ATAR Notes: Forum
VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: liamh96 on May 04, 2014, 02:14:38 pm
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Hi all,
Just a question regarding circuits Q's on the 2011/12 mid years:
section 2, Q6, page 20-21 on the 2011 exam: http://www.vcaa.vic.edu.au/Documents/exams/physics/2011physics1-w.pdf
and section 2, Q3b, page 18-19 on the 2012 exam http://www.vcaa.vic.edu.au/Documents/exams/physics/2012/2012physics1-w.pdf
Q: How do you know which component to attach the switching circuit to? Referencing the examiners reports, in 2011 the switching circuit is attached to the variable resistor, not the thermistor, whilst in the 2012 e.r the switching circuit is attach to the LDR, not the standard resistor.
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In 2011, the switching circuit comes on when the voltage across it is above 4.5 V. We want the circuit to turn on when the temp is greater than 20 C. The resistance decreases as we increase the temperature past this point. So if we were to put the switching circuit across here, our voltage drop would decrease under 4.5 V at that switching point. If we put it across the other resistor, the ratio of it's resistance to the sum of the two will increase with temperature, so you can set it to start at 4.5 V and increase past that as the temperature rises above 20 C.
In 2012, we want this switch to happen when the light is 2.5 lux or less. If we start at this switching point and work backwards, our resistance is increasing instead of decreasing like before. So if we put it over the LDR then if the light decreases below that certain level and continues to decrease our resistance across it increases and we get a larger voltage output, which will be above that 6.0 V to activate the switching circuit.
So yeah, the difference is one year we desire the switching circuit to come on above a certain value, and the other year we desire the switching circuit to come on below a certain value.