Define energy (joules) vs power (watts): Worked examples

This page is a drill through every kind of question the energy–power relationship can throw at you. We start by mapping out all the case classes on a single grid, then work one example for each cell. If you have not yet met $P = E / t$ or $P = I V$ , read the parent first: the parent topic builds them from scratch.

Figure — Define energy (joules) vs power (watts)

Look at the picture: energy is the grey area of a rectangle whose width is time and whose height is power. Solving for the missing quantity is just "which side of the rectangle don't I know yet?"

The scenario matrix

Here is every class of problem this topic can produce. Each later example is tagged with the cell it fills.

#	Case class	What is missing / special	Example
A	Find energy from steady power × time	$E$ unknown, unit conversion of $t$	Ex 1
B	Find power from energy ÷ tiny time	$P$ unknown, huge rate (limiting)	Ex 2
C	Find time from energy ÷ power	$t$ unknown, answer in hours	Ex 3
D	Electrical route via $P = I V$	swap $P$ for $I \times V$ , then $E = P t$	Ex 4
E	Zero / degenerate input	$t = 0$ , or $P = 0$ , or infinite time	Ex 5
F	Comparison twist — low power beats high power	two devices, energy vs power confusion	Ex 6
G	Real-world word problem — the bill	kWh unit, money attached	Ex 7
H	Exam-style twist — varying power (area under graph)	power not constant, need the area	Ex 8

Notice there are no "negative" cases here: energy delivered, time elapsed, and power drawn by a device are all $\geq 0$ . That is the sign story for this topic — unlike angles or vectors, nothing goes below zero. We do still cover the boundary $= 0$ (Ex 5), because that is where formulas can quietly break.

Example 1 — Cell A: energy from power × time

Worked example Immersion heater

A $1500 W$ immersion heater runs for $4$ minutes. How much energy does it deliver?

Forecast: Guess the order of magnitude first — is it hundreds, thousands, or hundreds of thousands of joules?

Convert time to seconds. $t = 4 min = 4 \times 60 = 240 s$ . Why this step? The watt is joules per second. If you leave $t$ in minutes, every joule gets counted $60 \times$ too few times.
Apply $E = P t$ . This is the missing-corner rule: we have height (power) and width (time), we want the rectangle area (energy). $E = 1500 \times 240 = 360, 000 J = 360 kJ .$ Why this step? Power is steady, so energy is exactly the rectangle in the figure — no integral needed.

Verify: Units check — $W \times s = \frac{J}{s} \times s = J$ ✓. Sanity: $360 kJ$ is roughly the energy to heat a mug of water — sounds right for 4 minutes of a hot heater.

Example 2 — Cell B: power from a tiny time (limiting behaviour)

Example 3 — Cell C: time from energy ÷ power

Example 4 — Cell D: the electrical route ( $P = I V$ , then $E = P t$ )

Worked example USB charger, two steps

A USB port supplies $I = 3 A$ at $V = 5 V$ for $20$ minutes. Find (a) the power, (b) the energy delivered.

Forecast: Which do you compute first — power or energy? Why can't you jump straight to energy?

Get power from $P = I V$ . We use $I V$ (not $E / t$ ) because we were handed current and voltage, not energy. $P = I V = 3 \times 5 = 15 W .$ Why this step? Recall $E = Q V$ and $I = Q / t$ , so $P = E / t = Q V / t = (Q / t) V = I V$ — see Current as charge per second (I = Q/t) and Voltage as energy per charge (V = J/C). Power is the natural first result from $I$ and $V$ .
Convert time. $t = 20 min = 1200 s$ . Why this step? Watt is per second.
Get energy from $E = P t$ . $E = 15 \times 1200 = 18, 000 J = 18 kJ .$ Why this step? Now that we know the rate, multiply by how long it ran.

Verify: Units: $A \times V = \frac{C}{s} \times \frac{J}{C} = \frac{J}{s} = W$ ✓. Energy: $W \times s = J$ ✓.

Example 5 — Cell E: zero and degenerate inputs

Worked example The boundary cases

Answer three edge questions. (a) A device is switched on for $t = 0 s$ . Energy delivered? (b) A perfectly ideal switch that is off draws $P = 0 W$ for $2 h$ . Energy? (c) A $10 W$ nightlight left on forever. Energy?

Forecast: In which of these does a formula "break", and which just give zero?

(a) $t = 0$ : $E = P t = P \times 0 = 0 J$ . Why this step? The rectangle has zero width, so zero area — no time, no energy, regardless of how big $P$ is. No division involved, nothing breaks.
(b) $P = 0$ : $E = 0 \times 7200 = 0 J$ . Why this step? The rectangle has zero height. An off device transfers no energy however long you wait.
(c) $t \to \infty$ : $E = 10 \times t \to \infty$ . Why this step? A finite, non-zero rate over unbounded time grows without limit — energy diverges. This is why "always on" appliances dominate a bill even at low wattage.
The one that would break: $t = E / P$ with $P = 0$ gives $\frac{E}{0}$ — undefined. A zero-power device can never deliver a required non-zero energy, so "how long?" has no answer. Why this step? Division by zero is the topic's only genuine degeneracy — always guard against $P = 0$ before using $t = E / P$ .

