Electrical Question From A Student

[Jangee2]

I have a question that I am hoping someone can assist me with. It is a question I thought I got right but received my test back and some parts are incorrect. But - I believe I am right.

115 volt series circuit with 2 light bulbs. 1 bulb is 10 watts and the other is 150 watts.

Questions....
Which bulb is going to be lighted - answer 10 watt - correct
Why is the other bulb not lighted - answer - higher wattage - incorrect
What is gthe voltage across the terminals of the 150 watt lamp - answer 107 volts - incorrect

I have checked myself on a calculator - even an electrical calculator and believe I am right. Any suggestons or comments?

Sure would appreciate it!

 

[nakulak]

150 uses 1.30 A, R is 88.5 ohm
10 uses .087 A, R is 1322

current of bulbs in series is high enough to light 10, not enough current to light 150 W

current is approx .082 of bulbs in series

voltage accross 10 w is approx 108, voltage accross 150 w is approx 7

next time show your math so someone can show you math error (all based on E=IR)

 

[Jangee2]

Thanks

Thanks

Thanks for your help!

 

[LarryFine]

Welcome to the forum!

Why is the other bulb not lighted - answer - higher wattage - incorrect

The correct answer is "lower relative impedance." This is a characteristic of a higher-wattage bulb. The cause is the impedance; the wattage is a result.

In a series circuit (aka a voltage-divider), the relative voltage drop across each element in the circuit is directly proportionate to the relative impedance.

 

[zbang]

BTW, using lightbulbs is a trick, since the actual resistance varies with filament temperature. You have to approach it as nakulak did.

 

[LarryFine]

BTW, using lightbulbs is a trick, since the actual resistance varies with filament temperature. You have to approach it as nakulak did.

I believe he used normal-operating-temperature impedances in his calcs; the actual impedance difference should be even greater.

As you know, cold filaments have much lower impedances, which is why tungsten loads have such high inrush currents.

 

[gar]

091026-2303 EST

I am going to bet the instructor has the wrong answer. The 150 W bulb will have so little voltage drop across it that I can assume its cold resistance will apply and be about 1/10 of its full voltage resistance. Using nakulak's figures the 150 W is about 9 ohms and probably slightly less. A cold, room temperature, 100 W is about 9.2. So make the 150 W = 6 ohms. The voltage across the 150 W in series with the 10 W is about 115*(6/(6+1322) ) = 115*0.0045 = 0.5 V, and therefore the 10 W has 114.5 across it.

Instructors are making a big mistake using tungsten filament light bulbs in problems and assuming constant resistance independent of voltage. This totally misleads students relative to the actual characteristics of these devices. There is no reason the instructors can not just use the word resistor instead of light bulb.

See the photos of lamp inrush current at P1 and P2 on my web site at
http://beta-a2.com/EE-photos.html

.

 

[gar]

091026-2355 EST

Adding to my above comment.

I have a small portable heater I use for a resistive load for experiments. This has a resistance of 8.6 ohms at room temperature and 10.6 ohms hot at 118 V. Dim red glow of Nichrome wire. Provides approximately 12 A load at 120 V.

An Ohmite power resistor I checked recently had a much lower change of resistance because the wire is run much cooler at full rated load and may have a lower temperature coefficient of resistance than copper or Nichrome.

.

 

[charlie b]

The correct answers are
(1) Both, but the 150 watt bulb might not burn brightly enough for the human eye to discern it.
(2) Not applicable, see (1). But the reason the 150 watt bulb's light might not be visible is that its resistance is so low that the voltage drop across it is too small. With 115 volts across it, a 150 watt bulb will burn with the power of, well, of 150 watts. But with only 7 volts across it, a 150 watt bulb won't burn very brightly at all. That said, it still has current flowing through it, so some number of photons, however slight, will be emitted by its filament.
(3) 7. But I gave that away in (2).

 

[gar]

091027-1118 EST

charlie b:

No way is there 7 V on the 150 W bulb, not even close.

Take two bulbs of the wattages specified, connect them in series, and measure the voltage drop across the higher wattage one.

I do not have the specific bulbs specified in the original question, but my previous calculation is probably fairly close for the 150 W bulb's voltage.

