Electrical work power and Joule-Lenz law

When a current flows through a conductor, the electric field does work moving charges. If a charge $q = It$ passes through a cross-section during time $t$ at voltage $U$, the work of the current is $A = UIt.$ Power is work per unit time: $P = \frac{A}{t} = UI,$ and using Ohm's law $U = IR$ (or $I = U/R$) it is convenient to rewrite this in two other forms: $P = I^2R = \frac{U^2}{R}.$ In a stationary metal conductor all of this work turns into internal energy, that is, into heat, so $Q = A = I^2Rt$ — this is the Joule–Lenz law. In a motor, a battery, or a charger, part of the energy becomes useful work, so the useful power is less than the total power, and the difference $P - P_{\text{useful}}$ is dissipated as heat. Work is measured in joules ($1\ \text{J} = 1\ \text{V}\cdot\text{A}\cdot\text{s}$), while household energy consumption is conveniently counted in kilowatt-hours: $1\ \text{kW}\cdot\text{h} = 3{.}6\cdot 10^{6}\ \text{J}$.

An electric kettle operates at a voltage $U = 220$ V with a current $I = 5{.}0$ A for a time $t = 120$ s. Find the power of the kettle and the work of the current during this time.

A heating element with resistance $R = 24$ Ω is connected to a circuit, and the current through it is $I = 2{.}5$ A. How much heat is released in the element in $t = 5{.}0$ min? What is its power?

An electric motor is connected to a $U = 220$ V mains and draws a current $I = 3{.}5$ A. Its efficiency is $\eta = 0{.}85$. Find the useful mechanical power and the heat released in the motor in $t = 1{.}0$ min (assume all losses are thermal).

A safe self-study plan (do not take appliances apart, do not touch mains wires or sockets, and do not build homemade circuits at 220 V):

1. Nameplate data. Make a table for three household appliances: name, rated power $P$ from the nameplate, and typical operating time $t$ per day.
2. Energy calculation. For each appliance compute $A = Pt$, convert to kilowatt-hours using $1\ \text{kW}\cdot\text{h} = 3{.}6\cdot 10^{6}\ \text{J}$, and estimate the daily consumption. Compare the appliances: how many times more energy does the most powerful one use?
3. Simulation. In a virtual lab (PhET “Circuit Construction Kit” or similar), build a circuit with a source, a resistor, an ammeter, and a voltmeter. Measure $I$ and $U$ for two different resistances, compute $P = UI$, and compare it with $I^2R$; then double the current at the same $R$ and confirm that the power grows by a factor of 4.
4. Conclusion. Write in your notebook when $Q = I^2Rt$ describes the entire energy consumption of a device and when it describes only a part of it.

Show that in a purely resistive element all the work of the current becomes heat ($A = Q$), whereas in a motor only the fraction $(1-\eta)$ of the energy becomes heat, and that at constant $R$ the released power grows as the square of the current.

A two-panel diagram.

Left — an energy-flow (Sankey-style) diagram: a block “electrical energy $A = UIt$” with a wide arrow leading to a block “internal energy, heat $Q = I^2Rt$” labelled $Q = A$ (resistor). Below it, a reduced version for a motor: the flow splits into a wide arrow “useful work $A_{\text{useful}} = \eta A$” and a narrow arrow “heat losses $(1-\eta)A$”.

Right — a graph of $Q$ versus $t$: a straight line from the origin with slope $P = I^2R$; next to it, a dashed second line for a doubled current in the same circuit — it is 4 times steeper, illustrating the quadratic growth.

A kettle is plugged into a $U=220\ \text{V}$ mains supply; its heating element has a power of $P=2\ \text{kW}$, so the current through it is $I=P/U=2000/220\approx9.1\ \text{A}$. In $t=4\ \text{min}=240\ \text{s}$ the current does work $A=UIt=220\cdot9.1\cdot240\approx480\ \text{kJ}$; with $c=4200\ \text{J}/(\text{kg}\cdot^{\circ}\text{C})$ that is enough to heat $1.5\ \text{L}$ of water by about $76\ ^{\circ}\text{C}$.

Here the work of the current is calculated as $A=UIt$ and the power as $P=UI$. The element's resistance $R=U^2/P\approx24.2\ \Omega$ rewrites the same power in Joule form $P=I^2R$, and the Joule–Lenz law $Q=I^2Rt$ shows that nearly all the work of the current becomes heat in the element itself, not in the supply wires.

A charger delivers $U=5\ \text{V}$ at a current $I=2\ \text{A}$, i.e. a power $P=UI=10\ \text{W}$. A battery of capacity $4000\ \text{mA}\cdot\text{h}$ at $3.7\ \text{V}$ stores $W=3.7\ \text{V}\cdot4\ \text{A}\cdot\text{h}=14.8\ \text{W}\cdot\text{h}\approx53\ \text{kJ}$. A full charge takes about $2\ \text{h}=7200\ \text{s}$, and during that time the mains supplies $A=UIt=10\cdot7200=72\ \text{kJ}$.

This is a direct calculation of the work of the current $A=UIt$ and the power $P=UI$ over a long time. The difference $72-53\ \text{kJ}$ goes into heating the phone and losses in the converter, so the charging efficiency is $\eta=53/72\approx0.74$; part of that power is dissipated according to the Joule–Lenz law, which is why the body warms up.

A trolleybus motor runs at $U=600\ \text{V}$ and a current $I=200\ \text{A}$, so its power is $P=UI=120\ \text{kW}$, and in $t=30\ \text{s}$ the work done is $A=Pt=3.6\ \text{MJ}$. The current flows through the overhead wire and the rails, whose combined resistance is $R\approx0.34\ \Omega$, so by the Joule–Lenz law they dissipate $P_{\text{loss}}=I^2R=200^2\cdot0.34\approx13.6\ \text{kW}$ — about $11\,\%$ of the motor power.

The same power is written in two ways: $P=UI$ for the useful work of the motor and $P=I^2R$ for the useless heating of the wire. This is exactly the content of the Joule–Lenz law: at a given current the losses grow as the square of the current, which is why transmission lines raise the voltage and lower the current.