A number of identical light bulbs are connected in parallel to a120V AC power so

Question

A number of identical light bulbs are connected in parallel to a120V AC power source. Each bulb dissipates an average powerof 60W. The circuit has a fuse that burns out when thecurrent in the circuit reaches 9.0 A. What is the maximumnumber of bulbs that can be added in parallel without burning outthe fuse?

I know that the voltage of 120 is the RMS value, so Vrms = 120V
I also know that the average power (Pav), equates to Irms^2 x R
I'm confused as to the current that is given. Should I setthis as the maximum current? Or would I use it like a rmsvalue?

Explanation / Answer

current through the circuit I = 9 A voltage across the circuit V = 120 V   total power of thecircuit     P = I V                                               = 9 *120   = 1080 W    equivalent resistance of the circuit  R = V2 / P                                                                = 1202 / 1080                                                             = 13.33      the identical bulbs are connected in parallelcombination current through each bulb is same resistance of each bulb is   r =   I2   / P                                                 = 81 / 60 = 1.35            equivalentresistance   R = resistance of eachbulb  * number of bulbs                              13.33 = 1.35  N                  there fore   N = 9.89                                        ˜ 10 bulbs                                                 

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