Please Advise. Malibu(TM) lamps are onneted to an ideal 12.0 V power supply alon
ID: 1356372 • Letter: P
Question
Please Advise.
Malibu(TM) lamps are onneted to an ideal 12.0 V power supply along a able onsisting of two parallel 18-gauge opper wires (one for outgoing urrent, one for inoming, separated by insulation). The lamps are onneted to the able at one-foot (30.5 m) intervals via ampire taps"|prongs whih piere the able insulation on eah side to make ontat with eah wire. Thus the lamps form the rungs of a resistane ladder, suh as we used in lass to model the axon in the eletrotonus ase, exept that in this problem there are no in nitesimal quantities. All quantities must be treated as nite. When onneted individually to the power supply with wires of negligible resistane, the lamps put out 4.00 W of light power. But the resistane of 18-gauge wire is 0.110 m /m, i.e., 3.35 m /ft. Assume the array of lamps is in nite," i.e., very long, with many sores of lamps. Assume also that the resistane of eah of the lamps in operation is the same as when they are operated individually. Although atual Malibu(TM) lamps are operated with alternating urrent, treat this as a DC iruit. CAUTION: A previous student downloaded the Installation Guide for Malibu(TM) lamps from the manufaturer's website. That does not ount as a solution to this problem!
a. Calulate the total resistane of the array.
b. Calulate the total power onsumed by the array in operation.
c. Calulate the number of lamps in the array whih put out 2.00 W of light power or more.
Explanation / Answer
(1) Here it is given that two copper wire are in parallel and one lamp is connected thereafter.
So the resistances of the copper wire is 0.110 m /m and cable is of 30.5 m
therefore the resistance will be = 0.110*30.5 = 3.355 ohm
when the two same resistances will be in parallel then net resistance will be half of the resistance
therefore net resistance of the parallel wire = 3.355 /2 = 1.6775 ohm
this resistances is connected to the lamp which have resistance = V2/P = 122/4 = 36 ohm
(a) the net resistance of the complete circuit = 36+1.6775 = 37.6775 ohm
(b) power =V2/R = 122/37.6775 = 3.82 W
(c) Power = n*power of one lamp
n = 3.82 /2 = 1.91
so approx we can say 2