A Physics Practical with a Faulty Ammeter
Grade 12 students are measuring the resistance of a wire using a battery of nominal 6 V, an ammeter, a voltmeter and a rheostat. The wire is 1.0 m long with a stated resistivity of 1.1 × 10⁻⁶ Ω·m and a cross-sectional area of 0.50 mm². The students record: current 0.40 A, voltage 2.4 V. A second group using the same wire records current 0.38 A and voltage 2.3 A. The teacher notices that the first group's ammeter reads 0.05 A even when the circuit is open, and that the voltmeter is connected in series with the wire rather than in parallel. The class must decide what resistance value, if any, to report.
1. Compute the resistance from each group's readings and from the wire's stated properties, then explain the discrepancies.
Read the worked answer
Group 1: R = V/I = 2.4 / 0.40 = 6.0 Ω. Group 2: R = 2.3 / 0.38 ≈ 6.05 Ω. From stated properties: A = 0.50 mm² = 0.50 × 10⁻⁶ m². R = ρL/A = (1.1 × 10⁻⁶ × 1.0) / (0.50 × 10⁻⁶) = 2.2 Ω.
The two groups agree with each other (≈6 Ω) but both differ from the stated value (2.2 Ω) by a factor of nearly three. That pattern points to a systematic error shared by both groups, not random scatter.
The teacher's observations explain it. First, an ammeter that reads 0.05 A on an open circuit has a zero error; the true current is about 0.40 - 0.05 = 0.35 A, which raises the computed resistance to 2.4 / 0.35 ≈ 6.9 Ω, making the gap worse, so the zero error is not the main cause. Second, and decisively, a voltmeter connected in series does not measure the voltage across the wire; it measures nearly the whole battery voltage while its own high resistance throttles the current. The readings are therefore not a valid V-across-the-wire and I-through-the-wire pair at all.
2. What should the class report, and what correction must be made before any value is accepted?
Read the worked answer
The class should not report 6.0 Ω or 6.05 Ω as the resistance of the wire. Those numbers come from an invalid circuit: with the voltmeter in series, the measured voltage is not the potential difference across the wire, so Ohm's law has not been applied to the wire.
The correction: reconnect the voltmeter in parallel across the wire, zero the ammeter (or subtract the 0.05 A offset), and repeat at several currents to check linearity. Only then compute R = V/I.
The judgement call is about honesty in reporting. The students should state that the first set of readings is invalid and why, and present the corrected measurement. If the corrected value still differs from 2.2 Ω, the class should discuss plausible causes: the stated resistivity may be for a different material or temperature, the area may be nominal rather than measured, and contact resistance at the terminals adds to a 1 m wire. A good report distinguishes a measurement error (fixable by reconnecting) from a genuine physical difference (to be explained, not hidden).