A-LevelPhysicsD.C. circuits9702/52 · Feb/Mar 2017 · Q215 Marks
A student is investigating the potential difference in a circuit. The circuit is set up as shown in Fig. 2.1. E P Q V [Figure 2.1] Two resistors P and Q are connected in series to a power supply of electromotive force (e.m.f.) E and negligible internal resistance. Resistor P has resistance P. The potential difference V across resistor P is measured. The experiment is repeated for different values of P. It is suggested that V and P are related by the equation V=(P/(P+Q))E where Q is the resistance of resistor Q. The value of Q is kept constant.
- (a)A graph is plotted of 1/V on the y-axis against 1/P on the x-axis. Determine expressions for the gradient and the y-intercept. gradient = y-intercept =[1]
- (b)Values of P and V are given in Fig. 2.2. [Table data - P/Ω, V/V] Calculate and record values of 1/P /10⁻³Ω⁻¹ and 1/V /V⁻¹ in Fig. 2.2. Include the absolute uncertainties in 1/P.[3]
- (c)(i)Plot a graph of 1/V⁻¹ against 1/P /10⁻³Ω⁻¹. Include error bars for 1/P.[2]
- (c)(ii)Draw the straight line of best fit and a worst acceptable straight line on your graph. Both lines should be clearly labelled.[2]
- (c)(iii)Determine the gradient of the line of best fit. Include the absolute uncertainty in your answer. gradient =[2]
- (c)(iv)Determine the y-intercept of the line of best fit. Include the absolute uncertainty in your answer. y-intercept =[2]
- (d)(i)Using your answers to (a), (c)(iii) and (c)(iv), determine the values of E and Q. Include appropriate units. E = Q =[2]
- (d)(ii)Determine the percentage uncertainty in Q. percentage uncertainty in Q = %[1]
This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2017). The mark or mark type is shown in brackets.
- [1] gradient = Q/E
- [1] y-intercept = 1/E
- [1] First mark for all first column correct either 2 and 3 significant figures or 3 and 4 significant figures.
- [1] Second mark for all second column correct.
- [1] absolute uncertainties from 0.4 to 0.1
- [1] six points plotted correctly must be within half a small square
- [1] error bars in 1/P plotted correctly all error bars to be plotted
- [1] line of best fit drawn
- [1] worst acceptable line drawn steepest or shallowest possible line mark scored only if all error bars are plotted
- [1] gradient determined with a triangle that is at least half the length of the drawn line
- [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] y-intercept determined by substitution into y = mx + c
- [1] uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line or uncertainty = ½ (steepest worst line y-intercept – shallowest worst line y-intercept).
- [1] E determined with correct unit using y-intercept E = 1 / y - intercept
- [1] Q determined with correct unit using gradient and given to two or three significant figures penalise power of ten errors correct substitution of numbers must be seen Q = E x gradient / y - intercept
- [1] percentage uncertainty in Q correct substitution of numbers must be seen %uncertainty E + %uncertainty in gradient or %uncertainty in y-intercept + %uncertainty in gradient
- Maximum/minimum methods
- Max Q = max gradient × max E = max gradient / min y - intercept
- Min Q = min gradient × min E = min gradient / max y – intercept
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