Circuit Calculations Mismatched

Yes, at room temperature. When the filament gets white-hot, its resistance is significantly lower.

--
St. John

D'oh! HIGHER! I meant higher.

--
St. John

Not only that, but the filament is a COIL, which has inductive reactance to AC. Your multimeter uses DC to measure resistance. Not the same. Trust the calculations. They tell the truth.

Dave snipped-for-privacy@hotmail.com

Thanks for the help guys.

Another inquiry. I'm confused about something. Say I have 2 watts resistor. If I plug this directly to the ac outlet with voltage of 110 volts. What would happen.

What does this mean. If I plug the resistor directly into the socket. Does it mean it will only absort 0.0182 Ampere from the power line.

How do you tie it to V=IR?

What I want to calculate is the wattage that the resistor must be in order that it won't explode. Also what must be the resistance?

Last. Are there variable resistor that I can use in the power line?

This is just for theoretical understanding and no specific applicationin mind. Tnx for the assist.

Manny

No, no, no. The 2 Watts is a maximum power rating which should not be exceeded. There is no fixed power which a resistor will operate at, which is how you are treating this spec. It will operate at 2 W only at a certain voltage (& current), not necessarily 110 V.

You need to know what R is to begin with.

The resistor will fail (not necessarily explde) at 2 W, or somewhat above this if a safety margin has been used in the spec.

--
That seems awfully high.

--- What calculations???

It's essentially an air-wound coil with an inductance of less than a microhenry, so the difference between the resistance and the impedance of the lamp due to the inductive reactance at 60Hz is going to be so insignificant that it can be ignored.

Even if it was a microhenry, we'd have, for the reactance:

Xl = 2pi f L = 6.28 * 60 * 1E-6 ~ 3.77E-4 ohms

and for a filament with a measured DC resistance of 0.1 ohm, an impedance of:

Z = sqrt (R² + Xl²) = sqrt (0.1² + 3.77E-4²) = 0.1000007 ohms

-- John Fields Professional Circuit Designer

That depends on the value of the resistance R which you don't mention.

Well once you have decided on R try inserting it in I=V/R.

The value of the resistor R will depend on what you are trying to achieve.

Lets try an example.

Suppose we play with 9V DC as it's safer...

Late say you have a 1W light bulb rated for 6V that you would like to connect to the 9V supply. What you need is a resister in series with the lamp to drop the voltage from 9V down to 6V (eg by 3V). Here is how you calculate the value and power rating for the resistor....

First calculate how much current the lamp will draw at 6V....

Power (W) = Current (I) * Voltage (V) so...

I = W/V

Now W=1 and V=6 so I=0.1667 Amps.

Now in the new circuit on 9V with a dropping resistor.... We want the Resistor (R) to drop 3 Volts (V) when the same current (I) is flowing through it and the lamp...

V=I*R or R=V/I

V = 3 I = 0.1667 so

R = 18 Ohms approx.

Now what power rating does the resitor need to have...

The power (Wr) burnt in the resistor is given by

Wr = V*I

V = 3 I = 0.1667 so

Wr = 0.5 Watts

Best use a 1W resistor because a 0.5 watt resistor would be operating at it's limit and that needs good airflow.

For tungsten, the cold resistance is roughly 1/15th of the operating hot resistance.

1.44 ohm / 15 is close to 0.1 ohm. So the 0.7 ohm measurement is actually too high for a room temperature resistance. Weird.

Mark

I didn't subtract the resistances in my multi-meter leads. Is there a resistance there. But when I short it, there reading says 0 ohms. I just directly measure the positive and negative lead of the 12 volts halogen lamp and it gives out 0.7 ohms. Where did I get it wrong.

About tungsten. There is something I wanna know. Thick tungsten has low resistance while thin tungsten has high resistances. This is regarding the thickness per section. If I make the thick tungsten longer in length, can it increase the resistance?? Or is thickness the main factor in resistances considerations?

Also suppose I want a certain resistance. How do I know what optimum thickness of the tungsten to use with good allowance for heat dissipation. If I use too thin, there may be so much heat and it may burn. If I use too thick, I may have to make the length longer to increase the resistance. How do you know what thickness and length is the optimum (is length a factor at all or is it just the thickness?). This is just for the sake of theoretical understanding and I'm not building bulbs ok.

Thanks.

Manny

Manny wrote:

It's puzzling to me too. There might be a film or oxide layer on the lamp leads. You could try sanding a small area on each lead, until a patch of shiny metal is showing, and then remeasure.

Yes, length is a factor too. Twice the length gives twice the resistance. More generally:

R = rho x length / area

where

R = resistance of the wire area = CROSS-SECTIONAL area of the wire rho = the resistivity of the material (which changes with temperature)

That requires a lengthy reply ... :

To design a tungsten lamp of a certain wattage, you need to know what voltage will be used to power it (eg., 12V for automotive, 115V for U.S. households, etc.)

The power and voltage determine what the current should be: I = P/V And the voltage and current determine what the resistance should be from Ohm's law.

Once you know what the resistance R should be, be aware that there are infinitely many combinations of wire thickness and length that give the same resistance. A thicker wire may be made longer in order to maintain the same resistance.

One more design parameter is the temperature of the wire. If it's too hot, the bulb won't last very long. If it's too cold, most of the "light" coming out is actually unusable infrared radiation, not visible to the human eye. The best compromise seems to be to run somewhere between

2200 and 2800 Celsius (those numbers from my memory when I use to work for a lamp-making company).

Once the lamp designer has picked an operating temperature, they can specify what the total SURFACE area of the wire should be (to be distinguished from the CROSS-SECTIONAL area mentioned earlier). Basically, the temperature is determined by the total power divided by the total surface area. This power-per-surface area is related to temperature by the well- known (to some) blackbody radiation equation, which I won't get into here.

In short, choosing the power and temperature will then fix the surface area to be a certain value.

Of all the thickness & length combinations that produce the required resistance, only one combination will also give the correct surface area. So the wire length and thickness are uniquely determined, once you have specified the voltage, wattage, and operating temperature.

Mark

--
John Fields
Professional Circuit Designer

Be careful, John. A few years ago, in the course of some work, I had occasion to measure the cold resistance of several hundred 100W halogen lamps. About half of them actually appeared open circuit on a low voltage DMM, (300mV open circuit) but nonetheless lit normally at rated voltage.

Looking at samples using a curve tracer proved interesting. Most exhibited rectifying properties in both directions, with a barrier potential of about half a volt, which turned out to be due to the method of fixing the filaments to the seal wires using a crimp technique, rather than spot welding.

--
"Electricity is of two kinds, positive and negative. The difference
is, I presume, that one comes a little more expensive, but is more
durable; the other is a cheaper thing, but the moths get into it."
                                             (Stephen Leacock)

Thanks for posting that Fred. Nice example of how the real world is full of surprises.

Not quite what the plant's ATM guy said at the time :-)

-- "Electricity is of two kinds, positive and negative. The difference is, I presume, that one comes a little more expensive, but is more durable; the other is a cheaper thing, but the moths get into it." (Stephen Leacock)