# Capacitor charging and discharging

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u22-capacitor-charging-and-discharging/

This module covers the exponential behaviour of capacitors during charging from a DC source and discharging through a resistor. You will learn to calculate key quantities, interpret graphs, and solve common CIE A-Level exam problems.

**Prerequisites:** [Definition of capacitance](https://www.owlsprep.com/study/cie-9702-u22-capacitance-fundamentals/); [Ohm's law and Kirchhoff's circuit laws](https://www.owlsprep.com/study/cie-9702-u10-dc-circuits-fundamentals/)

## Learning objectives

- Describe the charging and discharging processes of capacitors in RC circuits
- Derive and use exponential equations for charge, voltage and current
- Calculate and interpret the time constant of an RC circuit
- Sketch and interpret graphs for charging and discharging processes

## The Capacitor Charging Process

When an uncharged capacitor is connected in series with a resistor and a DC battery, charge starts to accumulate on the capacitor plates. Current flows in the circuit until the potential difference across the capacitor equals the emf of the battery, at which point current drops to zero.

**Capacitor Charging** — Process where charge accumulates on capacitor plates when connected to a DC source, until the capacitor voltage matches the source emf

*Example:* A 10 μF capacitor connected to a 12 V battery through a resistor will charge until it has 12 V across its plates.

**Worked example:** A 10 μF capacitor is charged through a 50 kΩ resistor from a 12 V battery. Calculate the charge on the capacitor after 1 time constant, and the initial current at the start of charging.

1. First calculate the time constant $\tau = RC$:
2. $$\tau = (50 \times 10^3 \Omega)(10 \times 10^{-6} F) = 0.5 \text{ s}$$
3. Final charge when fully charged is $Q_0 = C V_0 = 10 \times 10^{-6} \times 12 = 1.2 \times 10^{-4} C$. For charging, charge at time $t$ is $Q = Q_0(1 - e^{-t/\tau})$. At $t = \tau$:
4. $$Q = 1.2 \times 10^{-4} (1 - e^{-1}) ≈ 0.63 \times 1.2 \times 10^{-4} = 7.56 \times 10^{-5} C$$
5. At the start of charging ($t=0$), all voltage is across the resistor, so initial current is:
6. $$I_0 = \frac{V_0}{R} = \frac{12}{50 \times 10^3} = 2.4 \times 10^{-4} A = 0.24 mA$$

> **tip**
>
> At the start of charging: $V_C = 0$, $I = I_{max}$. At the end of charging: $V_C = V_0$, $I = 0$.

## The Capacitor Discharging Process

When a charged capacitor is connected across a resistor, charge flows from one plate to the other through the resistor until the potential difference across the capacitor is zero. The rate of discharge is proportional to the remaining charge, leading to exponential decay.

**Exponential Discharge** — Discharge of a capacitor follows exponential decay, where the charge, voltage and current all decrease exponentially from their initial values to zero.

**Worked example:** A 20 μF capacitor is charged to 10 V, then discharged through a 100 kΩ resistor. Calculate the voltage across the capacitor after 2 seconds, and the current after 1 second.

1. Calculate the time constant:
2. $$\tau = RC = (100 \times 10^3 \Omega)(20 \times 10^{-6} F) = 2 \text{ s}$$
3. For discharge, voltage follows $V = V_0 e^{-t/\tau}$. Substitute $t = 2$ s:
4. $$V = 10 e^{-2/2} = 10 e^{-1} ≈ 3.68 V$$
5. Initial current is $I_0 = V_0/R = 10/(100 \times 10^3) = 1 \times 10^{-4} A$. At $t = 1$ s:
6. $$I = I_0 e^{-t/\tau} = 1 \times 10^{-4} e^{-0.5} ≈ 6.07 \times 10^{-5} A = 60.7 \mu A$$

> **note**
>
> Charge, voltage and current all decay exponentially during discharge with the same time constant.

> **Exam tip:** Always check whether the question asks for charging or discharging before selecting the correct equation.

## Time Constant and Graph Interpretation

The time constant $\tau = RC$ determines how fast charging or discharging occurs. A larger $\tau$ (bigger R or C) gives a slower process, while a smaller $\tau$ gives faster charge/discharge.

**Time Constant** — For discharge: time taken for charge to fall to $1/e ≈ 37\%$ of initial value. For charging: time taken for charge to rise to $≈63\%$ of final value.

*Notation:* \tau = RC

| Quantity | Charging (vs time) | Discharging (vs time) |
| --- | --- | --- |
| Charge $Q$ | Rises exponentially from 0 to $Q_0$ | Falls exponentially from $Q_0$ to 0 |
| Capacitor Voltage $V_C$ | Rises exponentially from 0 to $V_0$ | Falls exponentially from $V_0$ to 0 |
| Circuit Current $I$ | Falls exponentially from $I_0$ to 0 | Falls exponentially from $I_0$ to 0 |

> **mnemonic**
>
> 63 up, 37 down: After 1τ, 63% of final charge for charging, 37% of initial charge left for discharge.

