Two-point determination of e.m.f. and internal resistance of a used cell
A student is asked to find the e.m.f. E and internal resistance r of a single AA cell. She connects the cell in turn to carbon resistors of 10 Ω, 22 Ω and 47 Ω and records the p.d. across each resistor with a digital voltmeter of resistance 10 MΩ, waiting 30 s before each reading: (10 Ω, 1.32 V), (22 Ω, 1.44 V), (47 Ω, 1.50 V). A classmate says the first reading alone is enough, provided the resistor is accurate to 1%, and quotes r = 1.4 Ω. The cell has been used before and becomes noticeably warm during the experiment. The laboratory is at 20 °C; the manufacturer's data sheet gives E = 1.55 V and r = 0.30 Ω for a fresh cell at 20 °C. The voltmeter's last digit flickers by ±0.01 V.
1. Use all three readings to determine E and r, and explain why the classmate's value of 1.4 Ω is an underestimate rather than an equally valid answer.
Read the worked answer
For each resistor the current is I = V/R and V = E − Ir, so a graph of V against I has intercept E and gradient −r.
- (10 Ω): I = 1.32/10 = 0.132 A
- (22 Ω): I = 1.44/22 = 0.0655 A
- (47 Ω): I = 1.50/47 = 0.0319 A
Gradient from the outer points: (1.32 − 1.50)/(0.132 − 0.0319) = −0.18/0.100 = −1.80 Ω. Gradient from the middle and last points: (1.44 − 1.50)/(0.0655 − 0.0319) = −0.06/0.0336 = −1.79 Ω.
The two gradients agree, so a straight line is justified and r ≈ 1.8 Ω. Then E = 1.50 + 1.79 × 0.0319 = 1.56 V.
The classmate's error: treating the largest terminal p.d. (1.50 V) as the e.m.f. gives r = (1.50 − 1.32)/0.132 = 1.36 ≈ 1.4 Ω. But 1.50 V is measured with 32 mA flowing, so it is below the e.m.f. by Ir ≈ 0.06 V. A single point cannot separate E from r at all without an independent value of E; using three currents makes the separation possible and provides a consistency check.
2. Explain why r is much larger than the data sheet's 0.30 Ω, and predict what happens to the readings if the 10 Ω resistor is left connected for five minutes before the 22 Ω reading is taken.
Read the worked answer
Why r is larger. The data sheet value is for a fresh cell. In a part-discharged cell the electrolyte is depleted and discharge products build up at the electrodes, lowering the ionic conductivity of the electrolyte and adding a polarisation contribution: as current is drawn, ions pile up and an opposing polarising p.d. grows. Both effects appear as an increased internal resistance, and both depend on the state of charge and on temperature.
Effect of the heating. About 1.3 V × 0.13 A ≈ 0.03 W is dissipated inside the cell on the 10 Ω load, and the cell warms. Warming usually lowers the internal resistance of an alkaline cell, while continued discharge raises it, so the two effects oppose each other and 30 s is not long enough to reach a steady state. That is why the three points may not lie exactly on a line.
Prediction: after five minutes on 10 Ω the cell is more deeply discharged, so its internal resistance is higher. The 22 Ω reading will then be lower than 1.44 V for two reasons: the e.m.f. falls slightly and r has increased, increasing Ir. The plotted line for that point would fall below the earlier best-fit line.
3. Show quantitatively that the voltmeter's 10 MΩ resistance cannot explain the discrepancy, and calculate what voltmeter resistance would be needed to depress the 47 Ω reading by 5%.
Read the worked answer
Actual loading. The current through the voltmeter is V/R_V = 1.32/1.0 × 10⁷ = 1.3 × 10⁻⁷ A, compared with 0.13 A in the resistor - a fractional change of about 10⁻⁶. The measurement is effectively unaffected, so the voltmeter is not the explanation.
Required voltmeter resistance for a 5% error. The voltmeter is in parallel with the resistor, so the load is R_p = RR_V/(R + R_V), and V = E R_p/(R_p + r). Take E = 1.56 V and r = 1.8 Ω from the three readings; a 5% drop requires V = 0.95 × 1.50 = 1.425 V:
1.425 = 1.56 R_p/(R_p + 1.8) 1.425 R_p + 2.565 = 1.56 R_p R_p = 2.565/0.135 = 19.0 Ω
Then 1/R_p = 1/R + 1/R_V gives 1/19.0 = 1/47 + 1/R_V, so 1/R_V = 0.0526 − 0.0213 = 0.0313 and R_V ≈ 32 Ω.
A voltmeter would have to have a resistance of only a few tens of ohms - comparable with the resistors themselves - to cause a 5% error. The 10 MΩ instrument is therefore vindicated, and the true explanation must be the cell's own r and its change with state of charge.