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

(a) A quantity delivered every second is a rate → power, unit watt (W). The value is $1500\text{W}$. (b) A total stored amount is energy, unit joule (J). The value is $40000\text{J}$. (c) A total used amount is energy again (the bill measures it in kWh, still an energy unit). Unit: joule, or in billing units, the kilowatt-hour. Rule of thumb: the words per second / rate / how fast ⇒ power; the words total / stored / used up ⇒ energy.

• $\text{J}$ = joule = energy.
• $\text{W}$ = watt = power.
• $\text{J/s}$ = joules per second = the definition of the watt = power.
• $\text{kWh}$ = kilowatt-hour = power × time = energy.
• $\text{N⋅m}$ = newton-metre = force × distance = energy (this is literally $1\text{J}$).
• $\text{W⋅s}$ = watt-second = power × time = energy (this is literally $1\text{J}$). The trick: any unit that is (rate) alone is power; any unit that is (rate × time) or (force × distance) is energy. See Electric charge and the coulomb and Work-energy theorem (mechanics) for where N·m comes from.

Convert time: $4\text{min}=4\times 60=240\text{s}$ (a watt is joules per second). $E=Pt=30\text{W}\times 240\text{s}=7200\text{J}=7.2\text{kJ}$ Picture: on the "energy = area" chart below, this is a rectangle of height $30\text{W}$ and width $240\text{s}$; its area is the joules.

$P=\frac{E}{t}=\frac{2\text{J}}{0.0005\text{s}}=4000\text{W}=4\text{kW}$ A tiny total of energy ($2\text{J}$, less than lifting a book) but in half a millisecond ⇒ a large rate. On the area picture this is a very thin, very tall rectangle: same small area, but the height (power) shoots up because the width (time) is nearly zero.

(a) Using the power law $P=IV$ we derived at the top (power = charge-per-second × energy-per-charge): $P=IV=5\text{V}\times 0.5\text{A}=2.5\text{W}$ (b) $t=10\text{min}=600\text{s}$: $E=Pt=2.5\text{W}\times 600\text{s}=1500\text{J}=1.5\text{kJ}$

$t=\frac{E}{P}=\frac{86400\text{J}}{12\text{W}}=7200\text{s}$ Convert: $7200\text{s}÷3600=2\text{h}$.

Compute each energy with times in seconds.

• Bulb A: $t=60\text{s}$, $E_A=100\times 60=6000\text{J}$.
• Bulb B: $t=8\times 3600=28800\text{s}$, $E_B=40\times 28800=1152000\text{J}$. Bulb B uses far more energy, despite lower power: $\frac{E_B}{E_A}=\frac{1152000}{6000}=192.$ What the figure shows: the bar chart below plots the two energies side by side. The orange bar (Bulb A, high power) is a sliver; the teal bar (Bulb B, low power) towers over it — a visual proof that a lower-power device left running long enough dwarfs a brief high-power burst. The purple arrow marks the $192\times$ gap. This is the whole point: higher power does not mean higher energy — time is the deciding factor.

Same total energy, different times ⇒ different power.

• Charger X: $t=3600\text{s}$, $P_X=36000/3600=10\text{W}$.
• Charger Y: $t=600\text{s}$, $P_Y=36000/600=60\text{W}$. Meaning: both deliver the identical "how much" (the same bucket of energy), but Y pours it $6\times$ faster, so Y needs a beefier wire and runs hotter. Energy = the bill; power = whether the wire melts. On the area picture these are two rectangles of equal area — X is long-and-short, Y is short-and-tall.

Power depends on the square of voltage: $P=V^2/R$. If $V\to 2V$: $P_{\text{new}}=\frac{(2V{)}^2}{R}=\frac{4V^2}{R}=4\times \frac{V^2}{R}=4P.$ Doubling the voltage quadruples the power, so four times the heat energy is released each second. (Not double — the square is the catch.) See Ohm's Law and P = I²R = V²/R.

