What is the watt calculation for a series bulb?

Status
Not open for further replies.
P

panneerrajan

Member level 1
Joined
Jul 9, 2013
Messages
38
Helped
0
Reputation
0
Reaction score
0
Trophy points
6
Location
Trivandrum
Visit site
Activity points
350
Hi friends,

how to calculate the Total Watts for following connections.

200W 250V Bulb two and 100W 250V Bulb one is connected in series, so what is the total Watts taken?
Please explain how to calculate this?
 

Re: watt calculation for series bulb

Your voltage source is it AC or DC?
For 200watts 250V, its resistance is Vsquared/ P
the resistance is 312.5 ohm(for one)
for 100watts, the resistance is 625 ohm

- - - Updated - - -

since they are connected in series, the total resistance is 312.5+312.5+625=1250 ohms
the total power is P=V squared/ R
= (250*250)/1250
 
Re: watt calculation for series bulb

The assumption of voltage independent resistance isn't right for filament lamps.

You need to know the lamps V/I characteristics for an exact solution. I guess it's around 90 W.
 
Reactions: WimRFP

    WimRFP

    Points: 2
    Helpful Answer Positive Rating
Re: watt calculation for series bulb

FvM said:
The assumption of voltage independent resistance isn't right for filament lamps.

You need to know the lamps V/I characteristics for an exact solution. I guess it's around 90 W.
Click to expand...

Let me add to this comment. It is difficult to know what incandescent bulbs will do when connected in series. A single bulb at the rated voltage relies on a low initial resistance for enough power to heat the filament and "get started" or produce visible light. As it heats, the resistance increases until a steady state is reached which gives the rated power. Put enough bulbs in series and they will not "start" or get hot enough to produce visible light. Perhaps it can be thought of as a V/I/T characteristic. Not enough increase in T and not much happens (unless you have an IR camera).

Than there is the issue of all bulbs having equal power. Let's take the case of three bulbs of equal rating. If all three bulbs are identical and heat up the same, they may have the same steady state power. However, if one bulb heats up faster, it's resistance is higher than the others and they may not end up with the same steady state power. Bulbs with different ratings are more prone to this problem. There is the physics class problem of connecting a 40W bulb in series with a 100W bulb. The 40W bulb is brighter.
 
Last edited:
Reactions: j33pn

    J

    j33pn

    Points: 2
    Helpful Answer Positive Rating
Re: watt calculation for series bulb

panneerrajan said:
Hi friends,

how to calculate the Total Watts for following connections.

200W 250V Bulb two and 100W 250V Bulb one is connected in series, so what is the total taken Watts? and Please explain How to calculate this?
Click to expand...

I would go out on a limb and say 300W, 200W for Bulb two and 100W for Bulb One

Bulb one will need 0.4 amps and Bulb two 0.8 for total of 1.2 amps times 250v = 300W
 

Re: watt calculation for series bulb

I agree that the answer given in post 2 is not correct, but I suspect it's the answer that's wanted. This looks like a typical classromm exercise, where the student is expected (or told) to assume that light bulbs behave like perfect resistors.

STS4 said:
I would go out on a limb and say 300W, 200W for Bulb two and 100W for Bulb One

Bulb one will need 0.4 amps and Bulb two 0.8 for total of 1.2 amps times 250v = 300W
Click to expand...
That would be true if the bulbs were connected in parralel. However the question is about bulbs connected in series.
 

Re: watt calculation for series bulb

Assume you have magic bulbs with constant resistance and none of the characteristics in #3 and #4 above, determine the resistance for each bulb from it's power rating at the 250 volts. This gives 312 and 625 ohms for a total of 937 ohms (R=V^2/P). 250 volts across 937 ohms is 67 watts (V^2/R). However, the actual total power will be MORE because the power in each bulb is less than the rating. This lower power means the temperature of the filaments will be less. This lower temperature means the resistance of the filaments will be less than at the rating. Lower resistance means more power at the same voltage (again, V^2/R). Hopefully this gives a more intuitive understanding for the 90 watts given in #3 above.

Important! Because of the characteristics given in #3 and #4, you cannot know very well the power of the bulbs in series. Only an approximation or "ball park". More information would be needed about the bulbs (unless they are magic bulbs).

BTW, I think there are two bulbs. I was also a bit confused at first by the description. In any case, all the principles discussed still apply.
 
Last edited:

Re: watt calculation for series bulb


You're right,completely missed that, i though it was a trick question.
In series - 67W as Analog Ground said.
#2 just miscalculated the number of bulbs.
 