Verify: (a) $1500 \times 0 = 0$ ✓, (b) $0 \times 7200 = 0$ ✓, (c) grows past any bound (e.g. after $1 0^{6} s$ , $E = 1 0^{7} J$ ) ✓.

Example 6 — Cell F: the comparison twist (low power wins)

Worked example Which uses more energy?

Bulb X: $100 W$ , on for $1$ minute. Bulb Y: $40 W$ , on for $8$ hours. Which uses more energy?

Forecast: Higher wattage feels like "more". Trust that feeling or not?

Energy of X. $t_{X} = 60 s$ , so $E_{X} = 100 \times 60 = 6, 000 J = 6 kJ$ . Why this step? Power alone can't answer "how much" — we need $E = P t$ with its own time.
Energy of Y. $t_{Y} = 8 \times 3600 = 28, 800 s$ , so $E_{Y} = 40 \times 28, 800 = 1, 152, 000 J = 1152 kJ$ . Why this step? Same rule, but the long runtime multiplies the small power into a big total.
Compare. $E_{Y} = 1152 kJ ≫ E_{X} = 6 kJ$ . The 40 W bulb uses far more energy. Why this step? Energy $=$ power $\times$ time. Time can swamp power. Power $\neq =$ energy.

Verify: Ratio $E_{Y} / E_{X} = 1, 152, 000/6, 000 = 192$ — Y uses $192 \times$ more ✓. This is the parent's "40 W beats 100 W" mistake, made concrete.

Look at the two rectangles: X is tall and razor-thin; Y is short but enormously wide. Area = energy, and Y's area dwarfs X's.

Example 7 — Cell G: the electricity bill (real-world, kWh)

Worked example What the fridge costs

A fridge draws $150 W$ continuously for a $30$ -day month. Electricity costs $\$ 0.20$ per kilowatt-hour. What is the monthly cost?

Forecast: Guess the bill — cents, a dollar, or several dollars?

Power in kilowatts. $P = 150 W = 0.15 kW$ . Why this step? The billing unit uses kilowatts, so match it.
Time in hours. $t = 30 \times 24 = 720 h$ . Why this step? A kilowatt-hour multiplies kW by hours, so keep time in hours here (the one time we don't convert to seconds — because the unit itself already bakes in the hour).
Energy in kWh. $E = P t = 0.15 \times 720 = 108 kWh$ . Why this step? Same $E = P t$ rule, just in bill-friendly units. See Kilowatt-hour and electricity billing.
Cost. $\text{cost} = 108 \times 0.20 = \$ 21.60 $. * W h y t hi ss t e p ? * M o n ey$ = $e n er g y$ \times$ price-per-unit-energy.

Verify: Cross-check in joules: $108 kWh = 108 \times 3.6 MJ = 388.8 MJ$ , and $0.15 kW \times 720 \times 3600 s = 388, 800, 000 J$ ✓. Cost $\$ 21.60$ ✓.

Example 8 — Cell H: exam twist, power that changes (area under the graph)

Worked example A motor that ramps up

A motor's power rises linearly from $0 W$ at $t = 0$ to $200 W$ at $t = 10 s$ , then holds steady at $200 W$ from $t = 10 s$ to $t = 30 s$ . Total energy?

Forecast: You cannot just multiply one power by one time. Why not — and what replaces it?

Recognise: power is not constant, so energy is the area under the power–time graph, not a single rectangle. Why this step? From the parent, $E = \int P d t$ ; the integral is the area. When $P$ is steady you get a rectangle; when it varies you sum the shapes.
Phase 1 (ramp, a triangle). Base $10 s$ , height $200 W$ : $E_{1} = \frac{1}{2} \times 10 \times 200 = 1000 J .$ Why this step? A straight-line ramp encloses a triangle; area $= \frac{1}{2} \times base \times height$ .
Phase 2 (steady, a rectangle). Width $30 - 10 = 20 s$ , height $200 W$ : $E_{2} = 200 \times 20 = 4000 J .$ Why this step? Constant power → ordinary rectangle.
Add the pieces. $E = E_{1} + E_{2} = 1000 + 4000 = 5000 J = 5 kJ .$ Why this step? Total area = sum of the two regions.

Verify: Average power over the ramp is $\frac{0 + 200}{2} = 100 W$ for $10 s$ → $1000 J$ ✓; steady part $4000 J$ ✓; total $5000 J$ ✓.

The shaded region is the energy: a triangle (the ramp) glued to a rectangle (the steady part). Add the two areas — that is all "integrate the power" means here.

Recap: the whole matrix in one breath

Recall Which formula for which missing corner?

Missing energy → use $E = P t$ (area). Missing power → use $P = E / t$ or $P = I V$ . Missing time → use $t = E / P$ (guard $P \neq = 0$ ). Power varies → energy = area under the $P$ – $t$ graph.

Missing corner is energy

E = P t

— the rectangle's area.

Missing corner is power

P = E / t

, or

P = I V

if given current and voltage.

Missing corner is time

t = E / P

, and never allow

P = 0

Power changes over time

Energy is the area under the power–time graph (triangle + rectangle).

Connections

Voltage as energy per charge (V = J/C)
Current as charge per second (I = Q/t)
Ohm's Law and P = I²R = V²/R
Electric charge and the coulomb
Kilowatt-hour and electricity billing
Work-energy theorem (mechanics)