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Ohm's law

Curricula: KS3, GCSE, cycle 4, 2de, Sek I, HS, AP2, IB B.5

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Aim

Establish the relationship between the current through a resistor and the voltage across it.

The bench

A 5 V source, an ammeter and a 100 Ω resistor; a voltmeter is connected in parallel with the resistor.

Procedure

  1. Use the voltage knob to set the source to 0 V.
  2. Raise the voltage in steps of 1 V up to 10 V and at each step write down the ammeter and voltmeter readings.
  3. Plot a graph of current against voltage.
  4. From the graph find the resistance \(R = U / I\).
Expected result

At 5 V the ammeter reads about 49.7 mA and the voltmeter about 4.97 V. The points lie on a straight line through the origin: the current is proportional to the voltage. \(R = U / I\) ≈ 100 Ω at every step; at 10 V the current is about 99 mA.

Questions

  1. How will the slope of the graph change if you use a 200 Ω resistor?
  2. Why does the voltmeter read slightly less than the value set on the source?
Answers
  1. The slope of the graph is \(I / U = 1 / R\), so with 200 Ω it is half as steep: at every voltage the current is half as large, about 50 mA at 10 V instead of 99 mA.
  2. The source has an internal resistance of 0.5 Ω, and the ammeter and the cables add a little more. The current drops a few hundredths of a volt across them, so the resistor gets slightly less than the voltage set on the source: about 4.97 V instead of 5 V.