A 30 W incandescent light bulb contains a tungsten filament...

A 30 W incandescent light bulb contains a tungsten filament with a circular cross section of area $1.75 \times 10^{-9} \text{ m}^2$ and length 1.05 m (it is tightly coiled to fit this length inside the bulb). The resistivity of tungsten at room temperature ($T_0 = 293 \text{ K}$) is $5.25 \times 10^{-8} \Omega \cdot \text{m}$ and the temperature coefficient of resistivity for tungsten is $4.50 \times 10^{-3} \frac{1}{\text{K}}$. What is the resistance of this light bulb at room temperature? When a 120 V potential difference is applied to the bulb 0.25 A of current will pass through it. What is the temperature of the filament when it is plugged in? (Assume that the length and area do not change with temperature.)

Transcript

0:00 Hi there.

00:01 So for this problem we have an incandescent light bolt that uses a coil filament of 210 and we know that it has a length of 580 millimeters and we can write this in meters 0 .58 meters now it has a diameter.

00:33 We also have that this has a diameter and this has a diameter and of 46 micrometers when the temperature is, is of 20 celsius degrees, and has a resistivity for the top stand, and it is 5 .25 times 10 to the minus 8 oms per meter.

01:14 Now, this tells us that the temperature coefficient of resistivity is 0 .045 celsius degrees to the minus 1.

01:40 And this remains, okay, so we are also told that the temperature of the filament increases linearly with current from 20 degrees to from 20 degrees to from 20 degrees.

01:53 To 2 ,520 celsius degrees at a current of 1 amp.

02:10 Okay, so the first question is, the question a for this problem is, what is the resistance of the light bulb at 20 celsius degrees, when the temperature is 20 celsius degrees? so the first thing to identify is that this problem is about resistance of a light bulb filament and it can barriers with temperature.

02:43 We know that the expression for the resistance at a given value of temperature is given by the product between its resistivity times its length over the area, its exceptional area.

03:01 So in this case, we know the value for the resistivity when the temperature is of 20 celsius degrees is this in here.

03:10 So we just need to simply substitute that value.

03:14 We also note the length of it, which is 0 .58 meters.

03:19 And we also can obtain the area because it will be the area for a cylinder in this case.

03:37 So we plug all of those values in here.

03:41 So the resistivity has a value of 5 .25 times 10 to the minus 8 oms per meter.

03:51 The length of bits as i said 0 .58 meters.

04:05 And this over the area, which is pi over the area, the cross -septional area of this filament.

04:17 So we are given the, this in here is the diameter.

04:23 So we need to obtain the radius of it.

04:26 So we need to divide that value by 2.

04:28 So we will have 40x times 10 to minus 6 that's micro over 2.

04:36 This to the square.

04:37 So when we plug this into the calculator we obtained that the resistance has a value of 18 .3 oms so that's the resistivity of the filament when the temperature is of 20 celsius degrees now for part b of this problem we are asked what is the current through the light ball when the potential when the potential difference across its terminal is 120 volts.

05:17 So part of this problem is when the potential difference is 120 volts.

05:30 So what we need to obtain is the current when we have that value of the potential difference.

05:39 So we need first to find the temperature as a function of the current.

05:46 We know that.

05:47 That the graph of temperature versus the current is a straight line passing through 0 at 20 celsius degrees and 1 p.

06:00 The value of this given volume here, 2 ,520 celsius degrees.

06:07 So we can obtain the slope from that.

06:13 And so the slope is going to be 2 ,550.

06:19 500 celsius degrees per ampere and there is a t intercept at 20 celsius degrees.

06:36 So using the slope intersect equation of a straight line, the equation for that the temperature in function of the current is going to be of the following.

06:49 It's going to be that value 2 ,500 celsius degrees ampere.

06:58 So with this multiplied by the current, because we want that in function of the current, and this plus 20 celsius degrees.

07:08 Now, we know that the initial value when we plot in here the current equal to zero, we're going to attain 20 celsius degrees.

07:16 So we can obtain the resistance as a function of temperature so the formula for that is that the resistance in terms of the temperature is given by the resistance at 20 celsius degrees times one plus alpha t minus the initial 20 celsius degrees this and constant in here we are given that value from the beginning we know that this value is this, 0 .0045.

08:02 So we need to blow that value in there and we need to load the value for the resistance at 20 celsius degrees that we just obtained in the part a of this problem.

08:15 So we will have that the resistance at this temperature is going to be, so no, so what i'm going to do.

08:32 Do now is to combine these two equations.

08:36 To combine this two equations, we plot the temperature in here.

08:42 So what we are going to obtain is the following that the resistance in function now of the current because we are introducing one into the other is going to be this.

08:56 One plus alpha times that is expression for the temperature...

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