In: Physics
A circuit contains a 120V battery, and a single light bulb whose resistance is 240 Ω.
(A) Determine the power dissipated by the single bulb. Assume 5 additional bulbs, all identical to the first bulb are placed in parallel with each other and the first bulb.
(B) Determine the power dissipated by each bulb.
(C) If a 6.00 A circuit breaker (CB) is placed in series with the battery in the circuit with the parallel bulbs, will the CB trip when all 6 bulbs are being used simultaneously (justify your answer)?
A) Power = V^2/i = 120*120/240 = 60W
B) now bulbs are placed in parallel, so V across each is 120V and R of each is 240 ohm
power dissipated by each = 120*120/240 = 60W
C) effective resistance
1/Reff = [1/240 ]*6
Reff = 240*6 = 40 ohm
new i = v/reff = 120/40 = 3 A
Hence CB will not trip as i < 6A