Capacitor charge and discharge (A-level only) — AQA A-Level Physics
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Capacitor charge and discharge describes how a capacitor stores charge on its plates and releases it through a resistor, following exponential curves governed by the time constant RC. AQA A-Level Physics requires students to interpret charging and discharging graphs, use the equations Q = Q0 e^(-t/RC) and I = I0 e^(-t/RC), and understand that the time constant tau = RC is the time for charge to fall to 37% of its initial value.
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Capacitor charge and discharge (A-level only) explained
Capacitor charge and discharge (A-level only) covers 5 specification points, including Graphical representation of charging and… and Corresponding graphs for Q, V and I against….
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Graphical representation of charging and discharging of capacitors through resistors.
When a capacitor charges through a resistor, the current is initially large because the p.d. across the capacitor is zero, so the full supply p.d. appears across the resistor. As charge builds, the capacitor p.d. rises and the current falls, so the charging curve is steep at first and flattens towards the supply value. Discharging reverses the process: the capacitor drives current through the resistor, and both charge and current decay towards zero. The shape is exponential, with a constant time constant τ = RC. A useful method is to plot the variable against time, identify the initial value and the asymptote, then check that the curve falls to about 37% of its starting value after one time constant.
Corresponding graphs for Q, V and I against time for charging and discharging.
For a charging capacitor, Q and V both rise from zero towards maximum values, while I starts at its largest value and decays towards zero. The Q and V curves have the same shape because Q = CV, so they are proportional. The current curve is also exponential but starts high, because at the first instant the capacitor has no p.d. and the supply p.d. is across the resistor. On discharge, Q, V and I all fall exponentially from their initial values towards zero, and I is in the opposite direction to the charging current. A reliable method is to write the initial value and the asymptote for each quantity, then sketch all three curves on the same time axis so their relationships are clear.
Interpretation of gradients and areas under graphs where appropriate.
Graphs summarise how capacitor quantities change with time. A gradient is the rate of change of the vertical quantity with respect to the horizontal quantity, found by drawing a tangent and dividing the vertical change by the horizontal change. An area under a graph is found by counting squares or by approximating with strips, and it equals the product of the two axis quantities. In capacitor work, a graph of charge Q against time has gradient equal to current I, because I = ΔQ/Δt. A graph of current I against time has area equal to charge Q, because Q = IΔt. A graph of potential difference V against charge Q has gradient equal to 1/C, since V = Q/C, and its area equals the energy stored, since W = ½QV. Always read axis labels and units before choosing a method.
Time constant RC.
The time constant of a capacitor–resistor circuit is the product of resistance and capacitance, τ = RC. It has units of seconds, because an ohm multiplied by a farad gives a second. During discharge, the charge, current and potential difference all fall exponentially, and after one time constant each quantity has fallen to 1/e of its initial value, about 37%. After about five time constants the values are close to zero. The time constant is the reciprocal of the exponential decay constant, so a larger R or a larger C gives a slower change. For charging, the capacitor reaches about 63% of its final value after one time constant. The time constant can be found from a graph by reading the time at which the value is 1/e of its starting value, or from the gradient of a graph of ln V against time, whose magnitude equals 1/RC.
Calculation of time constants including their determination from graphical data.
The time constant τ of a capacitor circuit is the product of resistance and capacitance, τ = RC, and equals the time for charge, potential difference or current to fall to 1/e of its initial value during discharge, about 37%. Calculate it directly from component values, ensuring ohms and farads are used so τ is in seconds. From graphical data, plot ln V against time for discharge; the gradient is −1/RC, so τ is the negative reciprocal of the gradient. Alternatively read the time at which V falls to V₀/e. For charging, τ is the time to reach 1 − 1/e of the final value, about 63%. Always state units and interpret the physical meaning of the value obtained.
Your focus
- Sketch charging and discharging graphs for charge, p.d. and current through a resistor.
- Interpret the initial value, asymptote and time constant from a capacitor graph.
- Explain how resistance and capacitance affect the rate of charging and discharging.
Show all 15 objectives
- Sketch corresponding Q, V and I graphs for charging and discharging.
- Relate the shapes of the Q and V graphs using Q = CV.
- Distinguish the direction and initial value of current during charging and discharging.
- Draw a tangent to a curve at a specified point and calculate the gradient with correct units.
- Estimate the area under a graph by counting squares or summing strips and state its physical meaning.
- Relate the gradient or area of a capacitor graph to the appropriate physical quantity using the defining equations.
- Calculate the time constant of a capacitor–resistor circuit using τ = RC with correct units.
- Interpret exponential charge and discharge graphs to obtain a time constant.
- Explain how resistance and capacitance affect the rate of charging and discharging.
- Calculate the time constant of a capacitor circuit from resistance and capacitance values.
