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# I have purchased four 100W (120V rated) incandescent light bulbs..?

The resistance of these incandescent light bulbs was 8.90 ohms on the average (+/- 0.251 ohms if you want to split some hair):

a) How much current will the bulb consume on the average if I plug it in a 120 V circuit?

b) How much current will the bulb consume in a 12.0V circuit?

(Values are in RMS)

How do I do this problem?

Wiki eloquently states “Incandescence is the release of electromagnetic radiation, usually visible radiation, from a body due to its temperature.” This is the key to the question asked. It is not evident right the way however some of you were quick to point out that 8.90 ohms does not work well in P=V^2/R. R (@120V, 100W)=144 ohms, a 1:16 ratio between cold and hot filament [3]. The question is what makes the resistance rise and remain at that fixed level in our case that being 144 ohms. We have to consider black body (BB) radiation P=ebA(T^4 – Tc^4)

P - Power radiated

e- BB emissivity( 0<e<1)

b- Boltzmann’s constant = 5.670 400(40)×10−8 W/(m^2•K^4).

A – area trough which BB radiates

T- temperature of BB

Tc – temperature of the environment BB radiates

References

1. About incandescent light bulb http://physicsed.buffalostate.edu/pubs/TPT/TPTDec9...

2. The physics of an incandescent bulb http://www.sci-ed-ga.org/modules/materialscience/l...

3. Linear relationship between temperature and resistance http://www.sci-ed-ga.org/modules/materialscience/l...

4. Key black body radiation and ... http://www.millersville.edu/~physics/exp.of.the.mo...

### 10 Answers

- Anonymous1 decade agoFavorite Answer
use ohm's law

E = IR

E = volts

I = current

R = resistance

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- 1 decade ago
You might have already noticed that the measured resistance of your light bulbs is a lot lower than the 144 ohms expected from the voltage and wattage rating. I was puzzled at first when I discovered the same thing many years ago.

The answer to this paradox is that the resistance of the light bulbs at low voltage has been accurately measured but as the filament heats up, its resistance increases, in this case by a factor of about 16 between room and operating temperature of about 3,000 deg K. When you first turn on the light, it does indeed draw a large surge of current that quickly decreases to the operating current after a fraction of a second. This is why switches sometimes have a separate, lower current rating for operating incandescent lights.

The answer to your problem is therefore this:

a) at nominal rated voltage, your bulb will consume the nominal rated power; the resistance measurements of the filament at room temperature don't apply. The RMS current can therefore be computed from the equation

I = P / V, for a current of 0.83 amps.

b) At 12.0 volts, the filament will dissipate enough power to heat up significantly and raise the resistance noticeably but nothing like the factor of 16 that occurs at full operating voltage. The current will therefore be less than 12.0 V / 8.9 Ω

or 1.35 amps but a lot more than the 0.083 amps you would expect if the light offered the same resistance it did at full working voltage. My seat-of-the-pants estimate would be a current draw in the range 0.2-0.4 amps based on my somewhat limited experience with incandescents.

Do try this at home.

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- NumbatLv 61 decade ago
A simple way is to use the formula I=P/V (transposed from P=I*E). So 100/120=0.833 Amperes (per bulb). But that would ignore your figures. So using your figure of 8.9 ohms and the formula I=E/R, 120/8.9=13.483 Amperes. Also, for 12 Volts, 12/8.9=1.348 Amperes.

Why the anomaly? In real life, the resistance is not constant with temperature. With metal filaments the resistance rises with temperature. Hence the difference.

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- TaraLv 44 years ago
Yes. adding up all the power gives 3450 W total. using P=VI we find, 3450 = 120I I = 28.75 A when the current rating of a fuse is exceeded it breaks the circuit. The total current of the circuit exceeds 20 A and the fuse will blow.

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- Brian TLv 61 decade ago
First, please understand the difference between hot and cold resistance. To get the current used, your resistance has to be the hot resistance. Voltage in RMS/hot resistance = current. Second. The bulb will give a different hot resistance when activated by 12 volts. You must know this resistance because the tungsten filament is a non linear resistance and the 12 hot will be closer to the cold resistance.

Source(s): Common problem in electronics with driver circuits feeding bulbs. Must know the surge(starting current of cold bulb) to calculate the rating of the driver parts. Many early circuits biased the bulb drivers with a low value to avoid the surge current and lengthen the life of the bulbs. (Most burn-outs occur during starting) Buying 130 volt rated bulbs avoids this by providing a slightly higher resistance when cold and reducing the surge.- Login to reply the answers

- Anonymous1 decade ago
Of Course Raffaele1111 is right!

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- AlexanderLv 61 decade ago
(a)

Current = Power/Voltage = 0.83 Ampere

(b)

The resistor is metallic tungsten. Its rsistance depends on temperature.

http://hypertextbook.com/facts/2004/DeannaStewart....

So we need to estimate temperature of the whicker, when the bulb is plugged into 12V circuit. Lets assume for the lack of better information that all loss of heat is due to black body radiation.

Resistance of the whisker under normal operating conditions is

R(T_norm) = U²/P = 144 Ω

Which means that

R(T_norm)/R(T_room) = 16.2

Assuming that room temperature is T_room = 300K, high resistance corresponds to T_norm = 3000K according to the table.

Now we know power radited by the whisker:

P(T) = (T/T_norm)^4 x P_norm

So we need to solve equation

P(U) = U²/R(T) = (T/T_norm)^4 x P_norm

Lets make simplifying approximation of linearity of R(T):

R(T) = αT

Then

U²/(αT)= (T/T_norm)^4 x P_norm

U² = αT³ /T_norm^4 x P_norm=

T = ³√[T_norm^4/α x U²/P_norm]

R(U) = α ³√[T_norm^4/α x U²/P_norm]

I(U) = U/ ³√[T_norm^4 α² U²/P_norm]

Finally we see that power is proportional to U^(1/3). So the answer is

Current at voltage U is about I_norm x ³√(U/Unorm) ~ 0.4 A

>How do I do this problem?

Use ammeter

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- 1 decade ago
Edward,

Great question--I'd give it two stars if I could--and multiple good answers. Nothing more to add.

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- Yahoo!Lv 51 decade ago
a) R=P/I^2

R=8.9/144000

R=6.18 ohms

b) 0.061 ohms

I'm not sure it would be the right way to do it.

Edit:

Oops, what a silly answer, you are looking for current...Current!

Oh well, at least now I am sure this is not the right way to do it.

I would like to thank the kind people whom gave a blind thumb up to this totally ridiculous answer...I suspect the members of the chocolate cult.

Source(s): 110 1101 110 1001 111 0011 111 0011 111 0101- Login to reply the answers

- 1 decade ago
Wow!!!!!!

I am amazed at all the formulas

The only and proper formula is as follow hereto

V/W=A

V=volts

W=watts

A=amperes

With amperes been the Amp per hour used.

Then one needs to

A*X*Y=$

A=amperes

X=power per hour charges

Y=total hours used per usage

$=total costs per usage

Source(s): My Own Studies/Qualifications & Practical Experience as Electrician and Electrical Contractor- Login to reply the answers