I have an 11 W 130 V bulb and a 100 W 120 V bulb and with these in a series combination fed from 123 V I read 0.9 V across the 100 W bulb.

.

 

[charlie b]

I got about the same answers as Nakulak. Here is my math:

In a purely resistive circuit, power equals voltage times current, and also equals voltage squared divided by resistance.

For the bulb rated at 150 watts,
150 = (115)*2 / R150, so R150 = 88.1 ohms

Similarly, for the bulb rated at 10 watts,
10 = (115)*2 / R10, so R10 = 1,322 ohms

Total resistance of the two bulbs equals 88.1 + 1,322, or 1,410 ohms.

Put a total of 1410 ohms across a 115 volt source, and you get 115/1410, or .0816 amps

Run .0816 amps through a resistance of 88.1 ohms, and you get .0816 x 88.1, or 7.19 volts.

Now why do you say that 7 volts was ?not even close??

 

[charlie b]

Here is my math, using the light bulbs from your experiment:

For the bulb rated at 100 watts,
100 = (120)*2 / R100, so R100 = 144 ohms

Similarly, for the bulb rated at 11 watts,
11 = (130)*2 / R11, so R11 = 1,536 ohms

Total resistance of the two bulbs equals 144 + 1,536, or 1,680 ohms.

Put a total of 1680 ohms across a 123 volt source, and you get 123/1680, or .0732 amps

Run .0732 amps through a resistance of 144 ohms, and you get .0732 x 144, or 10.5 volts.

Why the difference, why an order of magnitude difference between the theoretical values and the experimental values? I do not know.

 

[gar]

091027-1156 EST

charlie b:

Run the experiment and you will see.

In my earlier post I gave an estimated resistance for a 150 W bulb with little voltage applied. My estimate was about 6 ohms extrapolated from the measured resistance of a cold 100 W bulb.

Assuming no change in the 10 W bulb because of the very little change in its applied voltage, and this is a valid assumption, whereas the assumption that the 150 W resistance is not different is invalid, my estimate is 6 ohms cold, then 115*(6/(6+1322) ) = 0.52 V.

.

 

[charlie b]

I don't have the facilities to run an experiment. But you have the facilities to check my physics and my math. Any issues there?

Perhaps the difference between cold measurements and hot conditions may be more severe than we think, and that may be what we are seeing.

Also, I do not understand what you are saying about two assumptions, one being valid and the other not, one being about a 10 watt bulb and the other about a 150 watt bulb.

 

[augie47]

I agree with gar in that we would be better off if the instructors didn't use lamps as their example.
This thread, as with all the similar ones, goes well beyond my lil brain's comprehension.
I always pity the student who is looking for a simple answer based on resistance only.
These threads remind me of the following link:
http://www.youtube.com/watch?v=rLDgQg6bq7o&feature=related

 

[gar]

091027-1209 EST

charlie b:

The reason is a positive temperature coefficient of resistance for most metals and over a wide temperature range. Tungsten is very hot in a incandescent bulb and thus between room temperature and normal excitation there is a large change in resistance.

Take a tungsten filament lamp. I used a 250 W clear face reflector flood type IR lamp. At room temperature it reads 4.4 ohms. Put it in the path of another similar lamp and you will see the resistance increase. Go in the sunshine and point to the sun and the resistance will increase very much.

We have a very hazy sky today. I get the same 4.4 ohms pointing to the sky. Pointing at the ground 4.3 ohms, and at the bricks in my breezeway 4.2 ohms. Fairly quickly 5.6 by pointing a halogen work light about 6" away at the IR lamp. After no excitation the IR sensor bulb returns to 4.5 or 4.4 ohms.

I am using a Fluke 27 and resistance resolution is 0.1 ohm.

.

 

[gar]

091027-1247 EST

charlie b:

On the two assumptions.

The 10 W bulb only sees an 0.5 volt change from 115 V to 114.5 V. This does produce a small change in resistance but insignificant for this problem in relationship to the change in resistance that the 150 W bulb sees in going from 115 V to 0.5 V.

