Calculation

[Hfalz1]

Can some one solve?

 

[infinity]

It's ends up being two different resistances in series. The voltage drop across one will be double that of the other.

 

[synchro]

Along the lines of what Infinity mentioned, the answer that they likely want is the voltage from 1 to 2 would be 240V x 2/3 = 160V, and from 2 to 3 would be 240V x 1/3 = 80V.

Now in reality, the resistance of a lamp will be nonlinear with voltage, with the resistance increasing at higher voltages and higher subsequent currents because the filament is getting hotter. That will make an even greater disparity in the voltages they ask for. And with 160V and even more on a 120V 50W lamp, it might not take long before that filament opens up and the voltages requested go to 240V and 0V.
Now LED lamps are much more nonlinear than incandescent and vary a lot between brands and types, and so it's best to not even go there.

 

[Hfalz1]

How do you get the formula 240V x 2/3 and the other?

 

[synchro]

How do you get the formula 240V x 2/3 and the other?

The 50W lamp will have twice the resistance as the 100W lamp because P = V² / R, where P is power, R is resistance, and V is the nominal operating voltage of both lamps.
In the question, the resistances of the two lamps are in series and so they split the 240V between them. Because the same current flows through both lamps, the lamp with twice the resistance will drop twice the voltage as the other lamp. And so the 50W lamp will get 2/3 of the 240V or 160V, and the 100W lamp will get 1/3 of 240V or 80V.

 

[Hfalz1]

thank you sir!

 

[gar]

221228-2137 EST

Hfalz1:

This is a very poorly written question. Synchro provided a good answer.

For a tungsten filament bulb you can use it with any input voltage up to it's rated voltage, and a ways above that rating. Average lifetime of a bulb goes down quickly with voltage above its rating. The shortening of life is greatly accelerated above the bulb's rated voltage.

A 120 V 100 W bulb has a hot resistance of about 144 ohms from 100 / 120 = 0.84 A. At room temperature and no current the resistance of the 100 W bulb is about 10.7 ohms. A ratio of about 14 to 1.

With the two bulbs in series, and 240 V applied to the series string the lower wattage bulb will have a much higher voltage across it than the higher wattage unit.

As an experiment I used a 60 W bulb in series with a 100 W bulb, and with total voltage adjusted to put 120 V across the 60 W bulb the total voltage is 146 V. Or a 26 V drop on the 100 W bulb.

The person that created this question is ignorant and incompetent, and should not be teaching.

.

 

[LarryFine]

The moral of the story, boys and girls, is that questions like this should use fixed resistors, not lamps.

That really applies when discussing such questions here, where we pick apart everything to the quick.

 

[junkhound]

The person that created this question is ignorant and incompetent, and should not be teaching.

+1

Possibly a decent problem IF the author had included a statement to assume power proportional to V^1.6?

 

[junkhound]

re"

pick apart everything to the quick.

'perhaps' (hah) the question originator assumed LED lamps fed byan 88% 92% efficient constant current 24 to 277 V input switching power supplies. - and supplied an efficiency graph vs. input voltage <G>

 

[drcampbell]

... Now in reality, the resistance of a lamp will be nonlinear with voltage, with the resistance increasing at higher voltages and higher subsequent currents because the filament is getting hotter. That will make an even greater disparity in the voltages they ask for. And with 160V and even more on a 120V 50W lamp, it might not take long before that filament opens up and the voltages requested go to 240V and 0V. ...

You neglected transformer impedance.

... we pick apart everything to the quick.

Translation: We give due consideration to ALL the facts & variables that exist in the real world, because we're in the habit of designing real-world systems that will be safe, trouble-free, and long-lasting, not mere academic exercises.

 

[infinity]

This is an academic exercise so let's not over complicate it. Fact is with 160 volts across an incandescent lamp it will likely burn out before you can get you test probes on it.

 

[dkidd]

This is an academic exercise so let's not over complicate it. Fact is with 160 volts across an incandescent lamp it will likely burn out before you can get you test probes on it.

Formula for Incandescent Lamps​

(Actual Life/Rated Life) = (Rated Volts/Actual Volts)^13.

 

[Besoeker3]

It's ends up being two different resistances in series. The voltage drop across one will be double that of the other.

dkidd is right - you can't easily do that simple calculation.

 

[drcampbell]

Formula for Incandescent Lamps
(Actual Life/Rated Life) = (Rated Volts/Actual Volts)^13.

I wish more people understood this. I get so tired of seeing the "miracle" of the incandescent lightbulb that's been on continuously since 1941. Well, D'Oh! Its color temperature is so yellow that its V_r/V_a is almost 2.

 

[junkhound]

I wish more people understood this. I get so tired of seeing the "miracle" of the incandescent lightbulb that's been on continuously since 1941. Well, D'Oh! Its color temperature is so yellow that its V_r/V_a is almost 2.

Reminds me of the old lamp in my grade school, buitl circa 1910, torn down circa 1990. Pop was on the property management board ( a Luth school) and there was a well understood agreement from early 1950's to let the one original bulb still in place in a high hall ceiling fixture left turned on untill it failed. It was still operating when the school torn down. IIRC it was a dim yellow!
$700 power bill to run that lamp in today's dollars.
It looked similar to this one, still can see it in mind's eye, but had more of a taper to the envelope.

 

[ggunn]

The 50W lamp will have twice the resistance as the 100W lamp because P = V² / R, where P is power, R is resistance, and V is the nominal operating voltage of both lamps.
In the question, the resistances of the two lamps are in series and so they split the 240V between them. Because the same current flows through both lamps, the lamp with twice the resistance will drop twice the voltage as the other lamp. And so the 50W lamp will get 2/3 of the 240V or 160V, and the 100W lamp will get 1/3 of 240V or 80V.

That is assuming, of course, that power through the lamps will be constant across all voltages, which also of course isn't how it works. The point was to get the student to take power as a given and calculate the voltages, though, so it works as far as that goes. The real world solution would likely require the application of calculus; I don't think it was meant to be that complicated.

 

[kwired]

How do you get the formula 240V x 2/3 and the other?

a situation that sort of only applies to this particular set of lamps, or any set where one is twice the watts of the other. Put a 25 and a 60 watt lamp in the same drawing and the 1/3 and 2/3 approach won't apply.

I did not read the rest of the posts yet but bottom line is you do need to figure out what the resistance of each lamp is and then figure out what voltage drop will be across each one when they are placed in series across 240 volts.

 

[kwired]

The moral of the story, boys and girls, is that questions like this should use fixed resistors, not lamps.

That really applies when discussing such questions here, where we pick apart everything to the quick.

It sort of depends on what lesson you are trying to teach with the example. I remember when I was still in trade school the instructor giving an example of what happens with lost neutral having a similar drawing but with say a resistance heater on one side and a TV set on the other side.

Resistance heater was probably close enough resistance hot or cold it was negligible. Resistance of the TV set - I don't recall exact value he used but do remember him saying something to the effect "lest just assume it is XXX ohms". And likely was accurate enough to be within 10%, and still gave volts and current figures in the results that made you say that can't be good for that thing.

 

[roger]

; I don't think it was meant to be that complicated.

Thank you. Unfortunately some want to show their superior intellect and want to take an elementary level question showing basic ohms law to some rocket science level. What they fail to realize is everyone has to crawl before they walk.

To the OP, your question was answered by post # 5 and most anything beyond that was beyond what the course question was looking for.

To make the others happy maybe the question should have left "lamp" out of the equation but get over it.