Total Lamp Power

That is a home work problem if I ever saw one. Why don't you try to work it out and give your answer.

W = I x E W = I²R I = E/R. Now solve the problem

please solve it for me

What level of school are you in now? College, Training program ?

You have already been told - there is no simple answer - qualified engineer or not.

Be warned that the answer you calculate won't be the same as the result you'd get actually measuring the current & multiplying by the voltage.

With the lamps in series the filaments will be underrun & cooler than normal, so their resistance will be a bit lower.

The first part of the question isn't solveable by analysis as neither lamp will be operating at it's rated voltage when connected in series, they may well run too cool and therefore stay low resistance and blow the fuse. It is bad practice and a stupid thing to do/try/ask.

The second part just requires some understanding and arithmetic.

Here is a more sensible question.
If a squirrel runs at 12mph, a cat at 30 mph and a wombat at 15mph how fast will they run when tied together?

Actually the first part is useful if you mearure it practically as it will show how theory and practice are not always the same.

Del

Actually, the answer differs, depending on whether this is a homework problem, or whether this is a question about the real world...

For two resistances R₁ and R₂ in series, total resistance R = R₁ + R₂;

in parallel, 1/R = 1/R₁ + 1/R₂. I hope this helps without giving away too much.

Ok in theory, but lamp filaments are highly temperature dependant resistors in the real world.

Agreed. In series the lamps will run dim (= cool) and will have less than rated resistance. But once the resistance-at-temperature is known, the formulas will be correct. As others have noticed, this looks like homework. I wonder if the poser of the question has taken this into account.

BTW, incandescent lighting contactors do take into account the inrush current into lamps at the low-resistance starting condition.

Dear Mr. Ani.sa.o908,

For your problem, if the 100 watts bulb and 50 watts bulb is connected in series with 230 volts supply, the total resistance will be 1587 Ohms.

The Voltage drop across the 50 watt bulb will be 76.66 volts and the voltage drop across the 100 watts will be 153.33, current drawn in series circuit will be 0.1449 Amps.

The 100 watts bulb will give a power of 44.44 watts and the 50 watts bulb will give a power of 5.545 Watts, totalling to 50 watts.

I hope your question is answered.

DHAYANANDHAN.S,

INDIA.

quote DHAYANANDHAN.S "For your problem, if the 100 watts bulb and 50 watts bulb is connected in series with 230 volts supply, the total resistance will be 1587 Ohms. The Voltage drop across the 50 watt bulb will be 76.66 volts and the voltage drop across the 100 watts will be 153.33, current drawn in series circuit will be 0.1449 Amps.The 100 watts bulb will give a power of 44.44 watts and the 50 watts bulb will give a power of 5.545 Watts, totalling to 50 watts."

You need to check your figures again. R50 = 1033 ohms R100 = 540 ohms

Series amperage = .145 amps watt for 50 = I²R = 0.145 x 0.145 x 1033 = 22 w

Watt for 100 = .145 x .145 x 540 = 11 watts.

The lower wattage 50 will burn brighter than the high wattage 100. This assuming that the resistance is taken at full power. Actually the reisistance will be a lower figure.

YOU ARE RIGHT Mr. WAREAGLE, as regards the COMBINED POWER OF 100 WATTS and 50 WATTS Bulb in SERIES is concerned.

But as regards, the INDIVIDUAL resistances for 100 watts and 50 watts bulb, based on 230 volts, my calculation is right. the reason being POWER = VXI = VXV/R. i.e 100=230x230/R = 529 Ohms.

In the same way for 50 wattts bulb 50 = 230x230/R = 1058 ohms. is right. Total Resistance = 1058+529 = 1587 ohms.

Therefore the current in the series connection for the 100 watts + 50 watts bulbs = 230/1587 = 0.1449 Amps. Up to this point my calculation is correct.

But for the Power in series connection I made a mistake and you are right. For indivdual bulb the power is IxI/R and here my mistake is that I have interchanged the resistance for the bulbs and for 50 watts bulb the power will be 22.21 watts and for the 100 watts bulb the power will be 11.10 watts.

Thanks for your correction and there should be no error to be accepted, the REASON FOR THE ERROR IS INTER CHANGING OF RESISTANCES. The mistake is regretted.

