$ will remain unchanged when $B_1$ fuses.
$(ii)$ When $B_2$ fuses,the circuit branch containing $B_2$ becomes open. Thus,$A_2$ will show $0\,A$. The other two branches remain unaffected,so $A_1$ will show $1\,A$ and $A_3$ will show $1\,A$. The total current recorded by ammeter $A$ will be $1\,A + 0\,A + 1\,A = 2\,A$.
$(iii)$ The total power dissipated in the circuit is given by $P = V \times I$. Given $V = 4.5\,V$ and total current $I = 3\,A$,we have $P = 4.5\,V \times 3\,A = 13.5\,W$.