So the assumption of constant resistance for the 10 W bulb is valid in this problem, but assuming a constant resistance for the 150 W bulb is not.

When you apply power to a cold incandescent lamp at the peak of the AC voltage, then the instantaneous inrush current is the peak voltage divided by the cold resistance of the lamp. This is what my first photo, P1, previously referenced, illustrates. The 100 W lamp had a cold resistance of about 10 ohms. AC peak voltage is 120*1.414 = 170 V. Thus inrush should be 17 A. My scope measured 16 A. Good correlation. Go to P3 and the steady-state peak is 1.2 A.

.

 

[nakulak]

just a thought:

while it might be reasonable to dispute the the oversimplification of modern bulbs in the problem, the use of a bulb as a resistor does have some accuracy in a historical sense, and IMO it is indisputably a mind opening teaching aid to beginner students for the first step at understanding circuits which they have more likely than not been exposed to since childhood. It is no more innacurate to teaching middle schoolers newtonian physics, only for them to learn later that gravity is not constant, time dilates, atoms aren't neat little sphere's with spherical electron orbits, etc. I appreciate the astute insight that was provided by Gar, and the technical accuracy, but I think that in the context of the question, the answers that the teacher was looking for (and I think we know what they are) is what is most helpful for the student that was looking for it, not only in the sense that learning how to find the answer was part of the lesson, but also in the sense that it is especially difficult for some students to understand that in order to pass tests it is not always important to find the correct answer to the question - to pass the test it is necessary to find the answer that the tester is looking for. These two answers are often not the same, and while it may be intellectualy satisfying to find, and know that you know, the correct answer to a given question, it might not provide you with the test results that you would like. This is, IMO, an especially important tidbit of information that is extremely helpful to impart to individuals who may eventually need that information to put themselves on the winning side of a test.

(regardless, it would be interesting to see a plot of Gar's bulbs over the time period it takes for the bulbs to heat up, since i don't doubt that he is correct)

 

[gar]

091027-2054 EST

nakulak and others:

In all tests if you do not understand the question, make assumptions, include them in your answer, and show the details of your work.

My complaint with the original question is that it is blatantly obvious to anyone with knowledge of the characteristics of a tungsten filament lamp that its cold to normal operating point resistance has a very large change of resistance, about one order of magnitude, and therefore this question should not use a lamp bulb as a fixed resistor. The question should have said a resistor that dissipates 150 W at 115 V and a resistor that dissipates 10 W at 115 V.

I do think it is appropriate to use the lamp bulbs in the question if one wants to illustrate what does occur in reality. In this case the students should have been previously taught about this variation in resistance. It is quite obvious that many electricians lack knowledge about this resistance variation and why it causes a large inrush current when the lamp is turned on. I also think that the instructor that wrote the question was sloppy in writing the question, or the instructor is unaware of this variation.

If you want to see how the resistance varies with time use a low impedance DC source and plot current vs voltage on an oscilloscope. Or plot current vs time. The time constant of the curve will be a function of the mass of the filament. A small pilot lamp has a short time constant and a large lamp a long time constant.

Another experiment is to use a Variac to adjust the voltage to the lamp and take data at various voltage values.

There are a lot of questions asked on this web site that could be answered by the person asking the question if they would simply run some simple experiments.

My photo P3 of inrush current provides some idea of this warm-up variation of resistance. It looks like the time constant of a 100 W bulb is less than 1/120 second. The peaks of the half cycles give you some idea of what you would see with a DC source. Note: one time constant will have about a 63% change. When a time variation is looked at you see the rise of resistance correlating with the increase in the filament temperature.

If you look at my plot shown at P9 and reverse calculate from the voltage and power to resistance some of the points are 120 V 192 ohms, 90 V 162 ohms, 60 V 133 ohms. These are eyeball approximations. Also note the Edison carbon filament lamp, P11, has a negative coefficient of resistance with respect to temperature.

.

 

[dkarst]

For what it's worth, Gar's estimate is very good. I just took a quick measurement and my cold resistance value of an ordinary 150 W bulb (GE) was 7 ohms. I used my old ohm meter so the current through the bulb was very low although it would take a little digging to calculate the value.