PHYSICS APPLICATION , AGAIN!!!

[gentleman]

Please can anyone correct my steps and explain to me whether my conclusion is right or not? :good:
There are two qns that are nearly similar, about application:

1. Specimen paper 4, Qn 8
2. O/N 2009 41, Qn 10

I did the specimen paper qn and understood the concept
ok now i did the o/n 2009 paper qn and here are the steps:

At 16°C given that resistance of the thermistor is 2100 Ω and at 18°C the resistance of the thermistor is 1900 Ω
The question is to determine the change of states between R & G

At 16°C
Since the 'P' resistors have the same resistance the voltage is 1V for both, therefore the non-inverting voltage will be 1V
for the inverting voltage, the formula is V = IR, therefore V = [2/(2100 + 2000)] x 2000 ... which is = 0.976VS
ince V(+) is greater than V(-) the output will be (+) which will cause the diode (R) to be ON and diode (G) to be OFF.

At 18°C
Since the 'P' resistors have the same resistance the voltage is 1V for both, therefore the non-inverting voltage will be 1V
for the inverting voltage, the formula is V = IR, therefore V = [2/(1900 + 2000)] x 2000 ... which is = 1.026V
Since V(-) is greater than V(+) the output will be (-) which will cause the diode (G) to be ON and diode (R) to be OFF.

Therefore as temperature rises, diode R goes OFF and diode G goes ON

my question is: Compared with the specimen paper, for the voltage of the resistor and the thermistor, if V = IR, then the R we multiply with must be the R of the resistor, not the thermistor. Am I right ??? (in both the papers they user R of the resistor, not the thermistor!)

 

[workinghard]

umm i blv da resistance at 18o is i9oo ohm actually!
ll check it out n get back 2 u in a while!

 

[gentleman]

@working hard:yes it is 1900, my mistake

 

[Ferr]

Why do we use resistance of the fixed resister in V=IR? Why not of the thermistor?

 

[1357913579]

Please can anyone correct my steps and explain to me whether my conclusion is right or not? :good:
There are two qns that are nearly similar, about application:

1. Specimen paper 4, Qn 8
2. O/N 2009 41, Qn 10

I did the specimen paper qn and understood the concept
ok now i did the o/n 2009 paper qn and here are the steps:

At 16°C given that resistance of the thermistor is 2100 Ω and at 18°C the resistance of the thermistor is 1900 Ω
The question is to determine the change of states between R & G

At 16°C
Since the 'P' resistors have the same resistance the voltage is 1V for both, therefore the non-inverting voltage will be 1V
for the inverting voltage, the formula is V = IR, therefore V = [2/(2100 + 2000)] x 2000 ... which is = 0.976VS
ince V(+) is greater than V(-) the output will be (+) which will cause the diode (R) to be ON and diode (G) to be OFF.

At 18°C
Since the 'P' resistors have the same resistance the voltage is 1V for both, therefore the non-inverting voltage will be 1V
for the inverting voltage, the formula is V = IR, therefore V = [2/(1900 + 2000)] x 2000 ... which is = 1.026V
Since V(-) is greater than V(+) the output will be (-) which will cause the diode (G) to be ON and diode (R) to be OFF.

Therefore as temperature rises, diode R goes OFF and diode G goes ON

my question is: Compared with the specimen paper, for the voltage of the resistor and the thermistor, if V = IR, then the R we multiply with must be the R of the resistor, not the thermistor. Am I right ??? (in both the papers they user R of the resistor, not the thermistor!)

where is the Specimen paper 4?
can someone please share.