- Determine the time constant from a graph of potential difference or charge against time.
- Interpret the physical meaning of the time constant in charging and discharging processes.
Which command word is this?
You must use the given data and appropriate equations to find a numerical answer. Show all steps of your working, including rearrangement and substitution, and give your final answer with the correct unit and an appropriate number of significant figures. Marks are awarded for correct substitution and accurate arithmetic.
Capacitor charge and discharge (A-level only) exam tips
Topic Overview
This topic covers how capacitors store charge and energy, and how the charge, potential difference, and current change over time when a capacitor charges through a resistor or discharges through a resistor. You will study the exponential nature of these processes, define the time constant RC, and use equations such as Q = Q0 e^(-t/RC) and I = I0 e^(-t/RC) to solve problems. Understanding these graphs and equations is essential for practical work and for explaining the behaviour of circuits containing capacitors.
In the wider AQA A-Level Physics course, this topic builds on earlier work on capacitance, electric fields, and DC circuits. It links to later topics such as alternating current and magnetic fields, where capacitors are used in filters and timing circuits. Mastery of capacitor charge and discharge is also crucial for the required practical on capacitors, where you investigate discharge through a resistor and determine the time constant.
Key Concepts
- →The time constant tau = RC is the time for a discharging capacitor to lose 63% of its charge (fall to 37% of its initial value), or for a charging capacitor to gain 63% of its final charge.
- →During discharge, charge, current, and potential difference all decrease exponentially according to Q = Q0 e^(-t/RC), I = I0 e^(-t/RC), and V = V0 e^(-t/RC).
- →During charging, the charge and pd increase exponentially towards a maximum value, while the current decreases exponentially from an initial maximum value.
- →The rate of discharge is proportional to the charge remaining on the capacitor, which leads to the exponential relationship.
- →After a time equal to 5RC, a capacitor is considered fully charged or discharged for practical purposes, as it has reached more than 99% of its final value.
Marking Points
- Charging curves for charge and p.d. rise from zero and approach the supply value asymptotically, never overshooting it.
- Discharging curves for charge and p.d. fall from their initial values towards zero asymptotically.
- Charging current starts at a maximum value equal to supply p.d. divided by resistance and decays towards zero.
- Discharging current starts at a maximum value equal to initial capacitor p.d. divided by resistance and decays towards zero.
- The time constant τ = RC determines the rate: a larger resistance or capacitance gives a slower approach to the final value.
- After one time constant, a discharging quantity has fallen to about 37% of its initial value, and a charging quantity has reached about 63% of its final value.
- During charging, Q and V rise from zero and approach maximum values asymptotically.
- During charging, I starts at a maximum and decays towards zero.
- During discharging, Q, V and I all decay from their initial values towards zero.
- The Q and V graphs are proportional because Q = CV, so they have the same shape and the same time constant.
- The discharging current direction is opposite to the charging current direction.
- All three quantities share the same time constant τ = RC, so their graphs change at the same rate.
- Identifies the gradient as the rate of change of the vertical quantity with respect to the horizontal quantity, and states the unit as the vertical unit divided by the horizontal unit.
- Draws a tangent at the required point and reads two well-separated points on the tangent to calculate the gradient, rather than using two data points on a curve.
- Recognises that the area under a graph equals the product of the two axis quantities, and estimates it by counting squares or by summing strips.
- Applies the capacitor relationships: gradient of Q–t gives current I; area under I–t gives charge Q; gradient of V–Q gives 1/C; area under V–Q gives energy stored ½QV.
- Checks that the sign and magnitude of a gradient or area are physically sensible for the process, for example a discharging capacitor has a negative dQ/dt and therefore a negative current.
- Uses consistent SI units throughout, converting millicoulombs to coulombs, microfarads to farads and milliseconds to seconds before calculating.
- States that the time constant τ = RC and that it has the unit of seconds.
- Explains that after one time constant a discharging quantity falls to 1/e of its initial value, approximately 37%, and a charging quantity rises to about 63% of its final value.
- Uses the exponential relationships, such as Q = Q₀e^(−t/RC) and V = V₀e^(−t/RC), to calculate a quantity at a given time or to find the time taken to reach a given value.
- Determines the time constant from a graph, for example by reading the time at which the value is 1/e of its initial value, or from the magnitude of the gradient of a straight-line graph of ln V against t.
- Explains that a larger resistance or a larger capacitance increases the time constant and therefore slows the charging or discharging process.
- Converts units correctly, for example microfarads to farads and kilohms to ohms, before multiplying to find RC.
- States τ = RC and substitutes resistance in ohms and capacitance in farads to obtain a time in seconds.
- Explains that during discharge the quantity falls to 1/e of its initial value in one time constant, approximately 37%.
- For graphical determination, plots ln V against time and uses the gradient, since gradient = −1/RC, so τ = −1/gradient.
