A student investigates the discharge of a capacitor through a resistor as shown in Fig. 2.1. E C R A Fig. 2.1 The student initially closes the switch and charges the capacitor. The switch is then opened and a stop-watch is started. The capacitor discharges through the resistor. At different times t the current I is measured. It is suggested that I and t are related by the equation I = (E/R)e^(-t/(CR)) where E is the e.m.f. of the power supply, C is the capacitance of the capacitor and R is the resistance of the resistor.

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.

1. [1] Gradient = -1/CR
2. [1] y-intercept = In(E/R)
3. [1] Values: 3.83 or 3.829, 3.69 or 3.689, 3.53 or 3.526, 3.33 or 3.332, 3.18 or 3.178, 3.00 or 2.996
4. [1] Absolute uncertainties in In I from ± 0.04 to ± 0.1
5. [1] Six points plotted correctly. Must be within half a small square. Diameter of points must be less than half a small square.
6. [1] Error bars in In I plotted correctly. All error bars to be plotted. Total length of bar must be accurate to less than half a small square and symmetrical.
7. [1] Line of best fit drawn. Points must be balanced. Line must pass between (5.5, 3.75) and (8.0, 3.75) and between (56, 3.05) and (58, 3.05)
8. [1] Worst acceptable line drawn. Steepest or shallowest possible line that passes through all the error bars. Mark scored only if all error bars are plotted.
9. [1] Negative gradient determined with clear substitution of data points into Δy / Δx; distance between data points must be at least half the length of the drawn line.
10. [1] Gradient determined of WAL uncertainty = (gradient of line of best fit – gradient of worst acceptable line) or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
11. [1] y-intercept read from y-axis to less than half a small square, or y-intercept determined from substitution into y = m x + c.
12. [1] C determined using gradient and C given to two or three significant figures. Correct substitution of numbers must be seen, C = -1 / (150×10³ x gradient).
13. [1] E determined using y-intercept. Correct substitution of numbers must be seen, E = R x e^(y-intercept) = 150×10³ x e^(y-intercept) (×10⁻⁶) Or ln E = ln R + y-intercept
14. [1] C determined using gradient and E determined using y-intercept and dimensionally correct SI unit for C: F or s Ω⁻¹ or C V⁻¹ or A s V⁻¹ and E: V or A Ω.
15. [1] Absolute uncertainty in C. ΔC = (0.05 + Δgradient / gradient) x C OR Correct substitution for max/min methods max C = -1 / (142.5×10³ x minnumericalgradient) min C = -1 / (157.5×10³ × maxnumericalgradient)
16. [1] I determined from (d)(i) OR (c)(iii) and (c)(iv) with correct substitution and correct power of ten(s). Do not accept ecf for POT from (c)(iii), (iv) or (d). I = (E/R) x e^(-120/(CR)) OR I = e^(y-intercept) x e^(gradient×120) ×10⁻⁶ OR In I = 120 x gradient + y-intercept I = e^(120xgradient+y-intercept) ×10⁻⁶