A circuit contains a 120V battery, and a single light bulb whose resistance is 2

Question

A circuit contains a 120V battery, and a single light bulb whose resistance is 240 .

(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)?

Explanation / 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

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