Can you explain to me if you have 4 light bulbs with ratings 60W, 75W, 100W and 120W in series, connected to a 120V battery, which light bulb will shine the brightest, and why.
THANK YOU SO MUCH IN ADVANCE!
Question from Physics 2.
2007-10-24 19:34:46 · 7 answers · asked by stackinchips11 2 in Science & Mathematics ➔ Engineering
These wattage ratings are based on 120 V.
The respective resistances are 14400/P:
60W, 240Ω
75W, 192Ω
100W, 144Ω
120W, 120Ω
In a series connection, the brightness will be proportional to the resistance of the bulb, so the 60W bulb will be the brightest. It will use about 7.13 watts, so it will be quite dim, if you can see its glow at all.
2007-10-24 20:07:13 · answer #1 · answered by Helmut 7 · 2⤊ 0⤋
The four bulbs have decreasing value of resistance, the highest is with the 60-W bulb and the lowest is with the 120-W bulb. Since they are in series the current flowing is the same. But power is I^R; hence the 60-W bulb will shine brightest and the 120W bulb will shine dullest.
2007-10-24 20:33:52 · answer #2 · answered by Anonymous · 1⤊ 0⤋
Use rated wattage, voltage to find R=V^2 / W
Resistances are 240,192,144,120, for a sum of 696.
I=V/R = 120/696 = 172 mA.
Now compute actual power dissipated by each bulb using W=I^2 * R for each R: 7.1, 5.7,4.3, 3.6 W.
The 60 W bulb will be brightest, due to common current in a series circuit and the relation W = I^2 * R
2007-10-24 20:50:22 · answer #3 · answered by David F 7 · 0⤊ 0⤋
Well, incandescent bulbs never were that smart, but the fact that all of yours are in series means the higher wattage will have the lowest resistance. Let us give them an easy resistance to work with (not real, but it doesn't matter):
60W- 100 Ohm
75W- 75 Ohm
100W- 50 Ohm
120W- 40 Ohm
Power consumption will decide the winner.
100+75+50+40 = 265 Ohm total gives 120/265=453 mA.
.453x.453 = ~.20 for the I-squared part (I^2 R= power)
so the:
60W --> 20W
75W--> 15W
100W--> 10W
120W--> 8W
The 60 W bulb will shine the brightest.
See how to do it?
2007-10-24 21:23:01 · answer #4 · answered by Warren W- a Mormon engineer 6 · 0⤊ 0⤋
Assuming all the bulbs are designed to work with the same voltage, the highest wattage one (120W) will have the lowest resistance, since power = V^2 / R. When connected in series, all bulbs carry the same current, and the power is I^2 * R. The lowest resistance bulb will use the least power and be the dimmest. Therefore the 60W bulb will be brightest.
2007-10-24 19:54:24 · answer #5 · answered by gp4rts 7 · 1⤊ 1⤋
The 60W bulb will be the brightest. When connected in series, the bulbs will sum a total of 831 Ohms. That resistance, produces a current draw of 360 mA. The 360 mA, dissipates 48 Watts in the 60W bulb, and 31 Watts in the 75W Bulb, 17 Watts in the 100W bulb and 12 Watts in the 120W bulb. The higher the power dissipation, the higher the luminosity, so the 60W bulb should be brighter. The numbers are not precise, and assume a 110VAC voltage. Also, it assumes that the bulbs are 100% resistive in nature. So numbers are VERY close to reality, but imprecise.
Weird isn't it?
2007-10-24 19:53:28 · answer #6 · answered by Ricardo A 3 · 2⤊ 2⤋
relies upon on the wiring. i be conscious of in the event that they are parallel that the two mild bulbs will obtain a similar volume of present day yet whilst it is going one after yet another the 2d would be dimmer than the 1st one simply by fact mild bulbs additionally "act like resistors".
2016-10-13 23:50:20 · answer #7 · answered by ? 4 · 0⤊ 0⤋