Physics · Electricity and magnetism · Grade 8

Work and power of an electric current

Which appliance “eats” most of the monthly electricity bill?

The situation

At the end of the month the electricity bill arrives at the house: “412 kWh”. How much energy is that, and which appliance “eats” most of it — the kettle that boils several times a day, the air conditioner that runs non-stop in summer, or the lights that come on in the evening? Why does the meter disc sometimes turn slowly and sometimes very fast? Switch on appliances in the flat, then measure and calculate.

Topic

Work and power of an electric current. Kettle, air conditioner, light bulb: monthly kWh bill

Key ideas

Power P = UI
Energy, kWh
Joule–Lenz law

Formulas

P = U I, P_{\sim} = U I \cos φ

Electric power: a 2 kW kettle at 220 V draws a current $I = P / U \approx 9.1$ A. In alternating current $U$ and $I$ are effective (rms) values, and $\cos φ$ is the power factor.

A = P t, 1 kWh = 3.6 \cdot 10^{6} J

The work of the current (the energy consumed). The count is kept in kilowatt-hours.

Q = I^{2} R t

The Joule–Lenz law: the heat released in a conductor. The wires and the lamp filament both get hot because of it.

η = \frac{P_{u s e f u l}}{P_{i n}}

Efficiency: for a lamp the useful part is the light.

P = \frac{n}{C t}

Power from the meter: $n$ is the number of disc revolutions, $C = 600$ rev/kWh, $t$ is the time (hours).

Theory

Work and power

When a charge $q$ is moved under voltage $U$ , the current does work $A = U q = U I t$ . The work done in one second is the power: $P = U I$ . The “2 kW” written on an appliance is its power at 220 V. If the power is large, the current is large too: a kettle draws ≈ 9.1 A, an LED lamp only ≈ 0.05 A.

The kilowatt-hour and the monthly bill

The energy consumed is $A = P t$ . If an appliance of 1 kW runs for 1 hour, it uses 1 kWh = 3.6 MJ of energy. The monthly bill = the power of each appliance × its daily running time × 30 days. A powerful appliance that runs only briefly (a kettle) and a weak appliance that is always on can use the same energy — compare the table. Usually the biggest consumer is the air conditioner.

The Joule–Lenz law and LED lamps

When a current passes, a conductor gets hot: $Q = I^{2} R t$ . In wires this is a loss (in the flat’s wiring $I^{2} R$ is shown in the HUD), while in an incandescent lamp the light is obtained in just this way: the filament heats up to ≈ 2500 °C, but only ≈ 5 % of the energy becomes visible light. An LED lamp gives the same light (≈ 1000 lm) with 7–8 times less power: ≈ 100 lm/W compared with ≈ 14 lm/W. The Joule–Lenz law is used in the heating elements of kettles, irons and heaters (nichrome, with a high resistivity), in electric welding and in the fuse — with an excess current the thin wire melts and breaks the circuit.

The meter and the simplifications of the model

The aluminium disc of an induction meter turns at a speed proportional to the power: $C = 600$ rev/kWh — 10 revolutions per minute at 1 kW. The disc revolutions turn the counter drums. Model: the average power of each appliance is taken as constant (the refrigerator compressor runs for ≈ 1/3 of the day at 150 W, the washing machine averages 0.5 kW), and the resistance of the flat’s wiring is 0.25 Ω. The “wiring efficiency” $η = P / (P + I^{2} R)$ has the same meaning as the efficiency of a power source $η = R / (R + r)$ : part of the energy is turned into heat on the way (inside the source or in the wires) (in detail in the “Power sources, EMF, internal resistance” experiment). The tariff is only an example: take the real tariff from your own electricity bill.

Laboratory: finding the power of a lamp

The ammeter is connected in series with the lamp and the voltmeter in parallel with it; the power is calculated from $P = U I$ . For a 75 W lamp at 220 V we get $I \approx 0.341$ A and $P \approx 75$ W — the value written on the lamp. Reduce the voltage with the regulator and measure at several points: at 110 V the power falls not by 4 times but by about 3 times, because the resistance of the cooling tungsten filament also falls (model: $R \propto T^{1.2}$ , radiation $\propto T^{4}$ ). The same power can be found from the meter disc as well: $P = n / (C t)$ .

Work and power of alternating current

The voltage at home is alternating: $u = U_{m} \sin ω t$ . “220 V” is the effective (rms) value: a direct voltage of that size gives off the same heat in a resistor; the amplitude is $U_{m} = \sqrt{2} \cdot 220 \approx 311$ V. The instantaneous power $p = u i$ oscillates 100 times a second. In a kettle the current is in phase with the voltage and the mean power is $P = U I$ . In appliances with motors the current lags by an angle $φ$ : for part of the cycle $p < 0$ — energy returns to the grid, and the mean (active) power is $P = U I \cos φ$ . The meter counts active energy, but at a low $\cos φ$ the current for the same $P$ is larger and the loss $I^{2} R$ in the wires increases — check it with the “cos φ” slider.

Did you know?

1 kWh = 3.6 MJ: this energy is enough to lift a 1-tonne load about 367 m ( $h = A / m g$ ).
Isamu Akasaki, Hiroshi Amano and Shuji Nakamura received the Nobel Prize in Physics in 2014 for creating efficient blue light-emitting diodes — it is thanks to this invention that white LED lamps appeared.

Inside the experiment

The experiment starts from a real-life situation: you predict the result, change the controls, run the experiment and draw a conclusion. Inside are missions, theory, a quiz and a guided tour in which the camera moves and explains the key points.

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