Battery Internal Resistance with an Ammeter Systematic Error
A student tests a batch of 1.5 V zinc-carbon cells for a science-club project. She uses a resistor box, an ammeter of resistance 0.50 Ω, and a digital voltmeter of input resistance 10 MΩ. Because of where the leads reach, the voltmeter is connected across the resistor box, not across the cell. She records the voltage across the box, V, while the ammeter reads current I:
I / A: 0.10, 0.20, 0.30, 0.40, 0.50 V / V: 1.48, 1.39, 1.30, 1.21, 1.12
Each reading is taken 10 s after the current is switched on, at 20 °C. The cells are sold with the claim 'internal resistance ≤ 0.50 Ω'. The club plans to run a 2.0 A fuse-protected toy. A classmate says the straight-line graph 'proves the battery is fine'; another says a straight line says nothing about internal resistance. A least-squares fit gives V = 1.57 − 0.90 I.
1. Show that the fitted slope is r + R_A, not r, given the voltmeter is across the resistor box. Calculate the true internal resistance.
Read the worked answer
With the voltmeter across the resistor box, it reads V = I R_box. The cell sees R_box + R_A + r, so
E = I(R_box + R_A + r) ⇒ V = E − I(R_A + r).
Thus the fitted slope is R_A + r, not r. From the data, slope = (1.12 − 1.48)/(0.50 − 0.10) = −0.90 Ω, so R_A + r = 0.90 Ω and r = 0.90 − 0.50 = 0.40 Ω. The intercept 1.57 V is the emf; the ammeter resistance does not affect it.
2. Repeated fits give a slope between 0.88 and 0.92 Ω. Does the batch meet the ≤ 0.50 Ω claim? Discuss what the straight line does and does not show.
Read the worked answer
Repeated fits give slope 0.88–0.92 Ω, so r = slope − 0.50 lies between 0.38 and 0.42 Ω. The claim is r ≤ 0.50 Ω; the whole interval is below 0.50, so the batch passes with a margin of at least 0.08 Ω.
The uncertainty is dominated by the ammeter correction, so a poorly known ammeter resistance would matter more than the scatter in V. The straight line only shows ohmic behaviour over 0.10–0.50 A; it does not by itself validate the specification, which requires the corrected slope and its uncertainty.
3. A club member proposes briefly short-circuiting a cell to measure the short-circuit current directly on a 10 A range. Estimate I_sc from the fitted values and explain why this is a poor test.
Read the worked answer
From the fitted values, I_sc ≈ E/r = 1.57/0.40 ≈ 3.9 A. A 10 A meter could display it, but the test is poor: the terminal voltage collapses, the cell polarises, temperature and internal resistance change during the few seconds of the test, and a zinc-carbon cell can leak or overheat.
The reading is not the same r as in normal use, and a fuse-protected toy may simply blow its 2.0 A fuse when stalled, since 3.9 A is available. The fitted E and r from the V–I line are safer and more informative.