Electrical power and the two power equations, the energy transferred by a charge and by a device over time, and how to calculate the energy used and its cost using kilowatt-hours.

What this topic is asking

OCR wants you to calculate electrical power using both power equations, find the energy transferred by a charge and by a device over time, and work out energy use and cost in kilowatt-hours. This is topic P3.5 of the OCR Gateway Physics A (J249) specification.

Electrical power

Use $P = VI$ when you know the voltage and current; use $P = I^2 R$ when you know the current and resistance. The two equations are consistent: substituting $V = IR$ into $P = VI$ gives $P = (IR)I = I^2 R$, so they describe the same power in different terms. The $P = I^2 R$ form shows that the power wasted as heat in a resistance grows with the square of the current: doubling the current quadruples the heating. This is the key reason the national grid transmits power at a low current (and high voltage), so that very little energy is wasted heating the transmission cables.

Energy transferred

So a $60\,\text{W}$ lamp left on for $100\,\text{s}$ transfers $E = Pt = 60 \times 100 = 6000\,\text{J}$. The equation $E = QV$ links back to the meaning of potential difference: each coulomb of charge transfers $V$ joules of energy. The two energy equations agree, because $E = QV$ with $Q = It$ gives $E = ItV = (VI)t = Pt$, since $P = VI$. In a resistor this transferred energy ends up heating the component and its surroundings, which is why a high-power device such as a kettle or heater warms up quickly.

Energy and cost at home

For household bills, joules are far too small, so energy is measured in kilowatt-hours ($\text{kWh}$), the energy used by a $1\,\text{kW}$ device running for one hour. The energy in kilowatt-hours is:

with the power in kilowatts and the time in hours. The cost is then the energy in kilowatt-hours multiplied by the price of one unit (the cost per kilowatt-hour). Keeping the power in kilowatts (not watts) and the time in hours (not seconds) is the key to getting these calculations right.

Try this

Q1. A lamp transfers $3000\,\text{J}$ of energy in $60\,\text{s}$. Calculate its power. [2 marks]

• Cue. Rearrange $E = Pt$: $P = \dfrac{E}{t} = \dfrac{3000}{60} = 50\,\text{W}$.

Q2. Calculate the power dissipated in a $5\,\Omega$ resistor carrying a current of $2\,\text{A}$. [2 marks]

• Cue. $P = I^2 R = 2^2 \times 5 = 20\,\text{W}$.