We have already derived a relationship between energy and voltage. To move a charge between two locations with different potentials, the required work is
\[
W = qU
\]
To calculate the electric power, this formula must be differentiated with respect to time, which gives the following important relation:
\[
\boxed{P = UI}
\]
Due to collisions of electrons with atomic nuclei, all the work in a conductor is converted into heat, which causes the conductor to heat up. The release of heat from electric current is applied in many everyday devices. Stovetops and kettles are just two examples of devices that use resistors, in which electrical energy is converted into thermal energy.
Example: A voltage drop of 5 V is measured across a resistor. The current in the circuit is 200 mA. How much power is converted into heat? Solution: 1 W
Using Ohm’s law, one of the two quantities can be eliminated, and the power can, for example, be expressed as a function of current:
\[
\boxed{P = RI^2}
\]
A fuse, which burns out at too high a power consumption and interrupts the circuit to protect electrical installations and devices, depends only on the square of the current. In cars, such fuses are placed before every consumer to protect the battery and alternator from excessive currents. The quadratic dependence on current is also the reason why power transmission lines operate at very high voltages, sometimes up to several hundred kilovolts. To deliver the same power to consumers, the current can be kept low, which ensures that the power losses in the lines remain only a few percent, even over long distances.
Example: A toaster operating at 230 V in Central Europe has a heating power of 800 W. What is the approximate resistance of the heating wires? Solution: 44 Ω
It is also possible to eliminate the current. This gives the dependence on the voltage drop across the resistor:
\[
\boxed{P = \frac{U^2}{R}}
\]
The electrical work done in a circuit is generally the integral of power over time. For many devices, however, the power consumed is constant. In this case, the calculation simplifies to the product of power and time:
\[
W = Pt
\]
The SI unit of energy is the joule. However, in electrical applications, it is common to specify energy as the product of watts and seconds. One watt-second (Ws) is therefore equal to one joule. Electricity meters in buildings and laboratories instead measure energy in the larger unit of kilowatt-hour (kWh), so that the numerical values remain manageable.
Example: A 40-watt light bulb runs for 8 hours every night. How much will the homeowner approximately pay on the electricity bill if one kilowatt-hour costs 30 cents? Solution: 35 €