Choosing an Ammeter–Voltmeter Circuit for 60 Ω
A student must determine an unknown resistance R_x, believed to be about 60 Ω, using a 6.0 V battery of negligible internal resistance, a moving-coil ammeter of resistance 0.50 Ω and a moving-coil voltmeter of resistance 12 kΩ. Only one set of readings can be taken before the laboratory closes, and the student plans to calculate R_x as V/I from the meter readings. Two circuits are available. In circuit A the ammeter is in series with R_x and the voltmeter is connected across R_x only, so the ammeter also carries the voltmeter current. In circuit B the voltmeter is connected across the series combination of R_x and the ammeter, so the voltmeter current bypasses the ammeter and the ammeter reads only the current through R_x. A demonstrator warns that one arrangement is badly wrong for a small R_x, the other for a large R_x, and that the two percentage errors are equal at one particular value of R_x. The student also has a spare 5 Ω resistor and asks whether the same rule of thumb covers it.
1. Calculate R_x = V/I for both circuits when R_x = 60 Ω and decide which arrangement should be used.
Read the worked answer
Circuit A (voltmeter across R_x only). The voltmeter is in parallel with R_x, so the ammeter reads the total current:
R_measured = R_x R_V /(R_x + R_V) = (60 × 12000)/(12060) = 59.70 Ω.
Percentage error = (59.70 − 60)/60 × 100 ≈ −0.50 %.
Circuit B (voltmeter across R_x + ammeter). The ammeter reads only the current through R_x and the voltmeter reads the whole p.d.:
R_measured = R_x + R_A = 60 + 0.50 = 60.50 Ω, i.e. +0.83 %.
At 60 Ω, circuit A is the better choice (−0.5 % versus +0.83 %), because the voltmeter's shunting effect is still small while the ammeter's 0.5 Ω is already comparable with 1 % of 60 Ω.
2. At what value of R_x do the two arrangements give percentage errors of equal magnitude, and how should the choice be made in general?
Read the worked answer
The errors are approximately:
• Circuit A: −R_x/R_V (voltmeter loading, reading too low). • Circuit B: +R_A/R_x (ammeter drop included, reading too high).
Equal magnitudes: R_x/R_V = R_A/R_x → R_x = √(R_A R_V) = √(0.50 × 12000) = √6000 ≈ 77.5 Ω.
Rule of thumb:
• R_x < 77.5 Ω → use circuit A (keep the ammeter drop out of the measured voltage). • R_x > 77.5 Ω → use circuit B (keep the voltmeter from shunting R_x).
A useful check: at 60 Ω the predicted errors are −0.50 % (A) and +0.83 % (B), consistent with choosing A; at 1 kΩ the same formulas give about −7.7 % (A) and +0.05 % (B), so B wins by a wide margin.
3. The student tests the rule with the 5 Ω resistor. What will each circuit read, and what caution applies when R_x is very large?
Read the worked answer
Circuit A: R_measured = (5 × 12000)/12005 ≈ 5.00 Ω, error ≈ −0.04 %.
Circuit B: R_measured = 5 + 0.50 = 5.50 Ω, error = +10 %.
So for the 5 Ω resistor, circuit A is clearly correct and circuit B is unacceptable - the ammeter resistance is 10 % of the unknown.
Caution for large R_x: once R_x becomes comparable with R_V, circuit A's error grows without limit in practice (even 1 MΩ would read as roughly 12 kΩ), and circuit B is then only marginally better because the voltmeter still loads the circuit. In that regime the correct judgement is that neither simple arrangement is acceptable; a high-impedance measurement (or a substitution/ratio method) is required rather than picking the 'less bad' of the two circuits.