Resistor's (Question's)

vbimport

#1

I got an exam on monday and i’m still so lost. The notes are basically posted on the site, and even that we still have to look at other sites, although u’d think they’d be sufficient. Anyhoo these are my questions…hopefully someone can assist me! :bigsmile:

The formula is T = T(amp) + kP but don’t worry about k, just think of it as a constant for the moment

The temperature of a soldering iron tip is 300ºC above the ambient temperature. What is the change in temperature of the tip if the voltage supply is reduced by 50%?
Answer: 75 degrees celcius

The temperature of a soldering iron tip is 300ºC above the ambient temperature. What is the change in temperature of the tip if the power to it is reduced by 50%?
Answer: 150 degress celcius

I’m wondering why if power is halved, temp is halfed but if voltage is halved, temp is quartered.

Thanks


#2

Doesn’t this come back to Ohms Law?

Sorry I get confused with these things.


#3

Don’t think so…they didn’t teach us anythin for this subject! :frowning:


#4

Whats the site link?


#5

hmmmm…you want me to give away my uni username/password?? :bigsmile:


#6

The subject title is Prescott?


#7

don’t get ya! lolz…well…but the subject is design and innovation, and this is a part of the exam, resistors


#8

I think I have an answer, but an not completely sure if it is correct, because first of all it’s been a long time I worked with formulaes, and secondly, I have some doubts about how I use Ohm’s law.

From what I understand the relationship is T(tip) = T(room) + kP, with:

T = Temperature (Kelvin)
P = power (Watt)
k = constant

Let’s define these first:

V = voltage (Volt)
I = current (Amp)
(d) = delta (“the change in”)

OK, the change in temperature of the tip is:

T(tip) - T(room) = (d)T = k(d)P

To get rid of that constant, we can say that the change in temperature is proportional to the change in power:

(d)T ~ (d)P

So, if the power is halved, the temperature is also halved. Like in the first question. Now, for the second question:

The power can be calculated using the following formula:

P = V * I

So, you would say that when V is halved, P is also halved, right? No, that’s not the case. We can’t say anything about P unless we know something about I. And the problem is, we don’t know anything about I.

Now comes the trick: Ohm’s law states:

V = I * R

which rewrites to:

I = V / R

We can substitute this into the formula for calculating the power:

P = V * (V / R) = V^2 / R

(V^2 means “V square” or V * V)

Now we have a different formula for calculating P. But don’t we need to know something about R before we can say something about the change in P? Yes, and we know enough already:

Remember that a soldering iron generates heat becauses of high internal resistance. This value of R is so big, that it changes only little when varying V and I; therefore, this change can be neglected.

§P = §(V^2) / §R

becomes by neglecting the change in R:

(d)P = (d)(V^2) / R

which becomes

BP ~ (d)(V^2)[/B]

From this followes that when the voltage is halved, the power is quartered and with it the temperature of the tip.

Note that I’m not sure if I used Ohm’s law correctly here. Ohm’s law is used for calculating the resistance in an ideal conductor, and of course a soldering iron is far from ideal. However, since we don’t actually calculate the resistance here, I think it’s save to use it in this way.

Good luck with your exams.


#9

Maybe I learned all of that about 16-18 years ago but it’s new enough that P stands for power and v for voltage. :slight_smile:

Online encyclopedia’s a great tool.


#10

Hehe, wiki says that Ohm’s law holds for constant values of R. I guess that makes my calculations valid.


#11

Your post was easier to understand than my middle school science teachers. I spent a lot of time then wondering why so many names on science textbooks appeared to be foreign (Western names.) It helped me to open my eyes again.


#12

@tommyCP man gotta love the basics of ohm’s law…and good answer by the way


#13

thanks TommyCP…and yea…it’s celcius in aus!! :bigsmile:


#14

don’t know if i’ll get this answered in time but here goes anywayz. It applie’s to current as well, right? Cos there was an exact one but with current being reduced, and answer was exactly same (P = (I*I) R)…

I’m just thinking why this is reduced to 50degreess and not 12.5…cos well the current is reduced by 1/4 so overall power is reduced by 1/8?

When operating the voltage across a red light emitting diode always remains constant at 1.2V. The rated current throught the light emitting junction is 20mA which corresponds to the rated temperature of the light emmiting junction of 100ºC. The ambient room temperature is 20ºC. What is the temperature of the junction if the current through the LED is reduced to 5mA ?
choices
10ºC
20ºC
30ºC
40ºC
50ºC <-- the answer


#15

just post the whole exam …and see who will answer it all for ya…:wink:


#16

hahahah…very funny…it was from a quiz…


#17

well…how did ya do??? we wanna see ya post some grades dude


#18

well…it doesn’t count…i’m doin preparation for last exam…when i get my results for my exam…bout mid july…i’ll post them…u’ll be the first to get them, deal? :bigsmile:


#19

well…i’ll post one my subject’s being results…juz coursework!!!

SWLab1 Simple console calculator MNS E 6 89/100
HWLab1 Hardware lab 1 MNS E 6 100/100
SWLab2 GUI Vis Studio .NET Calculator MNS E 6 94/100
HWLab2 Hardware lab 2 MNS E 6 95/100
SWLab3 GUI or Console Calculator, your choice MNS E 6 98/100


#20

WOW i’m impressed !!! GREAT job