Power equations, which equation to use and why?

[Sarah0001]

Homework Statement

A light dependent resistor is connected across an ideal 12 V source
and placed in the open in the middle of a desert. When is the power
dissipated in the resistor highest?
A )dawn B) mid morning
C) noon D) midnight

Relevant Equations

P= IV --> P= I^2 R ---> P = V^2/R

Which equation do I use and why do I use it? I think it would be P = V^2/R, but why do we exclude current the from the equation?

 

[berkeman]

What is the resistance of an LDR versus illumination? Can you post a typical curve and say what you think that means about the answer to this question?

 

[Sarah0001]

What is the resistance of an LDR versus illumination? Can you post a typical curve and say what you think that means about the answer to this question?

A typical graph of LDR resistance against illumination would have a negative correlation.
I understand resistance of the LDR goes down with increase in Lux, as from my understanding this frees more valence electrons onto the conduction band, and so power is max when R is min. This is at noon when the illumination is greatest, but my question is why do we use P=V^2/R as opposed to the other power equations?

 

[haruspex]

why do we use P=V^2/R as opposed to the other power equations?

Because you are given voltage as a constant.
P=I²R is still true, but when R changes so does I, so you cannot reason that increasing R increases P.

 

[Sarah0001]

surely if say R is minmum , I therefore is maxmium so we can reason using P=I^2R that when R is minmum i.e at noon then P is max, as there is a factor of squaring.

 

[collinsmark]

surely if say R is minmum , I therefore is maxmium so we can reason using P=I^2R that when R is minmum i.e at noon then P is max, as there is a factor of squaring.

While that logic is correct, to be more clear you might want to go back and explore your original question about the $P = \frac{V^2}{R}$ expression. Perhaps if you derive it for yourself it might make it more clear.

Using the equations
$P = IV$
and
$V = IR$,
and a bit of substitution, you should be able to come up with that expression that eliminates $I$. This is useful for this problem since $V$ is a constant (due to ideal voltage source), allowing you to compare $P$ directly with the variable $R$.

 

[haruspex]

surely if say R is minmum , I therefore is maxmium so we can reason using P=I^2R that when R is minmum i.e at noon then P is max, as there is a factor of squaring.

No, that does not follow. It depends how swiftly I changes as R changes.
Suppose the relationship is $I=\alpha R^\beta$. Then$P=\alpha^2R^{2\beta+1}$.
$\frac{dP}{dR}=\alpha^2(2\beta+1) R^{2\beta}$.
Whether this is positive or negative depends on the sign of $2\beta+1$.
With V constant, $\beta=-1$, so $2\beta+1<0$.