`B_1,B_2 and B_3` are three identical bulbs connected as shown in (fig. 3.46). 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 readings 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 ?
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Explanation

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