Two Students, Two Voltmeters, Two Internal Resistances
Two students measure the emf and internal resistance of the same new 1.5 V alkaline cell. Student A connects the cell, an ammeter (resistance about 0.10 Ω) and a variable resistor in series, and a digital voltmeter across the cell's terminals; she reads 0.142 A at 1.44 V and 0.600 A at 1.26 V. Student B uses the same cell and meters but connects his voltmeter across the variable resistor; he reads 0.142 A at 1.43 V and 0.600 A at 1.21 V. Both plot V against I, both draw a straight line through their two points, and both obtain an intercept of 1.50 V. A reports r = 0.39 Ω and B reports r = 0.48 Ω, and each claims the other's circuit is wrong. The teacher confirms that both circuits are wired correctly and that both sets of numbers are honest.
1. Write the equation relating the voltmeter reading to the current for each circuit, and decide which student's value is the cell's internal resistance. Explain why both intercepts agree.
Read the worked answer
Let E be the emf, r the internal resistance, R_A ≈ 0.10 Ω the ammeter resistance and I the current, which is the same through the cell, the ammeter and the resistor (the voltmeter draws negligible current).
Student A (voltmeter across the cell). V_A = E − I r. Gradient = (1.26 − 1.44)/(0.600 − 0.142) = −0.18/0.458 = −0.393 Ω, so r = 0.39 Ω. Intercept = 1.44 + 0.142 × 0.393 = 1.496 ≈ 1.50 V.
Student B (voltmeter across the resistor). V_B = IR = E − I(r + R_A). Gradient = (1.21 − 1.43)/0.458 = −0.22/0.458 = −0.480 Ω, so r + R_A = 0.48 Ω. Intercept = 1.43 + 0.142 × 0.48 = 1.498 ≈ 1.50 V.
Decision. A is right: 0.39 Ω is the internal resistance of the cell. B's 0.48 Ω is the internal resistance plus the ammeter resistance, because his voltmeter misses the potential difference dropped across the ammeter. Subtracting 0.10 Ω from 0.48 Ω gives 0.38 Ω, agreeing with A to within rounding - and that agreement, not either single result, is the real evidence that both experiments are sound.
Why the intercepts agree. The intercept is the reading at zero current. With no current, no potential difference is dropped anywhere - not inside the cell, not across the ammeter - so either voltmeter reads the emf itself.
2. The students used only two points, taken one after the other over several minutes. Discuss what a full set of readings would show and how the cell's own behaviour could distort the result.
Read the worked answer
Two points hide everything except a straight line. Two points always fit a straight line exactly, so the experiment cannot test the model V = E − Ir at all; it can only produce numbers carrying whatever random error is in two readings (a 0.01 V error in one reading shifts r by about 0.02 Ω). Eight or ten readings, plotted with a line of best fit, average the random error and, more importantly, reveal curvature if the simple model fails.
Drift in the cell. The readings were taken over several minutes. As the cell delivers current its emf falls slightly (polarisation and depletion) and its internal resistance rises, especially at the high-current setting that warms it. A later point will therefore sit below the earlier straight line at the same current, making the graph look steeper and inflating r. Checks: take the readings quickly, in a random order, repeat the first reading at the end, and abandon the run if the repeat has drifted by more than about 0.01 V. Use a fresh cell, keep the maximum current modest, and switch off between readings.
If the graph were visibly bent. That would mean E or r changes with current - which real cells do - and the straight-line 'internal resistance' is then only an average over the current range used. The graph is the test of the model, and two points can never be.
3. Using student A's model, predict the ammeter and voltmeter readings if the variable resistor is set to 5.0 Ω, and the power converted inside the cell. Comment on whether the emf of 1.50 V is a fixed property of the cell.
Read the worked answer
Current. The series resistance is R + r + R_A = 5.0 + 0.39 + 0.10 = 5.49 Ω; the voltmeter's 10 MΩ is in parallel with the whole chain and can be ignored. I = E/(R + r + R_A) = 1.50/5.49 = 0.273 A, so the ammeter reads 0.27 A.
Voltmeter across the cell. V = E − Ir = 1.50 − 0.273 × 0.39 = 1.50 − 0.107 = 1.39 V. Across the 5.0 Ω resistor it would read IR = 1.37 V; the 0.03 V difference is the ammeter's share, which is exactly the distinction that tripped student B up.
Power inside the cell. P = I²r = (0.273)² × 0.39 = 0.0746 × 0.39 = 0.029 W. The cell delivers EI = 1.50 × 0.273 = 0.41 W in total, so about 0.37 W heats the resistor and 0.03 W heats the cell itself. That internal heating is why a low-resistance setting warms the cell and drains it quickly.
Is 1.50 V fixed? No. It is the emf at the time of the experiment. The emf falls as the cell discharges and with temperature, while the internal resistance rises as the cell ages or dries out. Repeating the experiment on an old cell would give a lower intercept and a steeper line - both a smaller emf and a larger internal resistance - and the simple straight-line model would then be a fit over a limited current range rather than a description of the cell.