Charging, discharging and time constant
Key idea: H2 Physics lessons on circuit representation, material behaviour, networks and RC transients.
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The core idea
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Big question: How does a capacitor approach its final charge over time?
In an RC circuit, charging current falls as capacitor p.d. rises; discharging current and charge decay exponentially. The time constant τ = RC sets the scale: after one τ, a decaying quantity is 1/e of its initial value, while a charging quantity has reached 1 − 1/e of its final change. Gradients link current to dQ/dt.
Explain charging from the loop equation
For a capacitor charging through R from a source ε, ε = IR + V_C. Initially V_C = 0, so I₀ = ε/R. As charge builds, V_C = Q/C rises, leaving less p.d. across R; current and dQ/dt therefore fall.
The solutions are Q = Cε(1 − e^−t/RC), V_C = ε(1 − e^−t/RC) and I = (ε/R)e^−t/RC. After one time constant τ = RC, charging quantities have completed 63.2% of their total change.
Check your understanding: What are V_C and I immediately after an initially uncharged capacitor is connected?
V_C = 0 and I = ε/R in the ideal circuit.
Read decay and gradients correctly
During discharge, Q, V_C and current magnitude fall as e^−t/RC. The current direction is opposite to charging, so its algebraic sign depends on the chosen reference. After one τ, a decaying quantity is 1/e = 36.8% of its initial value.
Current is dQ/dt, so the gradient of a Q–t graph gives signed current. A tangent is needed at a particular time. A graph is not 'complete' after one τ; ideal exponential change approaches its final value asymptotically.
Check your understanding: A discharging voltage falls to 25% in 8.0 s. Express the time in time constants.
e^−t/τ = 0.25, so t/τ = ln 4 = 1.386.
Key ideas to keep
- A capacitor is not fully charged after one time constant.
- Charging current is largest initially and tends to zero.
- Keep initial, final and changing differences clear in exponential equations.
See the reasoning
Worked example
Read one discharging state from the time constant
Question: A 40 μF capacitor discharges from 12 V through 1.0 MΩ. Find V and I at 20 s.
Step 1: Find the time constant
Why: Every transient time appears as the ratio t/RC.
Working: τ = RC = (1.0×10⁶)(40×10⁻⁶) = 40 s.
Step 2: Calculate voltage
Why: Discharging voltage follows V = V₀e⁻ᵗ/τ.
Working: V = 12e⁻²⁰/⁴⁰ = 7.28 V.
Step 3: Find current and state its sign
Why: Current magnitude is V/R but direction depends on the chosen reference.
Working: |I| = 7.28/1.0×10⁶ = 7.28 µA; it flows in the discharge direction.
Answer: τ = RC = 40 s. V = 12e⁻⁰⋅⁵ = 7.28 V. Current magnitude is V/R = 7.28 μA and decays with the same exponential; its conventional sign depends on the chosen reference direction.
Check: After half a time constant, 60.7% remains; 7.28/12 = 0.607.
Use a hint if needed
Practise with support
Try this
A discharging quantity falls to 0.25x₀. Express t in terms of τ.
Hint: Take the natural logarithm of both sides.
Check your answer
0.25 = e⁻ᵗ⁄τ, so t = −τ ln(0.25) = 1.386τ.
Now work without the hint
Practise independently
Your turn
Describe current, charge and capacitor p.d. during charging and discharging, including initial/final values and τ.
Check your answer
During charging, Q and VC rise as x₀(1 − e⁻ᵗ⁄τ), while current falls as I₀e⁻ᵗ⁄τ. During discharge, Q, VC and current magnitude fall as x₀e⁻ᵗ⁄τ. τ = RC; after one τ a rise is 63% complete and a decay retains 37%.
Avoid these traps
Common mistakes
Common mistake
Charging or discharging is complete after one time constant.
What is wrong with this reasoning?
Show better thinking
After one τ, a rise is 63.2% complete and a decay retains 36.8%; the exponential approaches its limit asymptotically.
Write for the examiner
Exam guidance
Mark the initial value, final asymptote and one-time-constant value on every transient graph.
Exam-style practice [5 marks]
A 25 μF capacitor discharges through 0.80 MΩ from 10 V. Find τ and V after 30 s.
Plan before you answer
- Calculate RC in seconds.
- Use the decaying exponential.
- Compare with the one-τ benchmark.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
τ = RC = 20 s. V = 10e⁻³⁰⁄²⁰ = 2.23 V.
Come back in three days
Check what stayed with you
Recall question
After one time constant, state the fractions reached during charging and remaining during discharge.
Check the answer
Charging reaches 1 − e⁻¹ = 0.632 of the final change; discharge retains e⁻¹ = 0.368 of its initial value.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Topic 16 states no explicit exclusions. I–V graph gradient language always names the plotted axes; resistance is V/I at an operating point rather than automatically a graph gradient. Metal temperature dependence is explained through drift velocity and NTC behaviour through carrier number density. Terminal p.d. is E − Ir while a source supplies current. Capacitor rules are not copied from resistor rules. The exponential forms assume constant R and C, an ideal source or isolated discharge loop, and τ = RC.
- GCE A-Level H2 PhysicsTopic 16(l) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027