Ammeter–voltmeter method: choosing the circuit before the measurement is made
In a first-year physics laboratory a student must determine an unknown resistance of about 50 Ω using a 6 V battery, a moving-coil ammeter of resistance 2.0 Ω, and a digital voltmeter of resistance 20 kΩ. The manual permits only two circuits. Circuit I: the ammeter in series with the resistor and the voltmeter connected directly across the resistor. Circuit II: ammeter and resistor in series, with the voltmeter connected across the series combination. The demonstrator warns that the wrong circuit can be several per cent out at 50 Ω, and that the same rule of thumb will fail later in the session when a 40 kΩ resistor is measured. The resistors are sealed inside modules, so the student cannot simply try both circuits and compare readings; the choice must be argued from the meter resistances before any measurement is taken, and the residual error must be reported honestly.
1. For the 50 Ω resistor, compute the value each circuit would report and its percentage error, then recommend one.
Read the worked answer
Let R = 50 Ω, R_A = 2.0 Ω, R_V = 20 000 Ω.
Circuit I. The voltmeter reads the true voltage across R, but the ammeter also carries the small current drawn by the voltmeter, so the ammeter over-reads. The parallel combination is what is actually being measured:
R_meas = V/I = R·R_V/(R + R_V) = (50 × 20000)/20050 = 49.88 Ω, error −0.25%.
Circuit II. The ammeter carries only the resistor current, but the voltmeter now reads the drop across resistor plus ammeter:
R_meas = R + R_A = 52.0 Ω, error +4.0%.
Recommendation. Use Circuit I. Its 0.25% error is acceptable for a three-figure result, whereas 4% (2 Ω on a 50 Ω answer) is not. The physical reason is that the ammeter's 2 Ω is inside the voltage-sensing loop in Circuit II, while in Circuit I the voltmeter's 20 kΩ is only a small shunt across a 50 Ω element.
2. Derive the resistance at which the two circuits are equally bad, and use it to state a rule that also covers the 40 kΩ resistor.
Read the worked answer
Circuit I underestimates by a factor R/(R + R_V); Circuit II overestimates by the additive term R_A/R. Set the fractional errors equal in magnitude:
R/(R + R_V) = R_A/R → R² − R_A R − R_A R_V = 0 → R = [R_A + √(R_A² + 4R_A R_V)]/2 = [2 + √(4 + 160 000)]/2 = 201 Ω.
Since R_V ≫ R_A, the crossover is essentially √(R_A R_V) = √(2 × 20 000) ≈ 200 Ω.
Rule. If the unknown is well below √(R_A R_V), use Circuit I; if well above, use Circuit II.
At 40 kΩ. Circuit I: R_meas = (40 000 × 20 000)/60 000 = 13.3 kΩ, an error of −67%, useless. Circuit II: R_meas = 40 002 Ω, error +0.005%.
So the choice genuinely flips, and the demonstrator's warning is correct: there is no single good circuit, only a circuit that is right for the size of resistance being measured.
3. The next module turns out to be about 150 Ω. What does each circuit give, and what should the student report or do?
Read the worked answer
Circuit I: R_meas = (150 × 20 000)/20 150 = 148.9 Ω, error −0.74%. Circuit II: R_meas = 152.0 Ω, error +1.33%.
Judgement. At 150 Ω the student is sitting almost on the 201 Ω crossover, so neither circuit is good; the best available is about 0.7% low, which is 1 Ω on a 150 Ω answer. That is barely acceptable and should be written as 148.9 ± 1.1 Ω, not as a bare four-figure value. The instinct to pick whichever reading looks more plausible is exactly the wrong move, because reversing the choice costs 1.3% and pushes the answer 2 Ω high.
Better options, in order. Use a voltmeter of 1 MΩ or more, which drops Circuit I's error to about 0.015%; use a substitution method or a Wheatstone bridge, which compares against a standard instead of relying on meter resistances; or, if the lab insists on these two meters, report the corrected value R = R_meas(R + R_V)/R_V with its uncertainty and say plainly that the residual is instrumental.