Voltage divider question.

[SDPHD]

I tried doing this question but I got lost in some really weird calculations.

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Any clue as to how to do it?

[SDPHD]

Nevermind, I figured out how to do it.  :uglystupid2:

V(1000 Resistor)=6-3.6=2.4v
I(1000 Resistor)=2.4/1000=0.0024A
I(4000 Resistor)=3.6/1000=0.0009A
I(Water Temp Sensor)=0.0024-0.0009=0.0015A
R(Water Temp Sensor)=3.6/0.0015=2400OHM

Then you read off the graph.

[Vincezor]

Nevermind, I figured out how to do it.  :uglystupid2:

V(1000 Resistor)=6-3.6=2.4v
I(1000 Resistor)=2.4/1000=0.0024A
I(4000 Resistor)=3.6/1000=0.0009A
I(Water Temp Sensor)=0.0024-0.0009=0.0015A
R(Water Temp Sensor)=3.6/0.0015=2400OHM

Then you read off the graph.

Yep, that would be correct. I guess it's a combination of voltage dividers and knowing kirchoff's law and that might get some people stumped

[macostar]

Nice question. Where is it from are there any more?

[Vincezor]

Nice question. Where is it from are there any more?

There is always a voltage divider question in the VCAA exams, all very similar. You could take a look there.

You will probably encounter this sort of question in every trial exam paper as well. It might not be as complex (however some are) but it's good practice to know how to do these inside out as for the reason mention above.

[macostar]

Thanks Vince. The majority of voltage divider questions are straight forward enjoyed this one because of the analysis required.

[DisaFear]

Thanks Vince. The majority of voltage divider questions are straight forward enjoyed this one because of the analysis required.

Indeed, picture saved, will have a crack at it later