# Power equations, which equation to use and why?

• Sarah0001
In summary, the resistance of an LDR decreases with increasing illumination due to the release of more valence electrons onto the conduction band. When comparing power equations, we use P = V^2/R because voltage is given as a constant, allowing for direct comparison with the variable R. Additionally, this expression eliminates the current, making it more useful in this problem. However, it is important to note that the change in resistance does not directly correspond to an increase or decrease in power, as it depends on the rate of change in current with respect to resistance.
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?

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?

phinds
berkeman said:
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?

Sarah0001 said:
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.

Sarah0001 said:
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 $R$.

Sarah0001 said:
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##.

## 1. What are power equations and why are they important in science?

Power equations are mathematical expressions that describe the relationship between power (ability to do work) and other variables such as force, velocity, and time. They are important in science because they allow us to quantitatively analyze and predict the behavior of systems and processes that involve power, such as energy production and mechanical systems.

## 2. How do I know which power equation to use for a specific problem?

The power equation you should use depends on the variables that are known and the ones you are trying to solve for. For example, if you know the force applied and the velocity of an object, you can use the equation P = Fv to calculate its power. It is important to carefully consider the given information and choose the appropriate equation.

## 3. Can power equations be used for both mechanical and electrical systems?

Yes, power equations can be applied to both mechanical and electrical systems. However, the equations may differ depending on the specific variables involved. For example, in an electrical system, the power equation would involve current and voltage, while in a mechanical system, it would involve force and velocity.

## 4. How can I use power equations to compare different systems or processes?

Power equations are useful for comparing different systems or processes because they provide a quantitative measure of the amount of work being done. By using the same power equation for each system or process, you can compare the power output and determine which is more efficient or effective.

## 5. Are there any limitations to using power equations?

While power equations are useful tools in science, there are some limitations to consider. One limitation is that they assume ideal conditions and do not account for factors such as friction or resistance. Additionally, they may not accurately represent systems or processes that are highly complex or involve multiple variables.

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