An experiment was undertaken to investigate one of the circuit properties of a capacitor. A capacitor C was connected via a switch S to a resistance R and a voltmeter V.

The initial potential difference across C was 12V. The switch S was closed and the potential difference V across R was measured at various times t. The data collected, along with error bars, are shown plotted below.

(i) On the graph opposite, draw a best-fit line for the data starting from t = 0. 

(ii) It was hypothesized that the decay of the potential difference across the capacitor
is exponential. Determine, using the graph, whether this hypothesis is true or not.

The time constant τ of the circuit is defined as the time it would take for the capacitor to discharge were it to keep discharging at its initial rate. Use the graph in (a) to calculate the 

(i) initial rate of decay of potential difference V.

(ii) time constant τ.

The time constant τ = RC where R is the resistance and C is a property called capacitance. The effective resistance in the circuit is 10 MΩ. Calculate the capacitance C.

(i) smooth curve; that passes through all error bars;

 Award [1 max] if an obvious straight line is drawn through first three points.

Award [1 max] if line touches time axis.

Do not penalize if line starts beyond zero.

Do not allow upward curve at high t in first marking point.

Do not allow double or kinked lines.

(ii) correctly identifies three points/intervals from own graph;

correctly processes these three using exponential/half-life/constant ratio/relationship;

to conclude that decay is exponential;

within uncertainty;

Award [0] for a bald statement.

Award [1] for a straight line or portion of straight line leading to conclusion that decay is not exponential.

Award [1] if uncertainties are not considered and conclusion that decay is not exponential.

(i) evaluates a gradient over a minimum of 5 s to give an initial rate for example, \(\left( {\frac{{12}}{{9.5}} = } \right)\)1.3 (Vs^-1) for graph above;  (allow ECF from the graph);

Vs^-1(unit paper mark)
Clear evidence of calculation of gradient must be seen.
Accept use of (0,12) (5,8) to give 0.8 (Vs^-1) .

Allow answer left as fraction (eg \(\frac{4}{5}\)).

Accept negative gradient.

(ii) obtains evidenced answer for t intercept;
Accept one of the following methods:

Drawing tangent to initial part of graph (yields 9.5±3s).
Extending the first two/three points to the time axis (yields 11 – 19).
Using answer to (b)(i) to calculate intercept.

\(C = \left( {\frac{{\left( b \right)\left( {ii} \right)}}{{10 \times {{10}^6}}} = } \right)1.0 \times {10^{ - 6}}\left( {{\Omega ^{ - 1}}{\rm{s/F}}} \right)\);
Expect to see 10⁶ in denominator. Award [0] for absence of 10⁶ unless unit is in terms of MΩ.

(i) Few candidates scored full marks. Too often examiners saw poor quality draughtsmanship and ruler-straight lines through the first three points. Most candidates were able to ensure that their lines stayed within the bounds of the error bars. Candidates are encouraged to read through the whole question before attempting to answer – had this been done then they might have gained additional clues from what followed. It should be noted that the skill being tested here was the ability of the candidates to ignore the points and draw a smooth curve through the uncertainty bars.

(ii) Good tests of exponential change were beyond many. Examiners expect to see a systematic test carried through accurately. A suitable test might include identification of halflife behaviour, constant ratio behaviour, or fitting to an exponential function. Each of these approaches could have scored full marks. Often there were vague and meaningless statements about the asymptotic behaviour of the graph.

(i) This was adequately done by about half of the candidates although there were few confident tangents seen by examiners. Errors were to omit the unit and to try to work out a gradient over the full 30s.

(ii) Examiners expected to see an evidenced solution. Candidates who wrote down the answer without explanation gained little credit.

The answer here had to use the answer to (b)(ii) and most candidates were able to do this satisfactorily. A substantial number failed to take account of the prefix to the unit in the resistance and were a factor of 10⁶ out in their answer.