9.5 Electrical energy and power  (Page 2/8)

The work done on the charge is equal to the electric force times the length at which the force is applied,

The charge moves at a drift velocity $v_{\text{d}}$ so the work done on the charge results in a loss of potential energy, but the average kinetic energy remains constant. The lost electrical potential energy appears as thermal energy in the material. On a microscopic scale, the energy transfer is due to collisions between the charge and the molecules of the material, which leads to an increase in temperature in the material. The loss of potential energy results in an increase in the temperature of the material, which is dissipated as radiation. In a resistor, it is dissipated as heat, and in a light bulb, it is dissipated as heat and light.

The power dissipated by the material as heat and light is equal to the time rate of change of the work:

With a resistor, the voltage drop across the resistor is dissipated as heat. Ohm’s law states that the voltage across the resistor is equal to the current times the resistance, $V=IR$ . The power dissipated by the resistor is therefore

If a resistor is connected to a battery, the power dissipated as radiant energy by the wires and the resistor is equal to $P=IV=I^{2}R=\frac{V^2}{R}$ . The power supplied from the battery is equal to current times the voltage, $P=IV$ .

Different insights can be gained from the three different expressions for electric power. For example, $P=V^2\text{/}R$ implies that the lower the resistance connected to a given voltage source, the greater the power delivered. Furthermore, since voltage is squared in $P=V^2\text{/}R$ , 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.