Click here to purchase the entire book in PDF format. Power and Watt's Law If we return to the example of a pump creating flow through a pipe, it's pretty obvious that this little system is useless for anything other than wasting the energy required to run the pump. If we were smart we'd put some kind of turbine or waterwheel in the pipe which would be connected to something which does work - any kind of work. Once upon a time a similar system was used to cut logs - connect the waterwheel to a circular saw; nowadays we do things like connecting generators to the turbine to generate electricity to power circular saws. In either case, we're using the energy or power in the moving water to do work of some sort. How can we measure or calculate the amount of work our waterwheel is capable of doing? Well, there are two variables involved with which we are concerned - the pressure behind the water and the quantity of water flowing through the pipe and turbine. If there is more pressure, there is more energy in the water to do work; if there is more flow, then there is more water to do work. Electricity can be used in the same fashion - we can put a small device in the wire between the two battery terminals which will convert the power in the electrical current into some useful work like brewing coffee or powering an electric stapler. We have the same equivalent components to concern us, except now they are named current and voltage. The higher the voltage or the higher the current, the more work which can be performed by our system - therefore the power it has. This relationship can be expressed by an equation called Watt's Law which is as follows: Power = Voltage * Current or $$P = V I$$ (3.2) where P is in watts, V is in volts and I is in amps. Just as Ohm's law defines the ohm, Watt's law defines the watt to be the amount of power consumed by a device which, when supplied with 1 volt of difference across its terminals will use 1 amp of current. We can create a variation on Watt's law by combining it with Ohm's law as follows: $P = V I$ and $V = I R$ therefore $$P = (I R) I$$ (3.3) $$= I^{2} R$$ (3.4) and $$P = V \frac{V}{R}$$ (3.5) $$= \frac{V^{2}}{R}$$ (3.6) Note that, as is shown in the equation above on the right, the power is proportional to the square of the voltage. This gem of wisdom will come in handy later.