# QOTD: Different Lamp Wattages in Series



## Footer (Sep 4, 2012)

QOTD care of JChenault. 

Last night I had to replace a lamp in a zip strip. For those of you who do not know, a zip strip connects 10 MR-16 12 Volt lamps in series across 120 volts.

The lamps in the unit were 75 watt lamps - but I only had a 35 watt lamp available so I put in in instead.

What happened and why? Full credit only if you show the math.

*As usual, standard QOTD rules apply. Students only for the first 7 days. *


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## Chris15 (Sep 5, 2012)

Note: This post was hidden for the student only period
The worked solution...

Let us assume for now that the lamps have a fixed resistance and neglect the complex parts of their impedance.

If we consider a 12V, 75W lamp, then by P = V^2/R, R = 1.92 ohms.
If we also consider a 12V, 35W lamp, then by the same maths, R = 4.11 ohms.

For a series connected string of 9 75W and 1 35W lamps, the total resistance = 9*1.92 + 4.11 = 21.39 ohms.

As they are connected in series, they must have the same current flowing through them.
That current can be given by Ohm's law as V= IR.
So I = 5.609A

This is less than the case of 10 75W lamps, where I = 6.25A, so at first approach, the whole string will be dimmer.

But that does not tell the full story. We need to examine what happens with the 35W lamp and what happens with each 75W lamp.

In the case of the 75W lamp, we multiply the 5.609A current by the 1.92 ohm resistance to obtain the voltage across the lamp. This gives us 10.77V. P=VI, so the power being dissipated by each "75W" lamp will be 60W. This matches what we predicted about the string being dimmer.

What is more interesting is to look at the case of the 35W lamp. Using Ohm's law again, V= IR = 5.609*4.11 = 23.08V. P=VI gives us 129W being dissipated by the "35W" lamp.
So this means that the 35W lamp is running at almost double its rated voltage. This is a problem...
If we consider the lamp formulas here: http://www.controlbooth.com/wiki/Collaborative+Articles:Mathematical+Formulas+for+Lighting, and plugging the numbers in, we find that the lumen output of the lamp will be 913% of the design, and the colour temperate increases by 31%.
But look at the lamp life. It's now a mere 0.02% of the design life.

So far we have only considered the steady state circuit, but the reality is that all lamps take a little bit of time to warm up, during which their resistance is much lower than when they warm up.
This thermal shock will mean that it's a certain bet that the moment this string of lamps is powered up, the 35W lamp will blow. It may fail catastrophically and throw glass everywhere. Or it may short circuit and increase the voltage through the remaining lamps. Or it may just be boring and blow boringly.

For the sake of completeness, the 75W lamps will now have 408% lamp life, 69% lumens output and lose 4.5% of their colour temperature.

While we see that practically you have a blown lamp and that's what matters, as an academic exercise, one should go back and reiterate the maths, because the resistance of a lamp changes depending on it's applied voltage (mostly because of heat dissipation). So the resistance of the 75W lamps in this case would go down, 1.76 ohms would be the second iteration of the numbers and the 35W lamp resistance will go up, 6.94 ohms would be the second iteration value. At this point, the voltage across the 35w lamp is 36.5 volts... By the 3rd iteration it's up to 50V, and you can see how it would continue...

One should also consider the inductance of the lamp filament and this too will increase the impedance, especially at the switch on peak, but I think we've seen enough to know the results...

So a long explanation of a very simple blown lamp


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## DuckJordan (Sep 5, 2012)

35w at 12v75w at 12 75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12  75w at 12 710w at 120v750w at 120v
The rest of the lamps were dimmer, seeing a lower wattage along the line (more resistance from the 35 watt lamp).


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## silicsound (Sep 6, 2012)

75 watt bulb = 1.92Ω (12²/75 = 1.92)
35 watt bulb = 4.11Ω (12²/35 = 4.11)

So a strand of 10 75w bulbs would have 19.2Ω (1.92*10=19.2)
And 19.2Ω at 120v consumes 750watts (120²/19.2=*750*)

The strand with 9 75w bulbs and 1 35 watt bulb would have 21.39Ω (1.92*9+4.11=21.39)
And 21.39Ω at 120v consumes 673watts (120²/21.39=*673*)


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## Chris15 (Sep 6, 2012)

I'm sorry Duck, but your maths is not right.

silicsound, you're on the right track, but what do these numbers practically mean?

My answer will become public once the week is up, but in this situation there will be something very specific that physically happens...


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## silicsound (Sep 6, 2012)

The strand with all 75w bulbs runs at 750w at 120v so it's current is 6.25A (750/120=6.25) [P/V=I]
The strand with the 35w bulb in it runs at 673w at 120v so it's current is 5.6A (673/120=5.6)

Since all the lights on the strand share the same amount of current, we can find how much energy each light dissipates.
Each light on the 75w strand would consume 75w each. (6.25²*1.92=75) [I²*R=P]

On the strand with the 35w bulb, the 75 watt bulbs would each consume 60w (5.6²*1.92=60)
while the 35w bulb would consume 129w (5.6²*4.11=129) 

So in other words the 35w bulb would glow two times brighter than the 75w bulb or it would break, catch on fire, explode or create a black hole because its consuming four times its rated wattage.


