RC Circuits (Charging & Discharging)

Key idea: Use τ = RC and exponential equations to describe and calculate how charge, current and capacitor voltage change in RC charging/discharging circuits (A Level Physics).

GCE A-Level H2 Physics 2027

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Learning objectives

Analyse charging and discharging in RC circuits using the time constant.

1. Definitions (Must Know)

A. RC circuit

An RC circuit is a circuit containing a resistor R and a capacitor C.

B. Time constant, τ

The time constant, τ, is:

τ = RC

Unit: s.

C. Capacitor voltage, V_C

V_C is the potential difference across the capacitor.

D. Final (steady) value

The final value is the limiting value as t → ∞. In the ideal exponential model, the capacitor approaches this value but does not reach it exactly in finite time.

2. Key Ideas (What Earns Marks)

The syllabus expects you to use equations of the form:

decay: x = x₀ e^(-t/τ)
growth: x = x₀[1-e^(-t/τ)]

For a capacitor charging from 0 through a resistor R from a d.c. source of e.m.f. ε:

V_C(t) = ε[1-e^(-t/RC)]
Q(t) = Cε[1-e^(-t/RC)]
I(t) = (ε/R)e^(-t/RC)

For a capacitor discharging through a resistor R from initial capacitor voltage V₀:

V_C(t) = V₀ e^(-t/RC)
Q(t) = CV₀ e^(-t/RC)
|I(t)| = (V₀/R)e^(-t/RC)

Key time-constant facts:

at t = τ: charging reaches 0.63 of final; discharging falls to 0.37 of initial
by t ≈ 5τ: the change is essentially complete (about 99%)

At t = τ, charging reaches 1-e⁻¹ = 0.63 of its final value and discharging falls to e⁻¹ = 0.37 of its initial value.

3. Detailed Explanations

A. Charging (what changes with time)

At t = 0 (uncharged capacitor):

V_C = 0
current is maximum: I₀ = ε/R

As time increases:

Q and V_C increase towards their final values (Q_f = Cε, V_C → ε)
current decreases towards 0

Scaled RC curves: charging rises as 1−e^(−t/τ) and discharging falls as e^(−t/τ). — At one time constant (t = τ), charging reaches about 0.63 of the final value and discharging falls to about 0.37 of the initial value.

Values and uncertainty for RC charging and discharging (voltage vs time, scaled)
Series	Time (t/τ)	Voltage fraction (arbitrary units)
Charging: Vc/ε = 1 − e^(−t/τ)	0	0
Charging: Vc/ε = 1 − e^(−t/τ)	0.5	0.393
Charging: Vc/ε = 1 − e^(−t/τ)	1	0.632
Charging: Vc/ε = 1 − e^(−t/τ)	1.5	0.777
Charging: Vc/ε = 1 − e^(−t/τ)	2	0.865
Charging: Vc/ε = 1 − e^(−t/τ)	3	0.95
Charging: Vc/ε = 1 − e^(−t/τ)	4	0.982
Charging: Vc/ε = 1 − e^(−t/τ)	5	0.993
Discharging: Vc/V0 = e^(−t/τ)	0	1
Discharging: Vc/V0 = e^(−t/τ)	0.5	0.607
Discharging: Vc/V0 = e^(−t/τ)	1	0.368
Discharging: Vc/V0 = e^(−t/τ)	1.5	0.223
Discharging: Vc/V0 = e^(−t/τ)	2	0.135
Discharging: Vc/V0 = e^(−t/τ)	3	0.05
Discharging: Vc/V0 = e^(−t/τ)	4	0.018
Discharging: Vc/V0 = e^(−t/τ)	5	0.007
At t = τ (markers)	1	0.632
At t = τ (markers)	1	0.368

Values and uncertainty for RC charging and discharging (voltage vs time, scaled)
Series	Time (t/τ)	Voltage fraction (arbitrary units)
Charging: Vc/ε = 1 − e^(−t/τ)	0	0
Charging: Vc/ε = 1 − e^(−t/τ)	0.5	0.393
Charging: Vc/ε = 1 − e^(−t/τ)	1	0.632
Charging: Vc/ε = 1 − e^(−t/τ)	1.5	0.777
Charging: Vc/ε = 1 − e^(−t/τ)	2	0.865
Charging: Vc/ε = 1 − e^(−t/τ)	3	0.95
Charging: Vc/ε = 1 − e^(−t/τ)	4	0.982
Charging: Vc/ε = 1 − e^(−t/τ)	5	0.993
Discharging: Vc/V0 = e^(−t/τ)	0	1
Discharging: Vc/V0 = e^(−t/τ)	0.5	0.607
Discharging: Vc/V0 = e^(−t/τ)	1	0.368
Discharging: Vc/V0 = e^(−t/τ)	1.5	0.223
Discharging: Vc/V0 = e^(−t/τ)	2	0.135
Discharging: Vc/V0 = e^(−t/τ)	3	0.05
Discharging: Vc/V0 = e^(−t/τ)	4	0.018
Discharging: Vc/V0 = e^(−t/τ)	5	0.007
At t = τ (markers)	1	0.632
At t = τ (markers)	1	0.368