**Worked example:** The half-life of discharge of a capacitor is 10 s. Calculate the time constant of the circuit.

1. Half-life $T_{1/2}$ is the time when $Q = Q_0/2$. Substitute into the discharge equation:
2. $$\frac{Q_0}{2} = Q_0 e^{-T_{1/2}/\tau} → \frac{1}{2} = e^{-T_{1/2}/\tau}$$
3. Take natural logarithms of both sides:
4. $$\ln(1/2) = -\frac{T_{1/2}}{\tau} → -\ln 2 = -\frac{T_{1/2}}{\tau} → \tau = \frac{T_{1/2}}{\ln 2}$$
5. Substitute $T_{1/2} = 10$ s:
6. $$\tau = \frac{10}{0.693} ≈ 14.4 \text{ s}$$

## Derivation of the Discharge Equation

CIE regularly asks for the derivation of the exponential discharge equation from first principles, so it is important to remember all steps.

**Derivation:** Derive the exponential equation for charge during capacitor discharge

*Starting from:* Kirchhoff's Voltage Law for a discharging RC circuit

1. For a charged capacitor discharging through a resistor, sum of potential differences around the loop is zero. $V_C = Q/C$, $V_R = IR = -R \frac{dQ}{dt}$ (negative because Q decreases with time).
2. $$\frac{Q}{C} - R \frac{dQ}{dt} = 0 → \frac{dQ}{Q} = -\frac{dt}{RC}$$
3. Integrate both sides, with initial condition $Q = Q_0$ at $t=0$:
4. $$\int_{Q_0}^{Q} \frac{dQ}{Q} = -\frac{1}{RC} \int_{0}^{t} dt$$
5. Evaluate the integrals:
6. $$\ln Q - \ln Q_0 = -\frac{t}{RC} → \ln\left(\frac{Q}{Q_0}\right) = -\frac{t}{RC}$$

*Conclusion:* Exponentiating both sides gives the final discharge equation: $Q = Q_0 e^{-t/RC}$

> **Exam tip:** Always state the initial condition and show full integration steps to get full marks for derivation questions.

## Common pitfalls

- **Wrong:** Using the discharge equation $Q = Q_0 e^{-t/\tau}$ for charging calculations
  - Why it fails: Confusing the functional form for charging and discharging
  - Correct: Use $Q = Q_0 (1 - e^{-t/\tau})$ for charging, which approaches $Q_0$ from 0 as $t$ increases.
- **Wrong:** Calculating time constant as $R + C$ instead of $R \times C$
  - Why it fails: Mistaking the definition of time constant, mixing units
  - Correct: Time constant $\tau = RC$, which has units of ohm × farad = second, as required.
- **Wrong:** Assuming half-life equals the time constant
  - Why it fails: Confusing the definition of half-life (time to halve) and time constant (time to fall to 1/e)
  - Correct: Use the relationship $T_{1/2} = \tau \ln 2 ≈ 0.69 \tau$ to convert between the two values.
- **Wrong:** Assuming voltage across the resistor is constant during charging/discharging
  - Why it fails: Treating the RC circuit as a steady-state DC circuit
  - Correct: Voltage across the resistor changes as current changes, proportional to the instantaneous current in the circuit.
- **Wrong:** Forgetting that current decays exponentially during charging, just like discharge
  - Why it fails: Only focusing on the change in charge/voltage, not current
  - Correct: Current starts at maximum and decays exponentially to zero for both charging and discharging processes.

## Cheatsheet

| Process | Charge Q | Voltage V | Current I | Time Constant |
| --- | --- | --- | --- | --- |
| Discharge | $Q_0 e^{-t/RC}$ | $V_0 e^{-t/RC}$ | $I_0 e^{-t/RC}$ | $\tau = RC$ |
| Charging | $Q_0 (1 - e^{-t/RC})$ | $V_0 (1 - e^{-t/RC})$ | $I_0 e^{-t/RC}$ | $\tau = RC$ |
| Half-Life |  |  |  | $T_{1/2} = \tau \ln 2 ≈ 0.69\tau$ |
| Key Rule | 1τ discharge: $Q ≈ 0.37Q_0$ | 1τ charge: $V ≈ 0.63V_0$ |  | 5τ ≈ 99% complete |

## What's next

Understanding capacitor charging and discharging is core to working with RC circuits, which appear across CIE A-Level Physics in alternating current topics, electromagnetic induction, and pulse circuits. The exponential behaviour you learn here also prepares you for other common exponential processes in physics, such as radioactive decay. Now that you have mastered this sub-topic, you can move on to related topics in capacitance, or extend your knowledge to alternating current circuits with capacitors.

- [Magnetic fields](https://www.owlsprep.com/study/cie-9702-u23-overview/)
- [Magnetic field concepts](https://www.owlsprep.com/study/cie-9702-u23-magnetic-field-concepts/)
- [Force on current-carrying conductor](https://www.owlsprep.com/study/cie-9702-u23-force-on-current-carrying-conductor/)

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