THANKS,

DHAYANANDHAN.S,

INDIA

Mr DHAYANANDHAN.S

We should be getting the same answer. I did it a little differently.

I x V = W I x 230 = 100 I = 100/230 = 0.42 amps I²R = W 0.42²xR = 100 R=566 ohm

I x V = W I x 230 = 50 I = 50/230 = 0.22 amps I²R = W 0.22² x R = 50 R = 1033 ohm

Total R = 566 + 1033 = 1599. I think the difference is in rounding off of the amps.

I rounded the amps and you did not. Should have the same answer.

I dont have all the time to solve this simple mathematics for u now,but i will give u all the formulars needed to get yr theoritical answers right.

V=IR; P=IV; P=IxIxR; P=VxV/R; R in series Rt=R1+R2; R in parallel 1/Rt=1/R1+1/R2 or Rt=R1xR2/R1+R2. P-Power, V-Voltage, I-Current, R-Resistance.

If u do yr calculation correctly,u will find from yr result that the total power of yr lamps above is much more smaller in series connection than in parallel connection.Aso the 230v will be shared between the two lamps in series(eg.say 100w-70v;50w-160v),while same current runs thru them both.In parallel,both lamps will hv 230v across them,while each lamp will draw different current value.The 100w will draw more current than the 50w.

In practice,if u do the connection in series,the 100w lamp will be dim,while the 50w lamp will glow more(much brighter than 100w),bcos 100w will drop more voltage than the 50w.

If u do the connection in parallel,both lamps will glow fully at their rated power.The 100w will glow more and will draw more current,while the 50w will also glow accordingly and draw its current which will be half or almost half that of 100w.

Thoeritically,100w will give-O.43Amps,while 50w will give-O.22Amps,but in practice,u may get different value for 100w and the value of 50w may vary from O.22Amps,but they would not be far from your theoritical values.I can assure you that the total power u will get in practice when connected in parallel will be up to 3 times or more greater than total power in series connection.Use yr clamp on ammeter to read off yr current and measure yr voltage both for series and parallel connection,then do yr calculation,u will understand more and appreciate the answers given to u so far.

Patrick Whowha

Hi all,

This is totally insoluble by calculation. He doesn't even say what kind of lamp. As already said, he could be talking about CFLs (which probably won't light at all), halogens, ordinary incandescent or even LEDs!

It has been said above that when cool the resistance is a little lower. It is not a little lower but a lot lower. I had two halogens to hand which gave the following;-

50W/ 230V = 102Ω actual value, calculated value 1,058Ω

500W/ 230V = 8.2Ω actual value, calculated value 106Ω

As can be seen the cold resistance is approximately 10% of the hot value. Normal incandescent lamps do not have such a large difference as they run cooler.

I have often put two lamps in series to form a temporary high power dropper resistor. I have never noticed the series resistance being lower than a single bulb. If I get a moment tomorrow I will run a test on the work bench. I might even get the apprentice to plot the curve for a typical 50W halogen. I will report tomorrow.

The slope is not at all linear, If I recollect correctly when V is plotted against I, the resistance rises rapidly as the element heats .

Regards

Chas

The answer is here

BTW: I am surprised that why some one has not reported a homework question.

And more surprised that the esteemed members are solving it for the OP. In the end are you doing any good for him?

No. And not for the goodness of the whole system

In the first couple of posts enough hints were given. And then the flood gates opened.

And yet, so far, there are various transpositions and imprecisions in the given answers; even when omitting the lower resistance of (incandescent) lamps running at part voltage. (The problem was pretty good as is, without taking that wrinkle into account.)

Here's power vs lamp current for a 12V 5W festoon incandescent (did this some years ago):

[Edit: 12V 5W is nominal; it's a car lamp, so probably rated at 14V.]

The total resultant power with both the lamps connected series will be around 33.3W.

Use W=(v)2/R (V square /Resistance) formula or R = (v)2/W. Calculate resistance of both the lamps and add resistance of both the bulbs and calculate power @ 230V.

Use above formula for parallel connected lamp.....Simple work........

Hemant

The total power of both lamps will be equal to or less than the applied power.

--and what is applied power?!

Here is the resistance characteristic of a 100W 240V incandescent lamp, as measured with the aid of a variac, clamp meter and voltmeter :

The resistance varies over the full voltage range, from a low of ~100Ω up to just short of 600Ω at 240V.