use equations of the form x = x0 e–(t / RC) where x could represent current, charge or potential difference for a capacitor discharging through a resistor

Discharging a Capacitor – A‑Level Physics (Cambridge 9702)

Learning Objective

Use the exponential decay equation x = x₀ e^‑t/RC to describe how the charge Q, voltage V and current I vary with time when a capacitor discharges through a resistor. Relate these results to the energy stored in the capacitor, the power dissipated in the resistor, and to the broader syllabus topics on capacitance, electromagnetic induction and energy.

1. Capacitance – Syllabus 19.1

• Definition: Capacitance C is the ability of a device to store charge per unit potential difference.
  

    C = Q / V

• Unit: farad (F) = coulomb per volt (C V⁻¹).
• Energy stored: The syllabus lists the result U = ½ C V². It follows from the work done in moving charge:

    U = ∫ V dQ = ∫ (Q/C) dQ = ½ C V².

• Sign convention: The positively charged plate carries Q > 0 and is at the higher potential.

2. Energy, Power and Dissipation – Syllabus 19.2

• Instantaneous energy during discharge:

    U(t) = ½ C V(t)² = ½ C V₀² e^‑2t/RC

• Instantaneous power in the resistor:

    P(t) = V(t) I(t) = \dfrac{V₀²}{R}\,e^‑2t/RC

• Energy check (AO2):

    \displaystyle\int_{0}^{\infty} P(t)\,dt = \frac12 C V₀^{2} – the total energy initially stored is completely converted into heat in the resistor.

3. Discharging a Capacitor – Syllabus 19.3 (Core)

3.1 Physical Situation

A capacitor of capacitance C is charged to an initial voltage V₀ (charge Q₀ = C V₀). At t = 0 a switch is closed, connecting the capacitor across a resistor of resistance R. The stored charge then flows through the resistor until the capacitor is (practically) empty.

3.2 Kirchhoff’s Loop Rule

For a closed loop with no external emf the algebraic sum of potential differences is zero:

∑ V = 0 (Kirchhoff’s first law)

Applied to the RC discharge circuit:

VC – VR = 0

3.3 Derivation of the Decay Law

1. Write the loop equation (from 3.2): VC – VR = 0.
2. Capacitor voltage in terms of charge: V_C = Q / C.
3. Resistor voltage in terms of current: V_R = I R.
4. Current is the rate of loss of charge: I = –\dfrac{dQ}{dt} (negative sign indicates flow out of the positive plate).
5. Substitute (2)–(4) into the loop equation:

     \dfrac{Q}{C} = –R\,\dfrac{dQ}{dt}.

6. Separate variables and integrate:

     \dfrac{dQ}{Q} = –\dfrac{dt}{RC}

   \;\Longrightarrow\;

   \ln Q = –\dfrac{t}{RC} + \ln Q₀.

7. Exponentiate to obtain the charge decay law:

     Q(t) = Q₀\,e^{‑t/RC}.

8. From V = Q/C and I = –dQ/dt the voltage and current follow the same exponential form:

     V(t) = V₀\,e^{‑t/RC}, 

     I(t) = I₀\,e^{‑t/RC}, where I₀ = V₀/R.

3.4 Time Constant τ = RC

• After one time constant (t = τ) the quantity has fallen to e⁻¹ ≈ 0.37 of its initial value.
• After 3τ it is ≈ 5 % of the start value.
• Practical “fully discharged” condition: t ≈ 5τ (≈ 0.7 % of the initial value).
• Non‑idealities: real capacitors have a small leakage resistance and resistors have tolerances (±1 % to ±5 %). These affect the exact value of τ but not the exponential form.

3.5 Graphical Interpretation (Semi‑log Plot)

• Plotting ln V (or ln I) against time yields a straight line.
• Slope = ‑1/τ, intercept = ln V₀. This provides a quick experimental method to determine the time constant.

