### 2.2.4 Ohm's Law, and Why We Care About Resistance

The device that we know as a toaster is remarkably simple. It consists basically of a wire through which current is passed. The wire heats up, toasting the bread. That's it!

But why does the wire heat up? The answer is because the wire has some resistance. When current is passed through a material with some resistance, the material heats up. That heat is actually the dissipation of some of the electric power being passed through the material in the first place. This dissipation of power in the form of heat is referred to as "losses" in the electric power grid.

The resistance of the material through which current is passed helps to determine the losses, but it's not the only factor. The voltage at which the power is passed through the material matters as well, as does the amount of current.

This relationship is summarized neatly in Ohm's law, which states that voltage is equal to the product of current and resistance, or $\text{V=I\xd7R}$ . Ohm's Law is used to determine the amount of voltage needed to move a given amount of current (I) across some material with a given resistance (R).

Meanwhile, remember our definition of power: $\text{P=I\xd7V}$ . This is essentially the amount of power delivered in a circuit like the one in our last exercise.

We can plug Ohm's Law into our definition of power to get:

$$\text{P=I}\times \text{V=I}\times \text{}\left(\text{I}\times \text{R}\right){\text{=I}}^{\text{2}}\text{}\times \text{R}$$

This equation describes the amount of power dissipated in a circuit. It also describes the quantity of the losses. Ohm's Law thus tells us that losses will increase with the square of the current. Thus, if we keep the voltage constant and double the current, losses will increase by a factor of four.

To see the importance of this, let's suppose that we pass 1,000 Amps of current through a circuit with a voltage drop of 100 V. So we have 100 kW of power. Losses in the circuit will be proportional to ${\text{I}}^{\text{2}}\text{\xd7R}$ , or ${\text{1,000}}^{\text{2}}\text{\xd7R}$ in this case.

But, if we wanted 100 kW of power, we could do it another way, by passing 100 Amps through the circuit at a voltage of 1,000 V. The resistance in the circuit doesn't change, but losses in the circuit will now be proportional to 100^{2} × R.

**Thus, by increasing the voltage (and decreasing the current) by a factor of 10 each, we have decreased our losses by a factor of 100. This explains the reason that we have an AC power grid instead of a DC power grid.** Remember that in Edison's DC power technology, the voltage at the source had to be close to the voltage at the point of consumption. But with the AC technology developed by Tesla and Westinghouse, power could be generated and transmitted at very high voltages, then reduced to lower voltages at the point of consumption. This had two big advantages - first, transmission losses could be reduced substantially, and second it was much safer for homes and businesses to be using electricity at low voltage rather than high voltage.