- Reads the time at which V = V₀/e from an exponential decay graph as an alternative to the logarithmic method.
- For charging, identifies τ as the time to reach 1 − 1/e of the final value, approximately 63%.
- Checks that the calculated or graphical value is consistent with the component values and includes the correct unit, the second.
Examiner Tips
- 💡Label the axes with the quantity and unit, and mark the initial value and the asymptote before sketching the curve.
- 💡Use the time constant to justify the shape: state that after one τ the value is about 37% or 63% of the way to its final value.
- 💡When comparing two circuits, refer to the product RC rather than only to R or C separately.
- 💡Sketch Q, V and I on separate axes but with the same time scale so the shared time constant is visible.
- 💡State the initial value of each quantity before sketching, especially for current during charging.
- 💡Use the relationship Q = CV to justify why the Q and V curves have the same shape.
- 💡Annotate the graph with the tangent or the counted squares so the examiner can see your method, and state the coordinates you used.
- 💡Quote the gradient or area with its unit, and comment on what the value means physically in the context of the capacitor.
- 💡When estimating an area, show the number of squares and the value of one square, then multiply, so partial credit is clear.
- 💡Show the substitution into τ = RC with converted units, then state the answer in seconds.
- 💡When reading a time constant from a graph, mark the 1/e value on the axis and draw across to the curve before reading the time.
- 💡Link any calculated time constant to the physical behaviour, such as stating that the capacitor will take about five time constants to discharge almost completely.
- 💡Show the rearrangement from gradient = −1/RC to τ = −1/gradient explicitly so the method is clear.
- 💡When reading a value from a graph, mark the construction lines on the graph and quote the coordinates used.
- 💡Give the final answer to an appropriate number of significant figures with the unit s, and comment briefly on whether it is physically sensible.
- 💡Always state the equation you are using and show substitution clearly. Many marks are lost through arithmetic errors or missing units.
- 💡When sketching graphs, label the axes with the correct quantity and unit, and indicate the initial value and the asymptote. For discharge, the curve should be a smooth exponential decay starting from the initial value.
- 💡In practical questions, remember that the time constant can be found from the gradient of a ln(V) against time graph, since ln(V) = ln(V0) - t/RC, giving a gradient of -1/RC.
Common Mistakes
- Drawing charging and discharging curves as straight lines: the correct curves are exponential, with a changing gradient.
- Showing the charging p.d. curve passing above the supply p.d.: the correct curve approaches the supply p.d. asymptotically from below.
- Confusing the current and p.d. curves during charging: the correct current curve starts high and decays, while the p.d. curve starts at zero and rises.
- Drawing the charging current curve as rising with the p.d. curve: the correct current curve starts at its maximum and decays.
- Making the Q and V curves different shapes: the correct curves are proportional and therefore identical in shape.
- Showing discharge current in the same direction as charging current: the correct discharge current flows the opposite way.
- Using two points on a curved graph to find a gradient instead of drawing a tangent at the required point; the correction is to draw a tangent that touches the curve at that point and to read two points on the tangent itself.
- Confusing the roles of gradient and area, for example treating the area under a Q–t graph as current; the correction is to check the units, since the area under Q–t has units C·s, which is not current.
- Forgetting to convert prefixes before calculating, such as using microfarads as farads; the correction is to convert every quantity to SI base or coherent units first, so that 47 μF becomes 47 × 10⁻⁶ F.
- Treating the time constant as the time for the capacitor to become fully discharged; the correction is that after one time constant the quantity has fallen only to about 37% of its initial value, and full discharge is approached only after several time constants.
- Adding resistance and capacitance, or multiplying incorrectly, when finding the time constant; the correction is to use τ = RC with both values in SI units.
- Using the natural logarithm of a ratio without the correct sign or base, for example confusing ln with log₁₀; the correction is to use natural logarithms consistently in the exponential decay relationship.
- Using resistance in kilohms or capacitance in microfarads without converting to ohms and farads, giving a time constant wrong by powers of ten; correct by converting all values to SI base units before multiplying.
- Taking the gradient of a ln V against time graph as τ rather than as −1/RC; correct by recognising that τ is the negative reciprocal of the gradient.
- Assuming the capacitor fully discharges in one time constant; correct by stating that one time constant leaves about 37% of the initial value, and full discharge is approached only after several time constants.
- Students often think that the current is constant during discharge. In fact, the current decreases exponentially because it is driven by the decreasing pd across the capacitor.
- Many believe that the time constant is the time to fully charge or discharge. It is actually the time to reach approximately 63% of the final value (charging) or to fall to 37% of the initial value (discharging).
- Some students confuse the charging and discharging equations, using Q = Q0 e^(-t/RC) for charging. For charging, the correct form is Q = Q0(1 - e^(-t/RC)).
