A cell is tested at steady currents. These (current in A, terminal voltage in V) pairs are measured: (0.2, 3.496), (0.4, 3.392), (0.6, 3.288), (0.8, 3.184). Treat the values as exactly collinear and use the model V = ε − Ir with constant emf ε and internal resistance r. Find ε and r. Then predict the current and useful load power when connected to a 5.2 Ω resistor. Explain one precaution needed to keep the assumed cell model consistent during the experiment.

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ε = 3.6 V; r = 0.52 Ω; current 0.6294 A; load power 2.06 W. Limit heating and discharge during measurements. The cell model is a straight line with terminal voltage on the vertical axis and current on the horizontal axis. Its vertical intercept is emf and its negative gradient is internal resistance. Calculate gradient = ΔV/ΔI = −0.52 V A⁻¹, so r = 0.52 Ω. Substituting any supplied pair into ε = V + Ir gives ε = 3.6 V. With external load R, the emf drives current through both R and r: I = ε/(R + r) = 0.6294 A. Using ε/R alone would ignore the internal voltage drop. The useful external power is I²R = 2.06 W. Internal heating is I²r, so the total electrical power εI is greater than the power delivered to the load. Use short measurement intervals with rests and avoid excessive currents so temperature and state of charge change little. Repeat an earlier reading to detect drift; otherwise a single constant emf and resistance may not describe the entire dataset.

Infer a cell's emf and internal resistance · set 6

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Cell terminal-voltage measurements

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Cell terminal-voltage measurements — supplied values
Current / A	Terminal voltage / V
0.2	3.496
0.4	3.392
0.6	3.288
0.8	3.184