An issue with Vout formula

[methodsboy]

hey all
For question 26 a), how do you know which resistance to put at the top (R1) of Vout = ((R1/ R1+2) x Vin) formula?
[IMG]http://img152.imageshack.us/img152/1912/vout.jpg[/img]

[TrueTears]

measures the voltage drop across the LDR

Let LDR be and Resistor be

[pHysiX]

hmmm

i disagree,

I'm saying that R1 is the 100ohm resistor. the connection is between Vout and 0V, hence passing through 100ohm resistor.

what's the answer? well that's my two cents anyway

[methodsboy]

measures the voltage drop across the LDR

Let LDR be and Resistor be

im sorry to say, but TT is wrong

[methodsboy]

hmmm

i disagree,

I'm saying that R1 is the 100ohm resistor. the connection is between Vout and 0V, hence passing through 100ohm resistor.

what's the answer? well that's my two cents anyway

yep, ur right.

[TrueTears]

o shit gg

o my bad, its meant to be over the 100 ohm resistor fk.

[pHysiX]

Vout does not exist. It's just like a "concept"

in practise, it just means an eaten up voltage. So from the circuit, we see that the LDR has taken some potential. That leaves the resistor.

Now, the main point is to use the voltage of the resistor (the two components are in series, same current, hence this equation exists). So we have to parallel Vout onto the resistor, one at the +ve end, and one to the -ve end. So that means that Vout is between the top of the 100ohm resistor and the end of the 100ohm resistor, which means R1 is 100hm =]

Another piece of logic is that we need a potential difference. If we hook Vout to the 12v rail, then sure we have a potential difference but that would mean that u get 12v (parallel). Not wat the main idea was.

Hope it helps =]

[TrueTears]

Yeah change mine one to

In other words the numerator always has the resistance which is across, in this case its the 100 ohm resistor.

[methodsboy]

k thnx guys =]