Revision Plan
- 1Day 1-2: Review the definitions of capacitance and the equations for charge, pd, and energy. Practise using C = Q/V and E = 1/2 CV^2 with numerical problems.
- 2Day 3-4: Learn the charging and discharging equations and the meaning of the time constant. Sketch graphs of Q, I, and V against time for both charging and discharging, labelling key points such as 0.37Q0 and 0.63Q0.
- 3Day 5-6: Work through calculations involving exponential decay, including finding time, resistance, or capacitance. Use natural logarithms to linearise data and determine RC from a graph.
- 4Day 7-8: Complete past paper questions on capacitor charge and discharge, focusing on practical analysis and graph interpretation. Mark your answers using the mark scheme and note common errors.
- 5Day 9-10: Revise the required practical on discharging a capacitor, including how to set up the circuit, collect data, and analyse it to find the time constant. Test yourself with active recall questions.
Exam Question Types
- 📋Calculation questions: These ask you to use the exponential equations to find charge, current, pd, time, resistance, or capacitance. Always show your working and give units. Be prepared to rearrange the equations and use natural logarithms.
- 📋Graph interpretation questions: You may be given a graph of charge, pd, or current against time and asked to determine the time constant, initial value, or the value at a specific time. You might also be asked to sketch the graph for a different resistance or capacitance.
- 📋Practical-based questions: These describe an experiment to investigate capacitor discharge and ask you to suggest improvements, identify sources of error, or explain how to determine the time constant from the data. You may need to explain why a logarithmic plot is used.
- 📋Explain questions: These require you to describe the physical processes during charging and discharging, using correct terminology such as 'exponential decay', 'time constant', and 'potential difference'. You might be asked to explain why the current decreases or why the process is exponential.
Command Word Expectations (AQA)
You must use the given data and appropriate equations to find a numerical answer. Show all steps of your working, including rearrangement and substitution, and give your final answer with the correct unit and an appropriate number of significant figures. Marks are awarded for correct substitution and accurate arithmetic.
You must provide a detailed justification that links cause and effect, using correct physics terminology. For example, explaining why the current decreases during discharge requires stating that the pd across the capacitor decreases, so the current (I = V/R) decreases, leading to a slower rate of discharge. Simply stating 'because it is exponential' is not sufficient.
You must use the information provided, often from a graph or table, to find a value. This may involve reading values from a graph, calculating a gradient, or using an equation. You should show how you obtained your answer, including any construction lines on a graph, and state the value with its unit.
How Students Lose Marks (Examiner Pitfalls)
Step-by-Step Worked Solutions
Question: A 470 microfarad capacitor is charged to 12 V and then discharged through a 2.2 kilo-ohm resistor. Calculate the time taken for the potential difference across the capacitor to fall to 3.0 V.
- 1.Step 1: Identify given facts: C = 470 x 10^-6 F, R = 2.2 x 10^3 ohms, initial pd V0 = 12 V, final pd V = 3.0 V.
- 2.Step 2: Calculate the time constant: tau = RC = (2.2 x 10^3) x (470 x 10^-6) = 1.034 s.
- 3.Step 3: Use the discharging equation V = V0 e^(-t/RC). Rearrange to t = -RC ln(V/V0).
- 4.Step 4: Substitute values: t = -1.034 x ln(3.0/12) = -1.034 x ln(0.25) = -1.034 x (-1.386) = 1.43 s.
- 5.Step 5: State final answer with unit: t = 1.4 s (2 significant figures).
Question: A student charges a capacitor through a resistor and records the potential difference across it every 5 seconds. The results are: 0 s, 0 V; 5 s, 3.8 V; 10 s, 6.0 V; 15 s, 7.5 V; 20 s, 8.4 V; 25 s, 9.0 V. The supply pd is 10 V. Determine the time constant of the circuit and explain how you obtained your answer.
- 1.Step 1: Recognise that the capacitor is charging and the pd follows V = V0(1 - e^(-t/RC)).
- 2.Step 2: The final pd is 10 V, so the time constant is the time for V to reach 63% of 10 V, which is 6.3 V.
- 3.Step 3: From the data, V = 6.0 V at 10 s and V = 7.5 V at 15 s. The 6.3 V point lies between 10 s and 15 s.
- 4.Step 4: Use interpolation or plot a graph of V against t and read the time at V = 6.3 V. Approximately t = 10.8 s.
- 5.Step 5: Alternatively, use the equation with two data points to solve for RC. Using V = 6.0 V at t = 10 s: 6.0 = 10(1 - e^(-10/RC)), so e^(-10/RC) = 0.4, giving -10/RC = ln(0.4) = -0.916, so RC = 10.9 s.
- 6.Step 6: State the time constant: RC = 11 s (2 significant figures).