9.3 Electric power and energy  (Page 2/7)

Different insights can be gained from the three different expressions for electric power. For example, $P=V^2/R$ implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in $P=V^2/R$ , the effect of applying a higher voltage is perhaps greater than expected. Thus, when the voltage is doubled to a 25-W bulb, its power nearly quadruples to about 100 W, burning it out. If the bulb’s resistance remained constant, its power would be exactly 100 W, but at the higher temperature its resistance is higher, too.

The cost of electricity

The more electric appliances you use and the longer they are left on, the higher your electric bill. This familiar fact is based on the relationship between energy and power. You pay for the energy used. Since $P=E/t$ , we see that

is the energy used by a device using power $P$ for a time interval $t$ . For example, the more lightbulbs burning, the greater $P$ used; the longer they are on, the greater $t$ is. The energy unit on electric bills is the kilowatt-hour ( $\text{kW}\cdot \text{h}$ ), consistent with the relationship $E=\text{Pt}$ . It is easy to estimate the cost of operating electric appliances if you have some idea of their power consumption rate in watts or kilowatts, the time they are on in hours, and the cost per kilowatt-hour for your electric utility. Kilowatt-hours, like all other specialized energy units such as food calories, can be converted to joules. You can prove to yourself that $\text{1 kW}\cdot \text{h = 3}\text{.}6\times {\text{10}}^6\text{J}$ .