9702 PHYSICS P4 SPECIMEN PAPER QN 8 (b)

[gentleman]

part (i) 2.
* mmm i need to know how we took 800 as the R and didnt consider the thermistor (2200)?
* for part 3. the output will be -9.0 because the inverting (-) voltage (4.5) is greater than the non-inverting (+) voltage 2.4. Ok in the application booklet it says that when the output exeeds the power supply voltage the maximum power supply 'll be the voltage (ie 9 or -9). Can anyone explain to me, [even if the the output does not exeed the power supply voltage, the output will be 9 or -9?? :fool:
thnxs

 

[beacon_of_light]

Mark scheme has directly found the p.d across thermistor...
If you consider thermistor, take 2200Ω and find p.d across the thermistor... {2200/(2200+800)} x 9 = 6.6V ... And this is the potential drop across the thermistor and you've to give the potential at point B after this drop across thermistor... 9v is the p.d across thermistor and resistor, if 6.6V is the drop across thermistor, 9-6.6=2.4V is the pd at point B ...

If the output is b/w +9V and -9V the amplifier is not saturated and gives the calculated output. However, if the output is beyond the supply voltage, amplifier saturates and the output pd = supply voltage ...

Hope that helps