4.
Power
When a battery creates a current, it is in effect moving a positive charge from a lower potential to a higher potential, and because positive charges like to move downhill, this increases their potential energy by $$\Delta PE = q\Delta V$$ This energy must be supplied by the batteries themselves, in the form of electrical work. If we imagine the battery operating over a certain time, then we can divide both sides of this equation by that time to get $${{\rm work\,done}\over {\rm time}} = {\rm charge\over \rm time}\Delta V$$Power
\(P\)
watts (W)$$P=I\Delta V$$
Power, as you'll recall is measured in units of joules per second, which are known as watts (W). We can see here that one watt is also equal to 1 ampere-volt.
This equation holds not just for batteries but for resistors too, or any time a current moves to a higher or lower potential. For resistors, the current always moves from higher to lower potential, and so currents release power $P=-I\Delta V$ in resistors.
Combining this equation with Ohm's Law allows us to write two other formulas for the power emitted by a battery:
$$P=I^2R \quad{\it and }\quad P={(\Delta V)^2\over R}$$
which may be useful depending on what is already given in a problem.
A "60W" incandescent light bulb is called that because when connected to a 120V power supply (the standard voltage of the plugs in US houses) it emits 60W of power (most of it as heat, which is why modern light bulbs are more efficient). To find the resistance of this light bulb, we take the second equation above and solve it for $R$, since we know $\Delta V$ and $R$ but we don't know $I$: $$P={(\Delta V)^2\over R} \implies R={(\Delta V)^2\over P}$$ In this case, $R={(120\u{V})^2\over 60\u{W}} = 240\u{\Omega}$. Notice that a less powerful light bulb, like a 40W light bulb, has a larger resistance: $R={(120\u{V})^2\over 40\u{W}} = 360\u{\Omega}$. The fact that high-powered devices (like vacuum cleaners and microwaves) have low resistances will become important later.