no tagsLecture 016: Be a Physicist Day (Lightbulb Edition)SteveSekula, 7 October 2010 (created 3 October 2010)Series and Parallel Lightbulb DemoConsider plugging a 40W lightbulb into a wall outout (V=110V). We can compute the resistance of the bulbknowing its power output (40W) when plugged into the wall:The current through such a circuit is .Consider doing the same thing with a 100W lightbulb. We can calculate its resistance in the same way:A higher-wattage light bulb actually has a lower resistance! This makes sense, if you think about what shortingyour house wiring will do - it will heat up the wiring so fast, due to the low resistance, that the wiring will glow,possibly melt, and definitely cause anything nearby to catch fire. A high-wattage lightbulb is closer to a shortthan a lower-wattage one. It outputs more light and heat, as a result.Now, consider plugging in two of the 40W lightbulbs in series, so that current has to pass from the wall,through the first light, then through the second, and then back to the wall. The circuit is in its steady-state - thelight output is constant and no charge is building up anywhere in the circuit. Thus, the current through bothlightbulbs must be the same.General Physics - E&M (PHY 1308) Lecture NotesGeneral Physics - E&M (PHY 1308) Lecture NotesR =P 03Ê 40= V240= 3I :36A 40= 0R =P 21Ê 100= V2100= 1General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...1 of 4 10/07/2010 11:00 AMLet's then consider what happens to the power that can be maximally dissipated through each of the lightbulbs.Since the current is the same through both resistors, we can calculate it from Ohm's Law. The total resistance isthe sum of the individual resistances:Thus:which is half the current through the original 1-bulb circuit.The voltage across each bulb is reduced from the total, but from conservation of energy the work per unitcharge done across the whole circuit by the wall outlet is equal to the work per unit charge done in eachresistor. Thus:which is half the voltage across a single lightbulb in the original 1-bulb circuit.The maximum power that can be dissipated across either bulb is thus:That's one-quarter the power that either lightbulb can output when hooked up to the wall by itself! This isbecause the voltage through each light is cut in half (due to their being in series), but also the current throughthe circuit is down by half (due to twice the resistance), so the power is down by a factor of 4.So we expect each light to output about 1/4 of its value when it's alone in the circuit, and this is basically whatwe observe with the demo.R 06Ê total= 6I =R :19A total= I1= I2= Vtotal= 0V = V1+ V2V R 5V 1= Itotal1= 5V R 5V 2= Itotal2= 5P V V 0W max= I1 1= I2 2= 1General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...2 of 4 10/07/2010 11:00 AMWhat about if we replace one of the 40W lightbulbs with a 100W lightbulb? Let's make a table of all the numbersin the circuit now.We see that the 40W bulb can maximally dissipate 20.5W in this setup - HALF of its value when plugged inalone. Thus we expect it to glow BRIGHTER when hooked up in series with a 100W (lower-resistance) bulb. The100W bulb, however, can only maximally dissipate about 8W of power - less than 1/10th of its rated value. Infact, this is even too little power to really light the filament. In a dark room, you can barely see it glowing.What happens if, instead, we hook up the 40W and 100W bulbs IN PARALLEL? Now the two bulbs are at thesame potential, but do not have the same current flow (current will flow preferentially along the path of leastresistance, so the 100W bulb should have MORE current than the 40W bulb).The maximum power that can be dissipated through the 40W bulb is:which is its rated power. What about the 100W bulb?Again, its maximum rated power. In parallel, both lights are are bright as if they were plugged in alone to thewall.V total110V R total= R40+ R100424Ê I total= I40= I1000:26A V 4079V V 10031V P 4020:5W P 1008:1W P =R 10V=303Ê 0W = V2= 1 = 4P =R 10V=121Ê 00W = V2= 1 = 1General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...3 of 4 10/07/2010 11:00 AMSo series bulbs = mood lighting. Parallel bulbs = study lighting.General Physics - E&M (PHY 1308) - Lecture Notes file:///home/sekula/Documents/Notebooks/PHY1308...4 of 4 10/07/2010 11:00
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