Work and power of current, Joule's law

Electric current does work: it lights a lamp, heats a kettle, turns a motor. In this topic we learn how to calculate the work and power of a current, Joule’s law (the Joule–Lenz law), the efficiency of a source and the cost of electrical energy. Every formula follows from Ohm’s law (I = U/R), so understand them rather than memorise them.

§1. Work done by a current

The electric field does work when it moves charge between points with potential difference U: A = qU. Since q = It:

Using Ohm’s law it can be written two more ways:

• A — work, in joules (J); U — voltage (V); I — current (A); t — time (s).
• A = UIt holds for any device, while I²Rt and U²t/R hold only for a purely resistive load where all energy turns into heat.
• In practice energy is measured in kilowatt-hours: 1 kW·h = 1000 W · 3600 s = 3.6·10⁶ J.

Worked example

A kettle on a 220 V mains draws 5 A for 10 minutes. How much work does the current do?

1. Time in seconds: t = 10·60 = 600 s.
2. A = UIt = 220·5·600 = 660 000 J = 660 kJ.

Worked example

An electric car (1200 kg, ε = 400 V) slows from 30 m/s to 5 m/s and returns the energy to its battery. How much charge flows back?

1. Change in kinetic energy: ΔE = ½·1200·(30² − 5²) = 600·875 = 525 000 J = 525 kJ.
2. From A = qε: q = 525 000/400 = 1312.5 C.

§2. Electrical power

Power is work per unit time:

The unit is the watt (W): 1 W = 1 J/s = 1 V·A.

Which formula to use depends on the connection:

• In series the current is the same, so P = I²R is convenient: the device with the larger resistance gets more power.
• In parallel the voltage is the same, so P = U²/R is convenient: the device with the smaller resistance gets more power.

The label on a device (e.g. "220 V, 100 W") gives rated values. From them you find resistance: R = U²/P. At a different voltage the power changes in proportion to U².

Worked example

Bulbs rated "220 V, 40 W" and "220 V, 100 W" are connected in series to 220 V. Which glows brighter?

1. Resistances: R₁ = 220²/40 = 1210 Ω, R₂ = 220²/100 = 484 Ω.
2. Current: I = 220/(1210 + 484) = 220/1694 ≈ 0.13 A.
3. Powers: P₁ = I²R₁ ≈ 0.01687·1210 ≈ 20.4 W, P₂ = I²R₂ ≈ 0.01687·484 ≈ 8.2 W.
4. Answer: the 40 W bulb is brighter — in series the larger resistance takes more power.

§3. Joule’s law (Joule–Lenz law)

As current flows, electrons collide with the ions of the crystal lattice and hand them energy — the conductor heats up.

The heat released is proportional to the square of the current, to the resistance and to the time. That is why power lines use high voltage and low current: cutting the current 10 times cuts the losses 100 times.

In heater problems write an energy balance:

where c is the specific heat capacity (for water 4200 J/(kg·K)).

Worked example

How long does a 1 kW heater with 80 % efficiency take to bring 2 kg of water from 20 °C to the boil?

1. Heat needed: Q = 4200·2·80 = 672 000 J.
2. Useful power: 0.8·1000 = 800 W.
3. t = 672 000/800 = 840 s = 14 min.

§4. Power in a complete circuit and source efficiency

For a source with internal resistance r: I = ε/(R + r). The total power of the source is P = εI, the useful (external) power is P_u = I²R = UI.

The external power P_u = ε²R/(R + r)² depends on the load and is greatest when `R = r`:

So maximum power and maximum efficiency do not occur together: as R grows, efficiency approaches 100 % but the power falls.

Worked example

ε = 12 V, r = 2 Ω. Find the maximum power in the external circuit.

1. Condition: R = r = 2 Ω, current I = 12/4 = 3 A.
2. P_max = I²R = 9·2 = 18 W (check: ε²/4r = 144/8 = 18 W).
3. Efficiency: η = 2/4 = 50 %.

§5. The cost of electricity

A meter records energy in kW·h. Method:

1. Convert power to kilowatts.
2. Convert time to hours.
3. Multiply E = Pt (kW·h) by the tariff.

Worked example

A 2 kW air conditioner runs 3 hours a day. The tariff is 1000 soums per kW·h. What is the bill for 30 days?

1. Per day: 2·3 = 6 kW·h; for 30 days 180 kW·h.
2. Cost: 180·1000 = 180 000 soums.

§6. The electric motor: useful work and heat

In a motor most of the electrical energy becomes mechanical work and a smaller part heats the windings. By conservation of energy:

Here R is the resistance of the windings. A spinning motor produces a back emf, so its current is much smaller than U/R. If the motor stalls (stops turning), the back emf disappears, the current rises to U/R and all the power turns into heat — the motor can burn out. That is why appliances with motors have fuses and thermal protection.

Worked example

A 12 V motor draws 2 A and its winding resistance is 1 Ω. Find the mechanical power, the efficiency and the speed at which it can lift a 10 kg load. (g = 10 m/s²)

1. Power drawn: UI = 12·2 = 24 W.
2. Heat: I²R = 4·1 = 4 W; mechanical power 24 − 4 = 20 W.
3. η = 20/24 ≈ 83 %.
4. P = mgv → v = 20/(10·10) = 0.2 m/s.

Check yourself

1. A current of 2 A flows through 10 Ω for 1 minute. How much heat is released?

2. Find the current and resistance of a "220 V, 60 W" bulb.

3. A 1.5 kW heater runs for 4 hours. Energy in kW·h and J?

4. Resistors of 2 Ω and 4 Ω are connected in parallel to 12 V. Power in each?

5. ε = 24 V, r = 1 Ω, R = 5 Ω. Current, useful power, efficiency?

6. A heater coil is shortened so its resistance halves (voltage unchanged). How does the power change?

7. A 220 V motor draws 5 A and its winding resistance is 4 Ω. Mechanical power and efficiency?

8. A 12 V motor with winding resistance 1 Ω stalls. Current and heating power?

Common mistakes

• Assuming the higher-resistance bulb is brighter in parallel too: in parallel P = U²/R, so the lower resistance is brighter.
• Treating rated power as fixed at any voltage: first find R = U²/P.
• Forgetting to convert kW·h to joules or minutes to seconds.
• Using A = U²t/R for a motor: that is valid only for pure resistance; for a motor use A = UIt.
• Thinking efficiency is greatest at maximum power: at R = r it is only 50 %.
• Calculating a motor’s current as I = U/R: in a running motor the back emf makes the current much smaller.