Power and the watt

1. The power of a component is the rate at which it transfers energy, which is the energy it transfers every second.
2. The unit of power is the watt, symbol $W$. A power of $1W$ means an energy transfer of $1J$ every second, so $1W=1{J\; s}^{-1}$.
3. A component of power $50W$ therefore takes $50J$ out of the electrical pathway during every second it is switched on.
4. Power says nothing at all about how long a device runs for. It is a rate, so a large power sustained for a short time and a small power sustained for a long time can transfer the same total energy.
5. The usual prefixes apply, with $1kW=1000W$ and $1MW=1\times 10^{6}W$, so $0.75kW=750W$ and $2400W=2.4kW$.
6. Values in a laboratory circuit run from about $0.10W$ for a small LED to a few watts for a filament lamp, while a mains heating element is measured in kilowatts.
7. In a circuit the energy is carried by the moving charge, so the power of a component depends on two things: how much charge passes through it each second, and how much energy each coulomb of that charge delivers.

Power from current and voltage

1. Each coulomb of charge passing through a component delivers a number of joules equal to the potential difference across it, measured in volts.
2. The current states how many coulombs pass through the component every second, measured in amperes.
3. Multiplying coulombs per second by joules per coulomb leaves joules per second, which is the power, so $P=IV$.
4. The same result follows from the energy relationship already established, because dividing the energy transferred by the time it took removes the time and leaves $P=IV$.
5. In this equation $P$ is the power in watts, $I$ is the current in amperes and $V$ is the potential difference in volts across the component.
6. The rearrangements are $I=\frac{P}{V}$ and $V=\frac{P}{I}$.
7. The current and the potential difference must both belong to the same component, because the equation gives the power of whatever component those two readings describe.

Power in a resistance

1. When the resistance of a component is known but the potential difference across it is not, a second form of the power equation is more direct.
2. The potential difference across a resistance $R$ carrying a current $I$ is given by $V=IR$.
3. Substituting that expression for $V$ into $P=IV$ gives $P=I\times IR$.
4. Collecting the two current terms gives $P=I^{2}R$, where $P$ is in watts, $I$ is in amperes and $R$ is in ohms.
5. Only the current is squared in this equation. The resistance is not squared, and the current must be squared before it is multiplied by the resistance.
6. The rearrangements are $R=\frac{P}{I^2}$ and $I=\sqrt{\frac{P}{R}}$.
7. This form is the natural one for a resistor, a heating element or a length of cable, because for those components it is the resistance that is quoted rather than the potential difference across them.

Doubling the current

1. Because the current appears squared in $P=I^{2}R$, a change in the current has a far larger effect on the power than the same fractional change in the resistance.
2. Doubling the current in a fixed resistance multiplies the power dissipated by $2^2=4$, so the power becomes four times as large.
3. Trebling the current multiplies the power by $3^2=9$, and halving the current reduces the power to $\frac{1}{4}$ of its earlier value.
4. Doubling the resistance at a fixed current only doubles the power, because the resistance is not squared.
5. This squared dependence is why the current is the quantity watched most closely wherever dissipation matters: a modest rise in current brings a steep rise in the power dissipated.
6. Squaring also removes any sign, so the power calculated is positive whichever direction the current is in.

Choosing the right equation

1. The data in a question decide which form to use, so the first step is to list the quantities that are known with their units.
2. A current together with a potential difference points straight to $P=IV$.
3. A current together with a resistance points straight to $P=I^{2}R$.
4. A potential difference together with a resistance is handled by finding the current first from $I=\frac{V}{R}$, after which either form can be used.
5. The two forms describe the same physical quantity, so a power found one way can be confirmed the other way, which is a fast check on the arithmetic.
6. Units are converted before substitution: a power in kilowatts becomes watts, a resistance in kilohms becomes ohms, and a current in milliamperes becomes amperes using $1mA=1\times 10^{-3}A$.
7. The answer is normally quoted in watts to two or three significant figures, and converted to kilowatts only when the question asks for that unit.