*Before you read this post, I suggest you read* post 17.44.

To move charge from a point with a low potential to a point with a high potential, we need to do work. This is the same as saying that moving a charge from a point with a low potential to a point with a high potential requires energy (post 16.21). But this energy isn’t stored, it is lost (dissipated), usually in the form of heat (post 16.35). Further explanation appears in post 17.44. We could summarise this by saying that *work is done on the charge*.

A charge moves spontaneously from a point with a high potential to a point with a low potential, leading to a release of energy, once again, usually as heat. We could summarise this by saying that *work is done by the charge*. This is rather like an object with a high gravitational potential energy spontaneously falling to the ground (lower potential energy) when it is dropped (see post 16.21).

So, whichever way the charge moves, it releases energy. This is another way of saying that energy is released by an electrical current, since current is flow of charge.

If 1 C of charge moves through a potential difference of *V*, the work done is equal, by definition, to *V* (post 17.44). If a charge *q* moves through the same potential difference, the work done must by *q* times as great, so the work done is given by

*W* = *Vq* = *VI*Δ*t*.

The final step comes from the definition of a steady current *I*, flowing for a time interval Δ*t* (post 17.44). According to Ohm’s law *V* = *IR*, where *R* is the resistance to movement of the charge (post 17.44). Substituting this expression for *V* into the equation above gives

*W* = (*IR*)*I*Δ*t* = *I*^{2}*R*Δ*t*. (1)

If *I* is measured in amps, *R* in ohms and Δ*t* in seconds, it follows, from the units used to define amps and ohms in post 17.44, that *W* will be measured in joules (J) (post 16.20) which is the SI unit of work and energy (posts 16.20 and 16.21).

The rate at which the charge does work is called *power* and is measured in watts (post 16.23). (Don’t confuse W, the SI unit of power, with *W*, the symbol I have chosen to use to represent work, see post 16.13). From this definition, electrical power, *P*, is given by

*P = W*/Δ*t* = *I*^{2}*R*. (2)

Unfortunately, people often talk about electrical “power” when they mean electrical energy. So, for example, a wind farm converts the kinetic energy (post 16.21) of moving air into electrical energy; its power is the rate at which it produces this energy. A “power station” produces electrical energy; the rate of energy production is its power.

A conventional electrical heater is simply a resistor (post 17.44) with a high value for *R*. According, to equation 1, a high value of *R* leads to release of more energy, mostly in the form of heat.

Let’s think about an ordinary electrical kettle, like the one in the picture above. Its purpose is to boil water. It is designed to hold water that surrounds a resistor called the *heating element*. My kettle has a power of 2.0 kW and draws current from a 220 V supply. (Giving a value to the domestic electrical voltage is a bit more complicated than I’m making it seem because it supplies what is called an *alternating current*, but this value gives a good guide to what is happening). I want to know (1) the resistance of the heating element and (2) the current drawn by the kettle. According to Ohm’s law, *V* = *IR* which is the same as *I* = *V*/*R* (post 17.44). Putting this expression for *I* into equation 2 gives

*P* = (*V*/*R*)^{2}*R* = *V*^{2}/*R*.

Multiplying both sides of this equation by *R* and dividing by *P* gives

*R* = *V*^{2}/*P*. (3)

Substituting the vales of *V* (220 V) and *P* (2.0 kW = 2.0 ×1 000 W, post 16.12) for my kettle into equation 3 gives

*R* = (220 ×220)/2000 = 24 Ω.

Here Ω is the standard symbol for the *ohm*, the unit of electrical resistance (post 17.44). If you wonder why I haven’t written the answer as 24.2 Ω, see post 16.7. We can now use Ohm’s law (see above) to calculate the value of *I* from the values of *V* and *R*; the result is

*I* = *V*/*R* = 220/24 = 9.1 A.

This is fortunate because the cabling that supplies my kettle can deliver currents up to 16 A.

Calculations, like the one above, are important for many aspects of electrical safety and circuit design. Once we understand the definition of an amp, a volt and an ohm (post 17.44), we can calculate energy and power requirements, and heating effects, for generators, domestic appliances and many other applications of electricity.

__Related posts__

17.44 Amps, volts and ohms

17.24 Fields and vectors

16.25 Electrical charge