The Hot Resistance Of the filament

Sullitp

Hey so im stuck on a 2part question, it goes as follows:

A tungstan filament in a light bulb is rated 75w for a 230v supply. The resistance of the filament at 20*C (*C means degrees celcius) is 68 ohm.
Assume a constant temperature coefficient of 0.005.

Calculate:
A) The hot resistance of the filament

Ok so I have no idea what temperature to use for "hot" so I used 100*C (I mean thats hot! )

Then used the equation R1=R0 (1+0.005x100) - Where
R1= Resistance at 100*C
R0= Resistance at 20*C
0.005= Temperature Coefficient
100= 100*C "Hot" temperature.

If you work this out it gives you a resistance of 102ohm @ 100*C.
I think that bit is right but is the 100*C right?

I have a second part to this so I'll put it in a different colour to help....

B) The operating temperature of the filament?

Well for this one I was going to use the formula (T1= R1/R0-1 All Divided By 0.005)

But turns out that gives you an operating temp of 100*C .... yea im lost and have googled looking for operating temperature formulas...Help!


Thanks People!!

Kurdt

You have two equations with two unknowns. Rather than try to solve one at a time try solving them both at the same time.

Sullitp

Ive got lost again - But hey I got a new resistance this time - The "Hot temperature resistance" maybe?

I used the R1= 68(1+0.005x20) = 74.8

But I still cant figure out how to find the operating temperature with that last part solved.

A formula for this would help LOTS!
....Hints are just plain annoying at this time of night (NZ 12:55am)
And yes I have left this assignment to the last minute...

Thanks!

Kurdt

Fair enough I can say nothing about whether the formulas are right because its so long since I've done anything with electrical circuits but assuming they are correct the mathematical way of solving them is by simultaneous equations. Try solving this:

R1-T1 = 1/k, where k=0.005 alongside this,

T1/R1 = 1/(k*R0-k)

Unfortunately i can only give hints as part of the rules of this forum, and with good reason because if you don't get there by yourself (albeit with a little help) then you won't have learned anything.

Hammie

The lamp is rated 75 watts at 230 volts. Do you know how to calculate the resistance of the lamp with these values (i.e., the hot resistance of the filament) ?

Hootenanny

From what I can remember the resistance of a conductor is given as;

[tex]R = R_{0}(1 + \alpha \cdot \Delta T)[/tex]

Where [itex]\alpha = k[/itex] in kurdt's example.

~H

Kurdt

If that is the case Hoot, I hope the second equation involves a delta T too!

Hootenanny

Perhaps hammie's suggestion of using;

[tex]P = I^2 R[/tex]

to calculate the running resistance would be a more logical step, in light of the lack of a delta T in the second equality.

~H

Hammie

I was thinking of

[tex]P = V^2 /R[/tex]

Hootenanny

I apologise, I thought the current was given . If we apply ohm's law to my original equation to eliminate I, we can obtain the above equation.

~H

Gokul43201

That's a thing you can NOT do! You are not given the temperature hot, and just choosing a random value is not the way to solve the problem. You need to look carefully at ALL the data provided and see how you can arrive at the required quantity from known things.

And just to let you know how wrong your assumed value (100 deg C) is, let me tell you that the typical filament temperature is about 2500 deg C. That's an error of about 3000% in the temperature change !

That is an incorrect equation. The correct equation is given in post #6. The temperature in the equation is not the "hot" temperature, but the temperature change. In any case, you do not use this equation in part A.

No, you should work this out using the method described by hammie.

Well, you assumed an operating temperature of 100 deg C to calculate the first part. Naturally, if you use the same equation as you used there, you will merely extract that same number.

If anything, this second part should have told you that you do not want to assume a value of the temperature in part A. Nevertheless, if you correctly solve part A as suggested you can use the equation in post #6 to find the temperature change.

Kurdt

Glad to see I was of zero assistance there . I'll leave my patheic knowledge of electrical circuits on the sidelines in future .

Hootenanny

I may come with you. Nevermind, we can eat all the half time oranges!

~H