B. Discharging (what changes with time)

At t = 0 (initially charged):

V_C = V₀
current magnitude is maximum: |I₀| = V₀/R

As time increases:

Q, V_C, and |I| all decrease exponentially towards 0

C. Why the graphs are exponential (qualitative)

As the capacitor charges, V_C increases, so the resistor voltage V_R = ε-V_C decreases. Since I = V_R/R, the current decreases with time, so the charging rate slows down.

For discharging, V_C decreases with time, so the current magnitude decreases with time.

Log-graph skill: straight line for discharge

From V_C = V₀ e^(-t/τ), take natural logs: ln(V_C/V₀) = -t/τ

So a plot of ln(V_C/V₀) against t is a straight line with gradient -1/τ.

A straight-line plot of ln(V/V0) against t/τ with gradient −1. — In real data, you plot ln(V) vs t (or ln(V/V0) vs t). The gradient gives −1/RC and the intercept gives ln(V0).

Values for Linearising RC discharge: ln(V/V0) vs time (scaled)
Time (t/τ)	ln(V/V0) = −t/τ
0	0
1	-1
2	-2
3	-3
4	-4
5	-5

Values for Linearising RC discharge: ln(V/V0) vs time (scaled)
Time (t/τ)	ln(V/V0) = −t/τ
0	0
1	-1
2	-2
3	-3
4	-4
5	-5

4. Common Mistakes

Using τ = R/C instead of τ = RC.
Using t/τ without checking t is in seconds.
Using the “growth” equation for discharging (or vice versa).
Forgetting that current direction may be opposite to your chosen sign convention (use magnitudes if unsure).

5. Exam Tips

Rearrangement you will use a lot:
- e^(-t/τ) = x/x₀ ⇒ t = τ ln(x₀/x)
Charging to a target voltage V_T:
- V_T = ε[1-e^(-t/τ)]
- t = -τ ln(1-V_T/ε)
Quote τ explicitly before substituting numbers.

6. Worked Examples

Problem

R = 100 kΩ and

C = 10 μF. Find the time constant

τ.

Study the worked solution

Convert to SI units
Method
Express resistance in ohms and capacitance in farads.
Reason
The product Ω F is seconds when SI prefixes are included correctly.
Working
R = 1.00 × 10⁵ Ω and C = 1.0 × 10⁻⁵ F.
Calculate the time constant
Method
Use τ = RC.
Reason
The circuit’s exponential time scale is the resistance–capacitance product.
Working
τ = (1.00 × 10⁵)(1.0 × 10⁻⁵) = 1.0 s

Problem

An initially uncharged capacitor charges from a

12 V supply with

τ = 1.0 s. Find

V_C at

t = 2.0 s.

Try this before viewing the solution

Capacitor p.d.

Unit: V

Hints

Hint 1: choose rise rather than decay

Charging from zero follows

V_C = ε[1-e^(-t/τ)].

Hint 2: use elapsed time constants

Here

t/τ = 2.0.

View solution step by step

Substitute into the charging equation
Method
Use the exponential rise from zero.
Reason
The final capacitor p.d. is the supply e.m.f.
Working
V_C = 12[1-e^(-2.0/1.0)]
Evaluate
Method
Obtain approximately 10.4 V.
Reason
e⁻² = 0.135, leaving 86.5% of the rise completed.
Working
V_C = 12(1-0.135) = 10.4 V

Learner claim

A capacitor discharges from

9.0 V to

3.0 V with

τ = 0.50 s. A learner uses

t = τ ln(V_C/V₀) and obtains negative time. Diagnose the rearrangement and find

Try this before viewing the solution

Positive-time form

t = τ ln(V₀/VC)t = τ ln(VC/V₀)

View solution step by step

Start from exponential decay
Method
Use V_C/V₀ = e^(-t/τ).
Reason
The remaining fraction decreases from one.
Working
3.0/9.0 = e^(-t/0.50)
Handle the negative exponent
Method
Use t = τ ln(V₀/V_C).
Reason
Multiplying the logarithm by -τ reverses the ratio.
Working
t = 0.50 ln(9.0/3.0) = 0.50 ln 3 = 0.55 s

Examination question

A capacitor discharges from

12 V through

R = 2.0 kΩ with

C = 220 μF. Find

τ, the current magnitude at

t = 0, and the magnitude at

t = τ. [4 marks]

Try this before viewing the solution

Time constant and two currents

View solution step by step

Find the time constant
1 mark
Method
Multiply resistance and capacitance.
Reason
τ = RC sets the decay scale.
Working
τ = (2.0 × 10³)(220 × 10⁻⁶) = 0.44 s
Find initial current
1 mark
Method
Use the initial capacitor p.d. across R.
Reason
At t = 0, its magnitude is V₀/R.
Working
|I₀| = 12/2000 = 6.0 mA
Apply one-time-constant decay
1 mark
Method
Multiply by e⁻¹.
Reason
Discharge current magnitude follows the same exponential.
Working
|I(τ)| = e⁻¹(6.0) = 2.2 mA
State the direction convention
1 mark
Method
Distinguish magnitude from signed current.
Reason
The algebraic sign depends on the chosen reference direction.
Working
Magnitude 2.2 mA; direction is the discharge direction.

