A2 Daily A Level Physics question
In a required practical, a student measures the terminal potential difference V of a cell for different currents I by varying the external load. A straight-line fit of V against I gives a slope of −0.40 V A⁻¹ and a V-intercept of 1.50 V. The cell is then connected to a single 2.0 Ω resistor. Which is closest to the percentage of the cell’s total power that is dissipated inside the cell?
Answer
The correct answer is C.
Correct: C — 17%. The slope magnitude gives the internal resistance r = 0.40 Ω and the intercept gives the open-circuit voltage 1.50 V; with a 2.0 Ω load, the fraction of the cell’s power dissipated internally is r/(R + r) = 0.40/(2.0 + 0.40) ≈ 0.17 ≈ 17%. A 20% results from taking r/R = 0.40/2.0 and ignoring that the total circuit resistance is R + r (or equivalently using I ≈ intercept/R and neglecting r when finding the power split). B 56% comes from incorrectly inverting the gradient to get r = 1/0.40 = 2.5 Ω, but the gradient already has units of ohms so r = 0.40 Ω, not 2.5 Ω. C is correct as shown. D 43% comes from misreading the intercept (1.50 V) as a resistance in ohms; the intercept is the open-circuit voltage, not the internal resistance.