(a) Current is charge per second. $t=2\text{min}=120\text{s}$: $I=\frac{Q}{t}=\frac{240\text{C}}{120\text{s}}=2\text{A}.$ (b) Energy is charge moved through a voltage (we derived $E=QV$ at the top, from $V=E/Q$): $E=QV=240\text{C}\times 5\text{V}=1200\text{J}.$ (c) Two routes: $P=\frac{E}{t}=\frac{1200\text{J}}{120\text{s}}=10\text{W},P=IV=2\text{A}\times 5\text{V}=10\text{W}.✓$ They match — because $P=E/t=QV/t=(Q/t)V=IV$ is the same statement seen two ways (exactly the derivation in the flow figure at the top).

(a) $P=IV=0.4\text{A}\times 9\text{V}=3.6\text{W}$. (b) Rearranging $P=I^{2}R$ gives $R=P/I^2$: $R=\frac{3.6\text{W}}{(0.4\text{A}{)}^2}=\frac{3.6}{0.16}=22.5\Omega .$ (c) $t=5\text{min}=300\text{s}$: $E=Pt=3.6\text{W}\times 300\text{s}=1080\text{J}.$ Because the resistor receives energy and turns it to heat, $P$ here is positive in our sign convention.

Set the lifting work equal to the kettle's energy and solve for height: $E=mgh\Rightarrow h=\frac{E}{mg}=\frac{180000\text{J}}{2\text{kg}\times 9.8{\text{m/s}}^2}=\frac{180000}{19.6}\approx 9184\text{m}.$ That is roughly the height of Mount Everest plus a kilometre — a striking reminder that boiling water is energy-hungry, and that one joule is one joule whether it lifts a laptop or heats water.

(a) $1\text{kWh}=1000\text{W}\times 3600\text{s}=3.6\times 10^6\text{J}$. So $E=60\times 3.6\times 10^6=2.16\times 10^8\text{J}=216\text{MJ}.$ (b) Stay in convenient units: energy $÷$ power. $t=60\text{kWh}/20\text{kW}=3\text{h}$. (The kWh/kW cancel to hours automatically — that's why the hour is baked into kWh. See Kilowatt-hour and electricity billing.) (c) $t=60\text{kWh}/150\text{kW}=0.4\text{h}=0.4\times 60=24\text{min}$.

(a) $t=10\text{min}=600\text{s}$, so $P=E/t=1500/600=2.5\text{W}$. (b) From $P=IV$, $I=P/V=2.5/5=0.5\text{A}$. (c) Required $0.5\text{A}>$ rated $0.4\text{A}$, so no — the wire is NOT safe: it must carry more current than it is rated for, and would overheat. You would need a thicker wire (higher current rating) or a higher supply voltage (which lowers the current for the same power, since $I=P/V$). This is exactly the "does the wire melt?" question that power/current — not total energy — answers.

(a) $P=E/t$. As $t\to \infty$ with finite $E$, the fraction $\to 0$: average power approaches zero. Any finite energy spread over forever is delivered infinitely slowly. (b) Same formula, opposite limit: as $t\to 0$ with finite $E$, the denominator vanishes and $P=E/t\to \infty$: power blows up without bound. This is why a camera flash or a capacitor pulse (L2.2) reaches enormous power from a tiny energy — squeeze the same joules into a shorter and shorter time and the rate climbs without limit. Delivering real energy in literally zero time is physically impossible precisely because it would demand infinite power. (c) $P=IV$ with $V=0$ gives $P=I\times 0=0\text{W}$. Even a large current dissipates no power if there is no voltage drop — nothing "falls through" a potential, so no work is done as heat. This is the whole appeal of superconductors. (d) $P=IV$ with $I=0$ gives $P=0\times V=0\text{W}$. No charge is moving, so no energy is being transferred right now — even though energy remains stored. Stored energy ≠ power being drawn.

1. State whether kWh is energy or power, and why ::: Energy — it is power × time (kW × h), a total, not a rate.
2. Why must time be in seconds for $E=Pt$? ::: Because a watt is defined as joules per second.
3. If voltage across a fixed resistor doubles, power does what? ::: Quadruples, since $P=V^2/R$ depends on $V$ squared.
4. Power dissipated in a superconductor ($V=0$)? ::: Zero, since $P=IV=I\times 0=0$.
5. As delivery time shrinks toward zero for fixed energy, power does what? ::: Grows without bound (tends to infinity), since $P=E/t$.
6. What does a negative value of $P$ mean? ::: The device is supplying energy outward, not receiving it — same size, opposite direction.