3.7.3.2FieldsStretch

Charging and Discharging Capacitors

A capacitor discharging through a resistor never quite reaches zero charge — it just gets closer and closer, forever, at an ever-slowing rate. This is $exponential decay$ , and a single number, the $time constant$ , tells you exactly how fast that decay happens for any resistor–capacitor (RC) combination.

40 min Video by Lascells Science 3.7.3 Capacitance

RC Circuit: Charging and Discharging capacitors — A-Level Physics PracticalWatch the full walkthrough before the notes below.

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What you'll be able to do

Use the discharge equation $Q = Q_{0} e^{- t / R C}$ (and the equivalent forms for V and I)
Define the time constant $τ = R C$ and interpret it physically
Use the charging equation $Q = Q_{0} (1 - e^{- t / R C})$
Use logarithms to find the time constant from experimental data
Calculate the time for charge to fall to half its initial value

Discharging a capacitor

When a charged capacitor discharges through a resistor, the charge remaining decreases $exponentially$ with time — falling quickly at first (when charge, and hence current, is large) and more and more slowly as charge (and current) diminishes. Because current, p.d. and charge are all proportional to each other in this circuit ( $V = Q / C$ , $I = V / R$ ), all three quantities follow exactly the same exponential decay shape.

Q = Q_{0} e^{- t / R C} V = V_{0} e^{- t / R C} I = I_{0} e^{- t / R C}

Exponential discharge of charge, p.d. and current (same time constant RC for all three).

1Use

Q = Q_{0} e^{- t / R C}

Q = 6.0 \times 1 0^{- 4} \times e^{- 8.0/4.0} = 6.0 \times 1 0^{- 4} \times e^{- 2.0}

e^{- 2.0} \approx 0.135

Answer

Q \approx 8.1 \times 1 0^{- 5}

The time constant

The $time constant$ , $τ = R C$ , is the time taken for the charge (or p.d., or current) to fall to $\frac{1}{e}$ (about 37%) of its initial value during discharge — or, during charging, the time taken to reach $(1 - \frac{1}{e})$ (about 63%) of its final value. A larger resistance means a slower discharge (less current at any given p.d.); a larger capacitance means more charge must flow away for the same drop in p.d. — both increase the time constant.

τ = R C

Time constant of an RC discharge/charge circuit.

1Use

τ = R C = 2.0 \times 1 0^{3} \times 500 \times 1 0^{- 6}

τ = 1.0

3By definition, the p.d. reaches 37% of its initial value after exactly one time constant.

Answer

τ = 1.0

s; the p.d. falls to 37% of

V_{0}

after 1.0 s.

Tip — After 5 time constants, a discharging capacitor has lost more than 99% of its original charge — in practice, this is usually treated as "fully discharged".

Charging a capacitor

When a capacitor charges through a resistor from a fixed-voltage supply, charge (and p.d.) build up quickly at first and level off as they approach their maximum values — the mirror image of the discharge curve. Current, by contrast, starts at its maximum (when the capacitor is empty and offers no opposition) and decays exponentially toward zero as the capacitor becomes fully charged.

Q = Q_{0} (1 - e^{- t / R C}) V = V_{0} (1 - e^{- t / R C})

Charge and p.d. rising toward their final values during charging.

1Use

V = V_{0} (1 - e^{- t / R C})

with

t = R C = 3.0

s, so

t / R C = 1

V = 10 \times (1 - e^{- 1}) = 10 \times (1 - 0.368)

Answer

V \approx 6.3

V (about 63% of the final value, exactly as the time constant’s definition predicts).

Finding the time constant experimentally

Taking natural logarithms of the discharge equation turns it into a straight-line equation: plotting $ln Q$ (or $ln V$ , or $ln I$ ) against time $t$ gives a straight line with gradient $- \frac{1}{R C}$ and $y$ -intercept $ln Q_{0}$ . This is the standard experimental method for finding a circuit’s time constant (and hence, if $R$ is known, the capacitance) from real discharge data.

ln Q = ln Q_{0} - \frac{t}{R C}

Linearised discharge equation: gradient = −1/RC.

1Gradient

= - \frac{1}{R C}

, so

R C = \frac{1}{0.25} = 4.0

2Rearrange

τ = R C

C = \frac{τ}{R} = \frac{4.0}{2.0 \times 1 0 ^{3}}

Answerτ = 4.0 s; C = 2.0×10⁻³ F (2000 µF).

Tip — The "half-life" of a discharging capacitor (0.69 RC) is shorter than the time constant (RC) itself — don’t confuse the two; the time constant corresponds to a fall to 37%, not 50%.

Equation recap

Q = Q_{0} e^{- t / R C}

Exponential discharge.

Q = Q_{0} (1 - e^{- t / R C})

Exponential charging.

τ = R C

Time constant.

t_{1/2} = R C ln 2 \approx 0.69 R C

Time for charge to halve.

Common mistakes to avoid

Confusing the time constant (fall to 37%) with the time for charge to halve.

The time constant RC corresponds to a fall to 1/e (≈37%); the time to reach exactly half is a different value, RC ln2 (≈0.69RC).

Thinking a capacitor fully discharges in exactly one time constant.

Discharge is exponential and never mathematically reaches exactly zero — after 5 time constants it’s conventionally treated as "fully discharged" (under 1% remaining).

Assuming current follows the same rising shape as charge and p.d. during charging.

Current starts at its MAXIMUM at t=0 and decays exponentially during charging — the opposite shape to the rising charge/p.d. curves.

Key takeaways

Discharge: Q, V and I all fall exponentially, Q = Q₀e^(−t/RC), with the same time constant.
Charging: Q and V rise toward their final values as Q₀(1−e^(−t/RC)); current instead decays exponentially from its initial maximum.
Time constant τ = RC is the time to fall to 37% (discharge) or rise to 63% (charge) of the relevant value.
Plotting ln(Q) against t gives a straight line, gradient −1/RC — the standard way to find RC experimentally.
Time for charge to halve = RC ln2 ≈ 0.69 RC.

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Charging and Discharging Capacitors

A capacitor discharging through a resistor never quite reaches zero charge — it just gets closer and closer, forever, at an ever-slowing rate. This is

exponential decay

, and a single number, the

time constant

, tells you exactly how fast that decay happens for any resistor–capacitor (RC) combination.

40 min Video by Lascells Science 3.7.3 Capacitance