3.6 Link to the Charging Process

The charging of a capacitor through a resistor obeys the same differential equation, but the sign of the exponent is opposite. The solution is

V(t) = Vs\,(1 – e^{‑t/RC}), where Vs is the source voltage. Recognising this symmetry prevents sign errors when switching between charging and discharging problems.

3.7 Experimental Context – Practical Tip

• Set up the RC circuit with a DC source, a switch, and a fast‑response oscilloscope (or a multimeter with a “record” function).
• Trigger the oscilloscope at the instant the switch closes; the voltage trace will display the exponential decay.
• Measure the voltage at two known times (e.g., t = τ and t = 2τ) and verify the ratios 0.37 and 0.14.
• Safety: never exceed the capacitor’s voltage rating; discharge high‑voltage capacitors through a suitable resistor before handling.

3.8 Relevance to Electromagnetic Induction – Syllabus 20.5

If a coil in a varying magnetic field is part of an RC circuit, Faraday’s law induces an emf ε = –dΦ/dt. The resulting current satisfies the same first‑order differential equation, with the induced emf playing the role of the source voltage. Thus the RC discharge solution is a special case of the general response of RL or RLC circuits to an induced emf.

4. Worked Example

Problem: A 10 µF capacitor is charged to 12 V and then discharged through a 4.7 kΩ resistor. Find the voltage across the capacitor after t = 0.020 s.

1. Calculate the time constant:

     τ = RC = (4.7 × 10³ Ω)(10 × 10⁻⁶ F) = 0.047 s.

2. Apply the voltage decay law:

     V(t) = V₀ e^{‑t/τ}.

3. Insert the numbers:

     V(0.020 s) = 12 V · e^{‑0.020/0.047}.

4. Evaluate the exponent:

     ‑0.020/0.047 ≈ ‑0.426.

5. Compute the exponential factor (calculator or table):

     e^{‑0.426} ≈ 0.653.

6. Final voltage:

     V(0.020 s) ≈ 12 V · 0.653 ≈ 7.8 V.

5. Common Mistakes & How to Avoid Them

• Sign of the current: During discharge I = –dQ/dt. The magnitude follows the exponential law; the negative sign only indicates direction.
• Confusing RC with a total discharge time: RC is the time constant that sets the scale of the decay, not the time required to reach zero.
• Unit conversion errors: Always convert µF → F (×10⁻⁶) and kΩ → Ω (×10³) before calculating τ.
• Using the charging formula for discharge (or vice‑versa): Remember the exponent sign – discharge: e^{‑t/RC}; charging: 1 – e^{‑t/RC}.
• Neglecting the energy check: Verify that \int₀^{∞}P(t)dt = ½ C V₀². This satisfies the AO2 requirement to “evaluate information”.
• Ignoring non‑idealities: Real circuits have resistor tolerances and capacitor leakage. Mentioning them earns marks for a broader understanding.

6. Summary Checklist (Exam‑Ready)

1. Write the loop equation: VC – VR = 0 (Kirchhoff’s first law).
2. Substitute VC = Q/C and VR = I R, then replace I with ‑dQ/dt.
3. Integrate to obtain Q(t) = Q₀ e^{‑t/RC}.
4. Derive V(t) = V₀ e^{‑t/RC} and I(t) = I₀ e^{‑t/RC} with I₀ = V₀/R.
5. Identify the time constant τ = RC and use it to estimate discharge progress (1τ ≈ 37 %, 3τ ≈ 5 %, 5τ ≈ <1 %).
6. Check units, sign conventions, and conversion factors.
7. Optional: Plot ln V vs. t to verify linearity and extract τ experimentally.
8. Remember the energy relation U(t) = ½ C V(t)² and the power dissipation P(t) = V(t)I(t); confirm that \int₀^{∞}P(t)dt = ½ C V₀².
9. Note real‑world effects: resistor tolerance and capacitor leakage may slightly modify the measured τ.