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Self-mark time constant, initial current, decayed current and direction statement.

Mark points

1 mark: Obtains 0.44 s.
1 mark: Obtains initial magnitude 6.0 mA.
1 mark: Obtains 2.2 mA at one time constant.
1 mark: Distinguishes magnitude from direction/sign.

Independent transfer

A capacitor charges from zero towards

ε = 10 V with

τ = 0.80 s. Find the time to reach

9.0 V and express it as a multiple of

τ.

Try this before viewing the solution

Rearrangement, time and τ multiple

Hints

Hint 1: isolate the remaining gap

For charging,

e^(-t/τ) = 1-V_C/ε.

View solution step by step

Find the remaining fraction
Method
Subtract the completed rise from one.
Reason
At 9.0 V, 10% of the final gap remains.
Working
e^(-t/τ) = 1-9.0/10 = 0.10
Solve for time
Method
Take the natural logarithm.
Reason
t/τ = - ln(0.10) = ln 10.
Working
t = 0.80 ln 10 = 1.84 s
State the time-constant multiple
Method
Give t = 2.30τ.
Reason
ln 10 = 2.303.
Working
t/τ = 2.30.

Next step

Attempt the 10%–90% charging interval independently.

7. Mind Stretchers

Mind stretcher 1: Time for a capacitor to charge from 10% to 90%Extension

Show that the time taken to go from 0.10ε to 0.90ε is t = τ ln 9.

Show Answer

Charging: V_C = ε[1-e^(-t/τ)].

At V_C = 0.10ε: 0.10 = 1-e^(-t₁/τ) ⇒ e^(-t₁/τ) = 0.90

At V_C = 0.90ε: 0.90 = 1-e^(-t₂/τ) ⇒ e^(-t₂/τ) = 0.10

Divide the two exponential factors:

e^((t₂-t₁)/τ) = 0.90/0.10 = 9

Therefore t₂-t₁ = τ ln 9.

Mind stretcher 2: Why is “about 5τ” considered complete?Extension

Explain why people often say an RC charging/discharging process is “essentially complete” after about 5τ.

Show Answer

For exponential decay, at t = 5τ: e^(-t/τ) = e⁻⁵ = 0.0067 So only about 0.67% of the initial value remains (or, for charging, it is within about 0.67% of the final value). That is “close enough” for many practical and exam purposes.

Mind stretcher 3: Optional (Enrichment)Extension

A. Where the exponential comes from (differential equation)

Charging:

ε = IR + V_C

With I = dQ/dt and V_C = Q/C:

ε = RdQ/dt + Q/C

This has the solution:

Q(t) = Cε[1-e^(-t/RC)]

Discharging:

0 = IR + V_C = RdQ/dt + Q/C

This has the solution:

Q(t) = Q₀ e^(-t/RC)

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Course and syllabus information

Course: GCE A-Level H2 Physics
Edition: GCE A-Level H2 Physics 2027

RC Circuits (Charging & Discharging)

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Learning objectives

1. Definitions (Must Know)

A. RC circuit

B. Time constant, τ

C. Capacitor voltage, V_C

D. Final (steady) value

2. Key Ideas (What Earns Marks)

3. Detailed Explanations

A. Charging (what changes with time)

B. Discharging (what changes with time)

C. Why the graphs are exponential (qualitative)

Log-graph skill: straight line for discharge

4. Common Mistakes

5. Exam Tips

6. Worked Examples

Find the time constant

Convert to SI units

Calculate the time constant

Charging to a given capacitor voltage

Substitute into the charging equation

Evaluate

Discharging to a given voltage

Start from exponential decay

Handle the negative exponent

Current during discharge

Find the time constant

Find initial current

Apply one-time-constant decay

State the direction convention

Time to reach 90% of final voltage (charging)

Find the remaining fraction

Solve for time

State the time-constant multiple

7. Mind Stretchers

Mind stretcher 1: Time for a capacitor to charge from 10% to 90%Extension

Mind stretcher 2: Why is “about 5τ” considered complete?Extension

Mind stretcher 3: Optional (Enrichment)Extension

A. Where the exponential comes from (differential equation)