PhysicsGrade 7–9

Ohm’s law

Turn up the voltage and the current rises, always in the same ratio. Point by point the characteristic curve appears, though not quite straight for a lamp.

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What is it about?

If a voltage UU is applied to a component, an electric current II flows. This is how you measure:

  • The ammeter is connected in series with the component – the whole current has to pass through it.
  • The voltmeter is connected in parallel with the component – it measures the voltage between its two terminals.

The electrical resistance tells you how strongly a component opposes the current:

R=UI[R]=1 Ω=1 VAR = \frac{U}{I} \qquad [R] = 1\,\Omega = 1\,\frac{\text{V}}{\text{A}}

Ohm’s law: For metal wires at a constant temperature, the current is proportional to the voltage, I∝UI \propto U. The resistance RR is then constant and U=R⋅IU = R \cdot I. The characteristic in an II–UU graph is a straight line through the origin – the larger RR, the flatter the line. In a UU–II graph the slope of the line equals RR.

Light bulb: The tungsten filament gets hotter than 2000 °C2000\,°\text{C}. As its temperature rises, so does its resistance – from about 3 Ω3\,\Omega (cold) to 48 Ω48\,\Omega at 12 V12\,\text{V}. Its characteristic is therefore curved; Ohm’s law does not apply to the bulb.

Resistance of a wire: It increases with the length ll and decreases with the cross-sectional area AA:

R=ρ⋅lAρconstantan=0.49 Ω⋅mm2mR = \rho \cdot \frac{l}{A} \qquad \rho_{\text{constantan}} = 0.49\,\frac{\Omega \cdot \text{mm}^2}{\text{m}}

The resistivity ρ\rho depends on the material. The resistance of constantan hardly changes when it warms up.

Try it

  1. Turn the knob of the power supply and watch the pointers. Record data points at different voltages – or start the automatic series.
  2. Measure different resistors one after the other and compare the lines in the graph. 47 Ω · 150 Ω
  3. Record the characteristic of the light bulb and watch the filament. Light bulb
  4. Switch to the U–I graph: now the slope of the line is the resistance. U–I graph
  5. Investigate the constantan wire: double the length, then the cross-section. 1 m · 2 m · double cross-section

Tasks

Task 1: Unknown resistor

Load task

Determine resistor B from a series of measurements. Draw the best-fit line yourself.

Show solution

For example at U=8 VU = 8\,\text{V}: I≈0.098 AI \approx 0.098\,\text{A}, so R=8 V0.098 A≈82 ΩR = \frac{8\,\text{V}}{0.098\,\text{A}} \approx 82\,\Omega. All data points lie on a straight line through the origin with this value. Show solution

Task 2: Predict the current

A voltage of 6 V6\,\text{V} is applied to a resistor of 100 Ω100\,\Omega. How large is the current?

Show solution

I=UR=6 V100 Ω=0.06 A=60 mAI = \frac{U}{R} = \frac{6\,\text{V}}{100\,\Omega} = 0.06\,\text{A} = 60\,\text{mA}. Check

Task 3: Set the voltage

Which voltage do you need so that exactly 0.2 A0.2\,\text{A} flows through a resistor of 47 Ω47\,\Omega?

Show solution

U=R⋅I=47 Ω⋅0.2 A=9.4 VU = R \cdot I = 47\,\Omega \cdot 0.2\,\text{A} = 9.4\,\text{V}. Check

Task 4: Resistance of the light bulb

Load task

Calculate the resistance of the bulb at 2 V2\,\text{V} and at 12 V12\,\text{V}. Why are the values different?

Show solution

At 2 V2\,\text{V} about 0.110 A0.110\,\text{A} flows, so R≈18 ΩR \approx 18\,\Omega. At 12 V12\,\text{V} it is 0.25 A0.25\,\text{A}, so R=48 ΩR = 48\,\Omega. At a higher voltage the filament is hotter, and the resistance of metals increases with temperature.

Task 5: Longer and thicker wire

A constantan wire with l=1 ml = 1\,\text{m} and A=0.1 mm2A = 0.1\,\text{mm}^2 has 4.9 Ω4.9\,\Omega. What is the resistance of a wire that is 2 m2\,\text{m} long with a cross-section of 0.2 mm20.2\,\text{mm}^2?

Show solution

Double the length doubles RR, double the cross-section halves it: again 4.9 Ω4.9\,\Omega. Check

Task 6: Current limit

Load task

For the wire (1 m1\,\text{m}, 0.1 mm20.1\,\text{mm}^2) the power supply is set to 12 V12\,\text{V}, but the voltmeter only shows about 9.8 V9.8\,\text{V}. Explain.

Show solution

At 12 V12\,\text{V} a current of 12 V4.9 Ω≈2.4 A\frac{12\,\text{V}}{4.9\,\Omega} \approx 2.4\,\text{A} would flow. But the power supply delivers at most 2 A2\,\text{A}, so it lowers the voltage to U=4.9 Ω⋅2 A=9.8 VU = 4.9\,\Omega \cdot 2\,\text{A} = 9.8\,\text{V}. That is why you always measure the voltage directly across the component.

Ideas for the classroom

  • Use: resistance, Ohm’s law and characteristics (grades 8/9), the wire resistance R=ρ⋅lAR = \rho \cdot \frac{l}{A} as an extension. Data are produced by reading the meters, like in the lab; with fit=0 students draw the best-fit line themselves. Since textbooks use both conventions, you can switch between the II–UU characteristic and the UU–II graph. Several series stay in the graph for comparison.
  • Meters: The analogue meters choose their range automatically (it is shown below the scale; small currents in mA). With real meters the range is set by hand – starting with the largest range.
  • Model: power supply 00 to 12 V12\,\text{V} with a current limit of 2 A2\,\text{A}; ideal meters. Bulb 12 V / 0.25 A12\,\text{V}\,/\,0.25\,\text{A}: balance between the electrical power supplied and the losses by radiation and heat conduction, resistance of tungsten R∝T1.2R \propto T^{1.2}. Heating of the constantan wire is neglected. The filament temperature is a model value.
  • Common stumbling blocks:
    • “R=U/IR = U / I is Ohm’s law.” – This is the definition of resistance and holds for every component. Ohm’s law says that RR is constant (I∝UI \propto U).
    • “A large resistance lets a lot of current through.” – It is the other way round.
    • “The ammeter can also be connected in parallel.” – In parallel it causes almost a short circuit.