Re: watt calculation for series bulb


Dear Sir,

Thank You for your help.
The voltage is AC (50Hz)
 

Re: watt calculation for series bulb

I tried to answer your question. Since I have had to write some equation and I'm not so familiar with the LATEX command, I've generated the attached pdf file.
If you will do some practical measurement I'm interested in knowing if my calculation is correct.

Sorry, I've read now the you have just two bulbs but at my first reading I've understood that you have two 100 W bulbs and one 200 W bulb so the pdf file is relative to this last configuration. In any case it is very easy to redo the calculation in your actual configuration that lead to a total power of 83.5W

However the method is applicable to any number of lamp in series.
 

Attachments

  • Lampade.pdf
    55.6 KB · Views: 160
Last edited:

Re: watt calculation for series bulb


The relationships in your calculations are very interesting. Could you post a link to the application note?
 

Re: watt calculation for series bulb

I also came up with an estimation of around 90W, based on published filament lamp V/I exponents https://en.wikipedia.org/wiki/Filament_lamp#Light_output_and_lifetime

The problem is however, that the formulas are said to be valid only for a "supply voltage V near the rated voltage of the lamp", which clearly isn't the case here. If you think about the physical fundament of lamp I/V characteristic (laws of convective and radiated heat dissipation, temperature dependent resistance of metals) it's clear why the formula changes for a wider range.

I can hardly believe that the excercise problem expects a calculation of I/V characteristics, in so far I agree with the assumption in post #6.
 

Re: watt calculation for series bulb

Yes, you are right.
In the short application of Sylvania (I've only the printed version,) is stated that the equations are valid near the nominal voltage and are averaged on many lamps, however the curves (on the same document) are taken for a variation from 40% to 160% of the rated voltage, so I expect my calculation shouldn't be so far from the reality.
I didn't check the equation and the graph superimpose over all the 40-150% but I'll do it.
 

Re: watt calculation for series bulb

There is a theory out there that bulb lifetime is shortened if it is turned on and off a lot. The theory goes that thermal cycling produces mechanical stresses in the filament. True or false? Any references?
 

Re: what is the watt calculation for a series bulb

I've found the article (is a part of an old catalog) by Osram Sylvania in a digital format. I attached it to this post.
Possibly, in its beginning, it also answer to the question if switching the lamp on and off the lifetime is shortened.
 

Attachments

  • Properties of Tungsten Filament Lamps.pdf
    142.7 KB · Views: 1,624
Reactions: FvM

    FvM

    Points: 2
    Helpful Answer Positive Rating
Re: what is the watt calculation for a series bulb

The said "keep alive" (preheating) technique is generally used in stage lighting systems.
 

Re: what is the watt calculation for a series bulb

Its called "black heat" , keep the bulbs at 70% of their ratings, no light output and a long life when they are switched on and off. Think of all those blinking lights on the front of 1960s main frame computers (long before the invention of LEDs).
Frank
 

Re: what is the watt calculation for a series bulb

panneerrajan said:
Hi friends,

how to calculate the Total Watts for following connections.

200W 250V Bulb two and 100W 250V Bulb one is connected in series, so what is the total taken Watts? and Please explain How to calculate this?
Click to expand...

I am not going to get involved with V-I curves for an incandescent bulb, because I don't have those curves. Therefore, I will treat them as a ideal resistors. The 200 watt bulb has the lowest resistance, call it R. The 100 watt bulb has twice the resistance of the 200 watt bulb, call it 2R. Put them in series for a total resistance of 3R. So if 1R gives a wattage of 200 watts, then 3R will give a total wattage of 200/3 watts, which is the dissipation of both bulbs in series. As an aside, the 100 watt bulb will dissipate twice the power as the 200 watt bulb when they are connected series because the 100 watt bulb has twice the resistance and the same current exists in both bulbs when connected in series.

Ratch
 

Re: what is the watt calculation for a series bulb

I am not going to get involved with V-I curves for an incandescent bulb, because I don't have those curves. Therefore, I will treat them as a ideal resistors.
Click to expand...
Like a politician saying in the budget debate "I don't know what the cost increase is, so assume there's none"?

It may be O.K. for a homework excercise, if ideal resistor behavior is explicitely specified. To avoid conflicting common knowledge, I would chose other technical resistors for the problem. But apparently, the OP doesn't care for the problem, so we must neither. The linear solution has been presented in post #7, problem solved.
 

Status
Not open for further replies.

Similar threads

Menu
Log in
Register
Cookies are required to use this site. You must accept them to continue using the site. Learn more…