Electrical power and energy, the power equations linking power to voltage, current and resistance, the energy equation, and using them to choose a suitable power rating for a component.

What this topic is asking

WJEC Eduqas wants you to calculate electrical power using the three power equations, calculate the energy transferred using $E = Pt$, and use a power calculation to choose a sensible power rating for a component so it does not overheat. These calculations recur throughout Component 1, especially when sizing resistors.

Electrical power

These two forms let you find the power from whatever values you have. Use $P = VI$ when you know the voltage and current; use $P = I^2R$ when you know the current and resistance. (A third form, $P = \dfrac{V^2}{R}$, follows by substituting $I = \dfrac{V}{R}$, useful when you know the voltage and resistance.) In a resistor this power is transferred to heat, which is why components and wires warm up when current flows.

Electrical energy

Energy is what the power delivers over time. Because $P = VI$, the energy can also be written $E = VIt$, but $E = Pt$ is the form on the equation list. Always work in seconds: convert minutes and hours first. This is the same energy idea used in physics, but in electronics it usually matters because the energy ends up as unwanted heat in a resistor or a transistor.

Choosing a power rating

Resistors are sold with standard power ratings such as $0.25\,\text{W}$, $0.5\,\text{W}$, $1\,\text{W}$ and higher. If a resistor will dissipate, say, $0.3\,\text{W}$, a $0.25\,\text{W}$ resistor would overheat, so you choose the $0.5\,\text{W}$ version, leaving a safety margin. Picking a rating that is too low is a real design fault that makes the component get hot and fail, so examiners often ask you to justify the rating you choose.

Try this

Q1. A heater element of resistance $20\,\Omega$ carries a current of $3.0\,\text{A}$. Calculate the power. [2 marks]

• Cue. $P = I^2R = (3.0)^2 \times 20 = 9.0 \times 20 = 180\,\text{W}$.

Q2. A $12\,\text{W}$ lamp is on for $2.0$ minutes. Calculate the energy transferred. [2 marks]

• Cue. $t = 120\,\text{s}$, so $E = Pt = 12 \times 120 = 1440\,\text{J}$.