The number on a light globe, a kettle or a phone charger is its power, the rate it turns electrical energy into light, heat or motion. Two short equations let you predict that rate from the voltage and current, and explain why a 2400 W kettle boils water so much faster than a 60 W lamp ever could.
Two heaters run from the same wall socket. One is rated 1000 W and the other 2000 W.
Before reading: which heater draws more current, and why? Write your prediction and a one-sentence reason.
Why power equals voltage times current
Voltage is energy delivered per unit charge and current is charge delivered per second, so multiplying them gives energy delivered per second, which is exactly power.
Recall that voltage is energy per charge ($V = E/q$) and current is charge per time ($I = q/t$). Multiplying the two:
$$P = VI = \frac{E}{q} \times \frac{q}{t} = \frac{E}{t}$$
So $P = VI$ is just energy per unit time, the definition of power. A device with a large voltage across it and a large current through it transforms energy quickly, which is why it has a high power rating.
Power is the rate of energy transformation, $P = E/t$, measured in watts. For an electrical component $P = VI$, because voltage is energy per charge and current is charge per second.
Pause, copy the highlighted statement into your book before moving on.
Did you get this? True or false: power is the amount of energy transformed, measured in joules.
From a rating and a voltage to the current drawn
We just saw that $P = VI$. That raises a question: how do we use it on a real appliance? This card answers it → rearrange for the unknown and substitute the appliance's rating and supply voltage.
Because $P = VI$ links three quantities, knowing any two lets you find the third, the same list-rearrange-substitute routine as Ohm's law.
Most appliances quote a power rating and run from a known supply voltage, so the usual task is to find the current: $I = P/V$.
A toaster is rated $1200\ \text{W}$ and runs from the $240\ \text{V}$ mains. What current does it draw?
Use $P = VI$ to relate power, voltage and current. Rearrange to $I = P/V$ to find the current an appliance draws, or $V = P/I$ for the voltage. A higher power rating at fixed voltage means a larger current.
Add the highlighted method to your notes before the check below.
Quick check: A globe draws $0.50\ \text{A}$ from a $12\ \text{V}$ supply. What is its power?
Pattern A, Calculate then compare
A $240\ \text{V}$ kettle is rated $2400\ \text{W}$ and a $240\ \text{V}$ phone charger is rated $25\ \text{W}$. In your book: (1) calculate the current each draws using $P = VI$; (2) using $R = V/I$, work out each one's resistance; (3) write one sentence explaining why the higher-power kettle has the lower resistance.
Fill the blanks: Complete the link between power and energy.
Power is the at which energy is transformed, measured in watts. The energy transformed by a device of power $P$ running for a time $t$ is $E = P \times$ .
Dot point: Relate power to energy transformations
Combining the power and Ohm's-law equations
We just learned to use $P = VI$. That raises a question: what if we know the voltage and the resistance but not the current? This card answers it → substitute $I = V/R$ into $P = VI$ to remove $I$.
Whenever you know two of voltage, current and resistance, you can rewrite the power equation to use only the quantities you actually have.
Starting from $P = VI$ and substituting $I = V/R$ from Ohm's law:
$$P = VI = V \times \frac{V}{R} = \frac{V^2}{R}$$
This form is useful when you know the voltage across a component and its resistance. It shows that, for a fixed resistance, power rises with the square of the voltage, halving the voltage cuts the power to a quarter.
Combining $P = VI$ with $V = IR$ gives $P = V^2/R$. For a fixed resistance, power is proportional to the square of the voltage: doubling $V$ multiplies the power by four.
Pause, copy the highlighted relationship into your book before the check below.
Last check: A $30\ \Omega$ heating element has $60\ \text{V}$ across it. Using $P = V^2/R$, what power does it dissipate?
Two truths, one lie: three of these statements about $P = VI$ are correct. Which one is the lie?
Dot point: solve problems using P = VI
A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct
Pick your answer, then rate your confidence, that tells the system what to drill next.
ApplyBand 4(4 marks) 1. A $240\ \text{V}$ heater has a heating element of resistance $24\ \Omega$. Calculate the power it dissipates and the current it draws, and explain which equation you used at each step.
1 mark: selects $P = V^2/R$ · 1 mark: $P = 2400\ \text{W}$ · 1 mark: uses $P = VI$ (or $I = V/R$) for current · 1 mark: $I = 10\ \text{A}$ with reasoning
ApplyBand 4(3 marks) 2. A $1200\ \text{W}$ heater is left running for $5.0\ \text{minutes}$. Calculate the energy it transforms, and state the main energy transformation taking place.
1 mark: converts $5.0\ \text{min}$ to $300\ \text{s}$ · 1 mark: $E = Pt = 3.6 \times 10^{5}\ \text{J}$ · 1 mark: electrical energy is transformed into thermal energy
Dot point: Relate power to energy transformations
ApplyBand 4(3 marks) 3. An electric kettle is rated $2400\ \text{W}$ on a $240\ \text{V}$ supply. Calculate the current it draws, and determine the current a $1000\ \text{W}$ toaster would draw from the same supply.
1 mark: rearranges $P = VI$ to $I = P/V$ · 1 mark: kettle draws $10\ \text{A}$ · 1 mark: toaster draws $4.2\ \text{A}$ (2 s.f.)
Dot point: solve problems using P = VI
MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set from the lesson bank.
Q1 (4 marks): Because the voltage and resistance are known, the power is found with $P = V^2/R = (240)^2 / 24 = 57600/24 = 2400\ \text{W}$. The current is then found from $P = VI$, rearranged to $I = P/V = 2400/240 = 10\ \text{A}$ (equivalently $I = V/R = 240/24 = 10\ \text{A}$). $P = V^2/R$ was used first because the unknown current was not yet available; once the power was known, $P = VI$ gave the current directly.
Q2 (3 marks): Power is the rate at which energy is transformed, so the total energy transformed is $E = Pt$. The time must be in seconds: $5.0\ \text{minutes} = 300\ \text{s}$. Then $E = 1200 \times 300 = 3.6 \times 10^{5}\ \text{J}$. The transformation is from electrical energy, delivered by the supply to the charge carriers, into thermal energy in the heating element, where collisions between the drifting electrons and the lattice raise its temperature. A small amount is also transformed into light if the element glows.
Q3 (3 marks): Rearranging $P = VI$ gives $I = P/V$. For the kettle, $I = 2400/240 = 10\ \text{A}$. For the toaster on the same supply, $I = 1000/240 = 4.17$, which is $4.2\ \text{A}$ to two significant figures. Because the supply voltage is common to both, the current each draws is directly proportional to its power rating, which is why the higher-rated kettle draws almost two and a half times the current of the toaster.
You can now find electrical power with $P = VI$ and $P = V^2/R$. Next lesson uses $P = I^2R$ to explain the energy lost as heat in the wires of a circuit.
↩ Back to focus-area hubAt the start you predicted which heater draws more current. The answer is the 2000 W heater: at the same voltage, $I = P/V$ so higher power means higher current. If you thought they draw the same current, revisit Card 2, current scales with the power rating when the voltage is fixed.
A short spaced-retrieval set, mixing this lesson with earlier ones. It gets harder to guess and easier to remember.
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