Question about voltage dividers

[Arkandea]

There is a question in my Physics textbook that I cannot answer, there is no back of the book answer given as well. I hope someone here can explain it:
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9     
A voltage divider circuit is set up as shown in the following figure (a). The thermistor has the characteristic curve shown in figure (b).

   
(a)
   

(b)
   
"Question"
   
What happens to the value of the output voltage as the temperature falls and the variable resistor remains at a fixed value? Explain your answer.
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I thought that as resistance in the thermistor decreases, so would current increase, therefore according to V=IR, voltage would remain constant.
Am I right or is there a different answer? Thank you in advance!

[Blakhitman]

Mind putting up the graph?

[shokstar]

U should be assuming that as temperature falls, resistance increases, from this curve  http://www.wecc.com/images/interH.gif , unless your graph is opposite, i cant see it. And u shouldnt be thinking in terms of V=IR, u should be thinking in terms of the voltage divider formula. Vout=R2/(R1+R2)*Vin. R2 is resistance of the thermistor and R1 the resistance of the variable resistor. Therefore, as R2 increases, voltage out will increase. I cant see your pictures, so i figured it out using an example. R1=50 R2=40, Vin=5. Work it out and itll be 2.22V. Chnage R2 to a higher value, 50, and work it out. Itll come to 2.5, so you can see that as temperature falls, resistance increases and voltage out increases.

Hope that helps.

[Arkandea]

Dang, when I previewed the post the images showed up just fine. I apologise for that, they worked when I looked at them...

Anyway thank you shokstar for that, helped alot.