A-LevelPhysicsD.C. circuits9702/52 · May/June 2018 · Q215 Marks
A student is investigating the charging of a capacitor. A circuit is set up as shown in Fig. 2.1. [Figure 2.1] The capacitor is initially discharged. A resistor of resistance R is connected between P and Q. When the switch is closed, the time t for the voltmeter reading to increase to a specific value V is measured. The capacitor is then discharged. The experiment is repeated with a different number n of resistors each of resistance R connected in series between P and Q. It is suggested that t and n are related by the equation 1 - V/E = e^(-t/(nRC)) where E is the electromotive force (e.m.f.) of the power supply and C is the capacitance of the capacitor.
- (a)A graph is plotted of t on the y-axis against nR on the x-axis. Determine an expression for the gradient.[1]
- (b)Values of n and t are given in Fig. 2.2. Each resistor has a resistance R of 4.7kΩ ± 10%. n t/s 1 15.8 2 34.8 3 50.8 4 66.8 5 83.8 6 97.2 Fig. 2.2
- (c)(i)Calculate and record values of nR/10³Ω in Fig. 2.2. Include the absolute uncertainties in nR.[2]
- (c)(ii)Plot a graph of t/s against nR/10³Ω. Include error bars for nR.[2]
- (c)(iii)Draw the straight line of best fit and a worst acceptable straight line on your graph. Both lines should be clearly labelled.[2]
- (c)(iv)Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer.[2]
- (d)(i)Using your answers to (a) and (c)(iii), determine the value of C. Include an appropriate unit. Data: E = 5.0V V = 4.0 V[3]
- (d)(ii)Determine the percentage uncertainty in C.[1]
- (e)The experiment is repeated using the same capacitor. Determine the resistance K of the single resistor that is required so that the time for the voltmeter reading to reach 90% of the e.m.f. of the power supply is 5.0 minutes. Include the absolute uncertainty in your answer.[2]
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Mark scheme
Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.
- [1] gradient = -C ln (1 - V/E) or -C ln(E / (E - V))
- [1] First mark for correct column heading and values of nR.
- [1] Second mark for absolute uncertainties in nR. Allow a mixture of significant figures.
- [1] Six points plotted correctly. Must be within half a small square. Diameter of points must be less than half a small square.
- [1] Error bars in nR plotted correctly. All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
- [1] Line of best fit drawn. Upper end of line should pass between (25, 90) and (26, 90) and lower end of line should pass between (10.5, 40) and (11.5, 40). Do not allow line from top point to bottom point unless points are balanced.
- [1] Worst acceptable line drawn (steepest or shallowest possible line). All error bars must be plotted.
- [1] Gradient determined with clear substitution of points from the line of best fit into Δy/Δx. Distance between points must be at least half the length of the drawn line.
- [1] Gradient of worst acceptable line determined.
- [1] uncertainty = gradient of line of best fit – gradient of worst acceptable line or uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)
- [1] C determined using gradient and given to 2 or 3 significant figures.
- [1] C determined using: -gradient / ln(1 - V/E) = -(c)(iii) / ln(0.2) = -1.609438
- [1] C determined correctly using gradient and with unit (F or sΩ⁻¹) and correct power of ten.
- [1] % uncertainty in C = % uncertainty in gradient.
- [1] K determined using C. Correct substitution of numbers must be seen.
- [1] Absolute uncertainty in K determined. Correct substitution of numbers must be seen.
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