Eight lights are connected in parallel to a 110-V source by two long leads of total resistance 1.6 ohms. If 240 mA flows through each bulb, what is the resistance of each, and what fraction of the total power is wasted in the leads?
Please show work so I will understand how to work similar problems.
2007-03-29 05:26:06 · 1 answers · asked by oceanchick4043 1 in Science & Mathematics ➔ Engineering
You need to know the current in the leads to work out the voltage on them and the bulbs, the leads carry the total current for the 8 bulbs so 8x240mA = 1.92A
So the voltage drop V = RI = 1.6*1.92 = 3.072v
So the bulbs see 110-3.072 = 106.928v
V=RI so R=V/I = 106.928/0.24 = 445.53 ohms.
P=VI so the power in the leads is 1.92*3.072 or I^2R which is 1.6x1.92^2 = 5.898W
Power per bulb is 0.24x106.928 = 25.663W. For 8 bulbs the total is 205.3W
Fraction of the total in the leads is 5.898/(5.898+205.3) = 2.86%. As a fraction? About 1/35.
2007-03-29 05:38:43 · answer #1 · answered by Chris H 6 · 0⤊ 0⤋