$$P=I\Dl V$$

in order to increase the charge's electric potential energy. Conversely, when current flows downhill, its charges lose potential energy, and power \(P=I\Dl V\) is released to the environment.
A current of 0.1A passes through a 3Ω resistor. How much power is released by the resistor into the environment?

Current flows downhill through resistors, and so power \(P=I\Dl V\) is released by the resistor. We know the current \(I=0.1\u{A}\), and the potential drop through the resistor is given by Ohm's Law: \(\Dl V=IR=(0.1\u{A})(3Ω)=0.3\u{V}\). Thus the power output is
$$P=(0.1\u{A})(0.3\u{V})=\boxed{0.03\u{W}}$$

We can combine the power equation above with Ohm's Law to write a specific power equation for resistors---two equations, in fact. Substituting \(\Dl V=IR\) into \(P=I\Dl V\) gives us $$P=I^2R$$ On the other hand, if we use the conductance version of the power law \(I=G\Dl V\), we can get $$P=(\Dl V)^2G {\quad\rm or\quad} P=\frac{(\Dl V)^2}{R}$$

Consider a 40 watt incandescent light bulb (which releases 40 watts of
power, mostly as heat) and a 60 watt incandescent
light bulb. Which bulb has the greater resistance?
*decreases* as you increase the resistance in this case.
The power is the product of an increasing number and
a decreasing number, and without knowing *how* they are
increasing or decreasing we can't say anything about what the power
does.
In short, when you have two apparently contradictory equations like
\(P=I^2R\) and \(P=(\Dl V)^2/R\), and you want to know how P and R are
related, choose the equation with the constant value.

A) 40W
B) 60W

The question is: is the resistance directly proportional to the power, or inversely proportional? And it looks like we have a contradiction: the equation \(P=I^2R\) suggests that they are directly proportional, while the equation \(P=(\Dl V\)^2/R\) suggests the opposite.

To solve this conundrum we must consider the context. Clearly,
a 40 watt light bulb doesn't *always* emit 40 watts of power, not
when it's sitting in a drawer or connected to a AA battery. It only
emits 40 watts when plugged into a wall socket, which provides a
specific potential difference of 120V. This potential difference is
the same for both light bulbs, so in the formula
$$P=\frac{(\Dl V)^2}{R}$$
increasing $R$ doesn't change $\Dl V$, and so $P$ must decrease. The
bulb which emits less power has the higher resistance.

Find the resistances of both light bulbs, if they are rated for a 120V
power source.

We know that
$$P=\frac{(\Dl V)^2}{R} \implies R=\frac{(\Dl V)^2}{P}$$
and so
$$R_{40W}={(120\u{V})^2\over 40\u{W}}=\boxed{360\u{\Omega}} \qquad
R_{60W}={(120\u{V})^2\over 60\u{W}}=\boxed{240\u{\Omega}}$$