Thermosistor Question

[methodsboy]

hey
once again, im stuck on this one:

[IMG]http://img152.imageshack.us/img152/4453/thermosistorquestion.jpg[/img]

Thanks guys 

[Gloamglozer]

Can anyone else see the question or is it just my computer starting to go F.U.B.A.R.?

[methodsboy]

i can see it fine.....

[Gloamglozer]

Is it an image in the post or an uploaded image?

[methodsboy]

i uploaded the image to imageshack and then pasted the link here

[danieltennis]

Can you see it now?

[pHysiX]

ok

We need a Vout of 4V
and the Resistance at 10 degress is 1000 ohms

Hence, the current flowing is 4x10^-3Amps

The remaining voltage for the variable resistor is 8 volts. with a current of 4x10^-3 Amps,

R = V/I = 2000Ohms

2000 Ohms for the variable resistor

[methodsboy]

^yes !
thankyou
the lame book says 1600 ohms!

[methodsboy]

this is how did it:
i used to the convenient Vout formula:
4 = 1000/ (1000+Rvariable resistor) x 12
solve for Rvariable resistor which yields: 2000 Ohms

[MattPritchard]

Use the same working out as other showed above but reading off the graph it appears to be closer to 800 Ohms not 1000 Ohms which once you sub that into the equations you will arrive at 1600 Ohms.

Vout = 4V
Resistance at 10 degrees = 800 ohms
V=IR
I=V/R
I=4/800
I=5.0 x 10^-3 Amps

Remaining Voltage = Total - Required at End = 12V - 4V = 8V
Current is 5.0 x 10^-3

V=IR
R=V/I
R=(/(5.0 x 10^-3)
R=1600 Ohms

[Gloamglozer]

(Image removed from quote.)

Can you see it now?

No.  This is what I see from the original post:




And that image was uploaded on Imageshack.

EDIT:  Hang on.  I can see the image now once I quoted Daniel's post.  What on Earth is going on?

[methodsboy]

Use the same working out as other showed above but reading off the graph it appears to be closer to 800 Ohms not 1000 Ohms which once you sub that into the equations you will arrive at 1600 Ohms.

Vout = 4V
Resistance at 10 degrees = 800 ohms
V=IR
I=V/R
I=4/800
I=5.0 x 10^-3 Amps

Remaining Voltage = Total - Required at End = 12V - 4V = 8V
Current is 5.0 x 10^-3

V=IR
R=V/I
R=(/(5.0 x 10^-3)
R=1600 Ohms

ah k. that makes sense