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## Chris15 (Sep 6, 2012)

silicsound said:


> So in other words the 35w bulb would glow two times brighter than the 75w bulb



Are you sure?
Mathematical Formulas for Lighting - ControlBooth


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## silicsound (Sep 6, 2012)

Chris15 said:


> Are you sure?
> Mathematical Formulas for Lighting - ControlBooth



I'm guessing that I would find the voltage drop across each lamp and then use volts^3.4
So the 75w bulb would have a drop across it of 10.75v (5.6*1.92=10.75) [A*I=V]
and the 35w bulb would have a drop of 23.02 (5.6*4.11=23.02)
the w35 bulb emits 42771 lumens (23.02^3.4)
the w75 bulb emits 3212 lumens (10.75^3.4)
So it would be 13 times brighter (42771/3212=13.3)
Thanks for the link, I'll have to save those lamp equations.


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## MNicolai (Sep 6, 2012)

Footer, I'm getting the impression you'd make a good question supplier for the ETCP Entertainment Electrician exam.


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## Footer (Sep 6, 2012)

MNicolai said:


> Footer, I'm getting the impression you'd make a good question supplier for the ETCP Entertainment Electrician exam.



I did not come up with the question. 

Sent from my Galaxy Nexus using Tapatalk 2


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## JChenault (Sep 6, 2012)

silicsound said:


> The strand with all 75w bulbs runs at 750w at 120v so it's current is 6.25A (750/120=6.25) [P/V=I]
> The strand with the 35w bulb in it runs at 673w at 120v so it's current is 5.6A (673/120=5.6)
> 
> Since all the lights on the strand share the same amount of current, we can find how much energy each light dissipates.
> ...



Just to be clear. There is only one circuit ( strand ). It has nine 75 watt lamps and one 35 watt lamp

Never mind- I reread your post more closely


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## MNicolai (Sep 6, 2012)

Footer said:


> I did not come up with the question.
> 
> Sent from my Galaxy Nexus using Tapatalk 2



Credit to JChenault. Reminds me of an ETCP exam sample question.


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## Lambda (Sep 8, 2012)

A higher wattage lamp consumes more current, and therefore has a lower resistance. A lower wattage lamp would have higher resistance since it consumes less current. Inserting a higher-resistance load into a series circuit will increase the resistance of the whole circuit. Therefore, less current will flow, and the result of that is that the entire string will have a reduced brightness. 
Or, at least I think. I didn't bother to do the exact math, I'm just generalizing here.


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## Chris15 (Sep 12, 2012)

So the standard seven days have now passed, anyone else want to have a crack before I post a worked solution?


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## epimetheus (Sep 12, 2012)

Chris15 said:


> So the standard seven days have now passed, anyone else want to have a crack before I post a worked solution?



Yes, without posting the detailed math, the entire string draws a slightly lower current due to the increased resistance of the 35W bulb. So the 75W bulbs will be dimmer than normal. The 35W bulb however will have an applied voltage almost double its nominal voltage. In other words, it will go supernova and blow as soon as the strip is powered up.


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## Chris15 (Sep 12, 2012)

Yeah - the 35W blows on power up.

See post #2 for the worked mathematical solution...


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## JChenault (Sep 13, 2012)

Chris15 said:


> Yeah - the 35W blows on power up.
> 
> See post #2 for the worked mathematical solution...



Sort of.

If you run the circuit up on dimmer, you note that at a low intensity, the 35W lamp starts glowing, and is brighter than the 75W lamps. At some point in the dimming curve, the voltage across the 35 W lamp exceeds 120 Volts by enough to make the lamp immediately fail.


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## Chris15 (Sep 13, 2012)

So I went to examine the maths and see if I could estimate the point at which the voltage across the 35W lamp would exceed 12V.
In doing this, I've discovered an error in the formula I was using for resistance, so until I work out what's going on, my maths posted above should be considered questionable...

As an aside, this is a really good reason to run a dimmer up slowly after you've changed something


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## DELO72 (Sep 21, 2012)

EGAD people! What's with all this math?? If I had wanted to do math I would have been an Engineer and had a $150,000 job out of college. I was a THEATRE MAJOR. Although...this might also explain why I blew up 3 ACLs my first year in Grad School prior to the M.E. running into the room and stopping me, pointing out to the "Bad PAR64 lamps I was testing" were actually "GOOD 12V. ACL lamps that I was frying." Live and learn... ;-)


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## jonliles (Sep 25, 2012)

DELO72 said:


> ? If I had wanted to do math I would have been an Engineer and had a $150,000 job out of college



If only it worked like that. As an engineer with 15 years (electrical power) industry experience I am no where near 150K. Granted, I do make considerably more than the majority of theatre majors fresh out of college. 

I like the math. it keeps us thinking and helps us to undertsand why things happen.


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## epimetheus (Sep 25, 2012)

jonliles said:


> If only it worked like that. As an engineer with 15 years (electrical power) industry experience I am no where near 150K. Granted, I do make considerably more than the majority of theatre majors fresh out of college.
> 
> I like the math. it keeps us thinking and helps us to undertsand why things happen.



Yeah, maybe remove that 1 in $150,000 and you're a lot closer to reality.


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## JonCarter (Sep 25, 2012)

This thread reminds me of a picture we were shooting in a steel mill. Our only power was 250 VDC taken from the crane rails, and everything had to be seriesed. This was OK for the arcs; brute with brute, 150 with 150, etc., but somebody on crew decided he could series 2 5 kW Moles with a tener. Should work, right, 10 kW in series with 10 kW? Well, yes, if they're all turned on. Ever seen a 5 kW go off like a flashbulb? Most impressive!


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## David Ashton (Jan 28, 2013)

As the resistance of the lamps changes by a factor of 10 from cold to rated voltage then to calculate the blow point would need calculus maths and a good deal more information, lamps also increase their resistance with age, so this question is practically unanswerable.


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