A cell gives (current in A, terminal p.d. in V) readings (0, 1.52), (0.5, 1.43), (1, 1.34), (1.5, 1.25), (2, 1.16). (a) Find e.m.f. and internal resistance. (b) At 1.50 A find external resistance, load power, internal heating and chemical power. (c) Verify conservation. (d) Find load for maximum power and its efficiency. (e) Explain why that operation is inefficient.

Question

Accepted Answer

E.m.f. 1.52 V; r = 0.18 Ω; at 1.50 A: R = 0.833 Ω, load 1.875 W, loss 0.405 W, chemical 2.28 W; maximum power at R = r with 50% efficiency. From V = ε − Ir, intercept gives ε = 1.52 V and gradient (1.16 − 1.52)/2.0 = −0.18 V/A gives r = 0.18 Ω. At 1.50 A and 1.25 V, external R = 1.25/1.50 = 0.833 Ω and load power VI = 1.875 W. Internal power I²r = 1.50² × 0.18 = 0.405 W. Chemical power εI = 1.52 × 1.50 = 2.28 W; 1.875 + 0.405 = 2.280 W. Maximum load power occurs at R = r = 0.18 Ω. Then terminal p.d. is ε/2 and equal powers go to load and internal heating, so efficiency is 50%. Maximum power is not maximum efficiency: half the chemical power heats the cell, wasting energy and potentially reducing life. Efficient supply normally uses R much larger than r.

Cell e.m.f., internal resistance and power

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Cell terminal characteristic

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Cell terminal characteristic — supplied values
Current / A	Terminal p.d. / V
0	1.52
0.5	1.43
1	1.34
1.5	1.25
2	1.16