A battery has to move current "uphill", and that requires energy. Lifting a charge \(q\) up a potential difference of \(\Dl V\) requires an increase in its potential energy of $$\Dl PE=q\Dl V$$ When discussing current, we're interested in the flow of charge per time, however. Dividing both sides of this equation by time gives us $$\rm\frac{energy}{time}=\frac{charge}{time}\Dl V$$ The fraction \(\rm\frac{charge}{time}\) is current, and \(\rm\frac{energy}{time}\) is power, a concept you learned about in mechanics class. Power is measured in watts; that is, joules per second. Whenever a current goes up a potential difference of \(\Dl V\), power must be supplied to the current
$$P=I\Dl V$$
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 $$P=(0.1\u{A})(0.3\u{V})=\boxed{0.03\u{W}}$$
For a wire or ordinary carbon resistor, this power is released as heat, but in other devices it may be released as light, sound, motion--- everything that electric devices do.

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?
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.

If we consider the formula $$P=I^2R$$ and increase the resistance, we can't assume that the power increases because the current isn't constant. In fact, the current 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.
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 $$R_{40W}={(120\u{V})^2\over 40\u{W}}=\boxed{360\u{\Omega}} \qquad R_{60W}={(120\u{V})^2\over 60\u{W}}=\boxed{240\u{\Omega}}$$