Two Circuits, One Resistor: Meter Placement Matters
A student measures an unknown resistor using a 0–20 V voltmeter with 100 kΩ input resistance and a 0–1 A ammeter with 0.50 Ω resistance. In circuit A, the voltmeter is connected directly across the resistor and the ammeter is in series with the parallel combination. The readings are 4.00 V and 0.380 A. In circuit B, the ammeter is in series with the resistor and the voltmeter is connected across both; the readings are 4.00 V and 0.400 A. The resistor is marked 10 Ω, 5%. The student computes R = V/I and gets 10.5 Ω for A and 10.0 Ω for B, then claims B must be correct because it matches the label. A lab partner says B includes the ammeter resistance and is therefore wrong. The power supply is adjusted so the resistor dissipates about 1.6 W.
1. Calculate the true resistance from each circuit, correcting for meter resistance. Which circuit gives the more accurate result for this resistor, and why?
Read the worked answer
Circuit A: the voltmeter reads V_R = 4.00 V and the ammeter reads the total current I = 0.380 A = I_R + I_V. The voltmeter current is I_V = 4.00/100,000 = 4.0×10⁻⁵ A, negligible, so I_R ≈ 0.37996 A and R = 4.00/0.37996 ≈ 10.53 Ω. The ammeter resistance is in series with the parallel combination and affects the total voltage, not V_R. Circuit B: the ammeter and resistor are in series, and the voltmeter reads the voltage across both. Thus R + R_A = 4.00/0.400 = 10.00 Ω, so R = 9.50 Ω. Circuit A is more accurate here because the voltmeter loading is only about 0.01%, while the ammeter drop in B is 0.50 Ω, or 5% of the resistor. B's apparent 10.0 Ω is not proof of accuracy; it is biased low by the ammeter resistance.
2. The student says the 5% tolerance means both answers are acceptable. Evaluate that claim. What would change if the unknown resistor were 100 kΩ instead of 10 Ω?
Read the worked answer
The tolerance describes the true resistor, not the measurement error. A 5% 10 Ω resistor may truly lie between 9.5 and 10.5 Ω, so a measured 10.5 Ω is at the edge and 9.50 Ω is also at the edge, but that does not excuse a known systematic bias. The measurement method should be corrected first, then compared with the tolerance. For a 100 kΩ resistor, circuit A would be poor because the 100 kΩ voltmeter is comparable to the resistor and bypasses a large fraction of the current. Circuit B would be preferable: the 0.50 Ω ammeter resistance is negligible compared with 100 kΩ, and the ammeter reads only the resistor branch current, so R = V/I − R_A is accurate to about 0.0005%.
3. The student wants to measure the battery's emf directly by connecting the voltmeter across it with no resistor. Explain what reading would be obtained and why it is not exactly the emf in a real circuit.
Read the worked answer
With no external resistor, the voltmeter itself is the only load. A 100 kΩ voltmeter draws a tiny current I = E/(r + 100 kΩ), so the reading is V = E(100 kΩ)/(r + 100 kΩ), slightly less than E. For a typical small battery with r of a few ohms, the difference is negligible-less than 0.01%-so the reading is an excellent approximation to the emf. The reading would not be exactly E because the voltmeter draws some current and because the battery's emf can drift with temperature, age, and recent current draw. To test whether loading matters, compare readings with two voltmeters of different input resistance; if the readings differ, correct for the finite meter resistance.