How do you calculate the resistance of 50 W and 100 W lamps and the current when they are connected in series across 240 V?
Assuming both lamps are rated for 240 V and using their hot (nameplate) resistances, the 50 W lamp is 1152 Ω and the 100 W lamp is 576 Ω, giving a total series resistance of 1728 Ω and a circuit current of about 0.139 A [#21686033] In real life, incandescent lamps are non-ohmic, so their resistance changes with temperature and the series current will not match the simple calculation exactly [#21686025][#21686026] Because of that, the simple Ohm’s-law result is only a theoretical answer unless the lamp characteristics are measured [#21686025][#21686026]
AI summary based on the discussion. May contain errors.
There was much disagreement to this question on another site: If, two lamps one is 50 Watts and the other is 100 Watts and they are in series with each other across 240 V. what would be the resistance of each lamp and the current for that circuit? Thanks in advance Fyi: I calculated 128 Ohms for the 50w and 256 Ohms for the 100w with a total current of .625a
If the wattage ratings are for a 240 V supply, that is both lamps would normally be in parallel across 240V, then the lamps in a series circuit will not be operating to spec, bulbs are non ohmic(nonlinear). They will have lower resistance than normal operating in a series circuit. Without measuring the lamp characteristics, don't expect to get sensible real world answers. If this a purely theoretical exercise, then your answer is correct. Article here https://www.electronics-notes.com/articles/ba...hy-is-filament-incandescent-lamp-nonohmic.php Cheers, Richard
Too many unknowns here. Are 50W and 100W the actual powers the bulbs are dissipating, or is that what they would dissipate if they were put across 240V (ie normal light bulbs)?
The other thing here is that light bulbs have a cold resistance which is significantly lower than the hot resistance. So if you put two in series like this the current will be limited by the 50W bulb and the 100W bulb may never get to its proper operating temperature and will remain colder and lower resistance. SO you might find that the 50W bulb has (say) 200V across it (close to normal) and the 100W bulb will only have 40V across it and may be only just lit.
You could do an experiment with a 5W and a 10W 12V bulb in series across 12V which would be a safe and easy way to verify what would happen.
FYI Power = V x I so I = P / V , and R = V / I so: a 50W 240V bulb will have 50/240 = 0.208A flowing through it so R = 240 / 0.208 = 1152 ohms a 100W 240V bulb will have 100/240 = 0.416A flowing through it so R = 240 / 0.416 = 576 ohms But these are hot resistances and cold resistance can be even 1/8 of hot resistance.
This property is often used in oscillators - a small light bulb is put in the feedback circuit to stabilise the output voltage. Hewlett Packards first product, the legendary 200A oscillator, used this method: https://en.wikipedia.org/wiki/HP_200A Hope this helps. Feel free to come back with more details!
This question was asked in a posts on a electrician site. The only known given was the Total Voltage (240V) and Watts (50W and 100W) of each lamp. Only the Voltage drop across each lamp was asked. I showed my work and did the answer as purely theoretical exercise. Many were telling me I was incorrect. I then tried my answers using different formulas and they worked and I got the same answer. I had the same current for each load which would be correct for a series circuit. The others worked the problem out but the series circuit current would not be the same for each lamp. THIS IS HOW THEY WORKED THE PROBLEM OUT: Wattage is a RESULT of current and voltage, it is NOT a constant P=IE So you'd have to calculate the resistance if an assumed 120v circuit across each bulb, since that's what they're rated for V=IR V=(P/E)*R R=V/(P/E) R=120v/(100w/120v) R=144ohms Do the same foe the other, R=120v/(50w/120v) R=288ohms Total resistance is r1+r2, 144+288 = 432ohms Now total current is V=IR I=V/R I=240v/432ohms I=0.555a Insert known values to calculate voltage drop across each load V=IR V=0.555*144ohms V=~80v for the 100w bulb V=IR V=0.555*288ohms V=~160v for the 50w bulb
I didn't think this answer to be correct.
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Need bulb rating basis before solving series circuit
Chris, the question appears to be theoretical in nature, perhaps it is some sort of academic exercise, from a school text book, or someone has decided to stir up the people out there like us who have too much time on their hands and like replying to such queries (and at times should know better) In light of that, no pun intended, the question is are the wattage figures supplied for bulbs individually rated at a nominal 240V out of the box, or for the power dissipated in the circuit after power has been applied and it is in a steady state condition. The latter results in a trivial solution, in the former case, as David said, there are too many unknowns to give any sort of definitive answer, the poster would need to reply before any more could be said, cheers, Richard
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Series-bulb calculation assumes hot resistance incorrectly
You say that "...an assumed 120v circuit across each bulb, since that’s what they’re rated for" (but that is not specified, like I say we need more information.....) and also "V=~160v for the 50w bulb"
So in fact no current will flow in the 50W bulb in this case, because if you put 160V across a 120V bulb it will burn out pretty quickly. Since they're in series, no current would flow in either bulb (except very briefly...)
As before, your above calculations assume the bulbs are both at full hot resistance, which they won't be.
> I didn’t think this answer to be correct. You're right!!
Chris, It seems like a theoretical inquiry, like something out of a school textbook, or maybe someone is just trying to get a rise out of the folks who, like us, have too much free time on their hands and enjoy responding to such questions (and at times should know better) Therefore, the question arises as to whether the wattage estimates provided for lights are for the power dissipated in the circuit once power has been connected and it is in a steady state condition (no pun intended) or for the power rated at 240V when the bulbs are first removed from the box. As David mentioned, the former leads to a simple answer, whereas the later requires the poster to respond before any further information can be provided due to the numerous variables at play.
Series lamp resistance calculation gives 139 mA current
The discussion addresses the theoretical calculation of resistance and current for two incandescent lamps rated at 50W and 100W connected in series across a 240V supply. Using the power rating and voltage, resistance for each lamp can be estimated by R = V² / P, yielding approximately 1152 ohms for the 50W lamp and 576 ohms for the 100W lamp. The total series resistance sums to about 1728 ohms, resulting in a circuit current of approximately 0.139 A. However, practical considerations highlight that incandescent lamps are non-ohmic devices with resistance varying significantly between cold and hot states. When connected in series, the lamps do not operate at their rated voltages or powers, causing uneven voltage drops and current flow, with the lower wattage lamp limiting current and the higher wattage lamp potentially not reaching full operating temperature. This makes real-world calculations complex and theoretical results approximate. The consensus is that such series connections are not typical for lamps designed for parallel operation at 240V, and experimental verification or detailed lamp characteristic data is necessary for accurate analysis. AI summary based on the discussion. May contain errors.
