`B_(1), B_(2), B_(3)` are three identical bulbs connected as shown in figure. Ammeters `A_(1), A_(2), A_(3)` are connected as shown. When all the bulbs glow, the current of 3A is recorded by ammeter A.
(i) What happens to the glow of the other two bulbs when 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 the three bulbs glow together?

Explanation

Resistance of combination of three bulbs in parallel
`(R)/(R_("eq"))= (V)/(I)= (4.5)/(3)=1.5Omega`
If R is the resistance of each wire, then
`(1)/(R_("eq"))= (1)/(R) + (1)/(R) + (1)/(R)`
or `(1)/(R_("eq"))= (3)/(R)`
or `R= 3, R_("eq") = 3 xx 1.5= 4.5 Omega`
Current in each bulb, `I= (V)/(R)= (4.5V)/(4.5 Omega)= 1A`
(i) When bulb `B_(1)` gets fused, the currents in `B_(2)` and `B_(3)` remain same `I_(2)= I)_(3) = 1A`, so their glow remains unaffected.
(ii) When bulb `B_(2)` gets fused, the current in `B_(2)` becomes zero and currents in `B_(1)` and `B_(3)` remains 1A.
Total current `I = I_(1)+ I_(2)+ I_(3) = 1+0+1 = 2A`
Current in ammeter `A_(1)` = 1 A
Current in ammeter `A_(2) = 0`
Current in ammeter `A_(3) = 1 A`
Current in ammeter A = 2A
(iii) When all the three bulbs are connected.
Power dissipitated `P= (V^(2))/(R_("eq"))= ((4.5)^(2))/(1.5)= 13.5W`