RC transients
Specialize the generic first-order framework to linear RC networks. Use capacitor-voltage continuity for the initial state, reduce the post-switch linear resistive network seen by the capacitor to a Thévenin equivalent when valid, derive τ=RthC and vC(t)=v∞+[v0−v∞]e^(−t/τ), obtain current and resistor voltage from the same solution, solve source-free and driven switching examples, and verify endpoints, slopes, energy and dependent-source handling without relying on separate charging/discharging formulas.
- Determine vC(0+) from the pre-switch state using capacitor-voltage continuity.
- Find the post-switch equilibrium vC(∞) from the t>0 DC network.
- Derive the Thévenin-reduced RC equation C dvC/dt+(vC−Vth)/Rth=0.
- Derive τ=RthC and the complete response vC(t)=v∞+[v0−v∞]e^(−t/τ).
- Derive capacitor current and resistor/source currents from dvC/dt with consistent reference directions.
- Solve source-free discharge, step charging and arbitrary nonzero-initial-condition cases with one formula.
- Compute Rth correctly in networks with dependent sources by retaining dependent sources and using a test-source method when necessary.
- Check initial slope independently from the post-switch circuit at t=0+.
- Track capacitor energy and explain where stored energy goes during passive discharge.
- State the limits of the ideal one-capacitor linear model when leakage, ESR, nonlinear capacitance or additional energy-storage elements matter.
A switch changes a resistor network connected to one capacitor. Instead of asking whether the capacitor is “charging” or “discharging,” ask four objective questions: what voltage did it have just before switching, what network exists after switching, what voltage will that new network impose after a long time, and what resistance controls the exchange of charge?
03.1Reduce the post-switch network, not the pre-switch network
For t>0, remove the capacitor temporarily and look into its two terminals. If the remaining network is linear, represent it by a Thévenin source Vth and resistance Rth:
Reconnect C with voltage reference vC oriented consistently with Vth. KVL gives
With iC=C dvC/dt,
Thus
03.2Initial value from continuity; final value from DC equilibrium
Under finite non-impulsive current, capacitor voltage is continuous:
If the t>0 sources are constant and the network is stable, at long time iC→0. No current flows through Rth, so no resistor drop remains:
The generic formula from chapter 02 becomes
Figure 3.1 — One method for every linear RC transient
The one question is: what information is sufficient to reconstruct vC(t) after switching?
Initial condition, post-switch topology, final value and equivalent resistance are sufficient to build the response of a linear one-capacitor circuit.
03.3Current is derived from the voltage—not memorized separately
Differentiate the capacitor voltage:
Then
At t=0+,
This is exactly the current predicted by replacing the capacitor at the switching instant by an ideal voltage source equal to its continuous initial voltage V0. The derivative and circuit viewpoints agree.
03.4Source-free discharge is just Vth=0
If the post-switch network contains only a resistor R across a capacitor initially at V0, then
The negative current sign, under a current reference into the capacitor’s positive-voltage terminal, means the capacitor is delivering energy to the resistor.
The capacitor energy is
Energy decays twice as fast in the exponent because it depends on voltage squared.
03.5Step charging from zero is just V0=0
For an ideal voltage source Vs charging C through R from vC(0)=0,
The familiar “charging formula” is therefore only the special case V0=0 of the universal final-value expression.
03.6Worked switching example with nonzero initial voltage
Pedagogical example: before switching, a 4 µF capacitor has settled at 8 V. At t=0 it is connected to a 2 V Thévenin source through Rth=3 kΩ.
Hence
Current into the positive capacitor terminal is
The capacitor initially delivers current because it starts above the new final voltage. At one time constant, vC=2+6/e≈4.207 V.
Independent slope check:
and i/C=(−2 mA)/(4 µF)=−500 V/s, consistent.
03.7Finding Rth with dependent sources
When the t>0 resistive network contains only independent sources and resistors, Rth can often be found by deactivating independent sources: ideal voltage sources → shorts, ideal current sources → opens.
Dependent sources are different: they remain active because their values depend on circuit variables. A robust method is:
- Remove the capacitor.
- Deactivate independent sources only.
- Keep dependent sources active.
- Apply a test voltage Vt or test current It at the capacitor port.
- Solve the network and compute Rth=Vt/It.
This is the same Thévenin discipline developed earlier in AoN and avoids the common but incorrect rule “turn off all sources.”
03.8Energy in a passive discharge
For source-free discharge through R, resistor power is
Integrate from 0 to ∞:
which equals the initial capacitor energy. The ideal resistor dissipates all stored energy and the capacitor tends to zero energy.
This energy check is independent of the waveform endpoint check and is therefore useful for validating a transient derivation.
03.9Model limits: when one exponential is not enough
The single-exponential result assumes one independent storage state and a linear time-invariant post-switch network. It can fail or require enrichment when:
- another capacitor or inductor contributes an independent state;
- capacitance varies strongly with voltage;
- leakage or ESR require extra resistive paths but may still remain first order if no new storage is added;
- dielectric absorption is modeled with additional RC branches, creating multiple time constants;
- switch resistance changes with voltage/current or the network is otherwise nonlinear/time-varying;
- distributed interconnect behavior becomes important.
A measured curve that is not a single exponential is evidence about model adequacy, not something to hide by forcing an RC fit.
03.10Mastery exercises
- A 10 µF capacitor initially at 6 V discharges through 20 kΩ. Find τ, vC(t), iC(0+) and energy at t=0.
- For exercise 1, find the time at which vC reaches 1 V.
- A capacitor with vC(0+)=−2 V is connected to a 5 V Thévenin source through 4 kΩ; C=2 µF. Write vC(t) and iC(t).
- In exercise 3, find the initial slope and verify it from i/C.
- A capacitor sees two resistors 6 kΩ and 3 kΩ in parallel after independent sources are deactivated. With C=5 µF, find τ.
- Why must a resistor disconnected at t=0 not be included in Rth for t>0?
- Why must dependent sources remain active in an Rth calculation?
- Show by integration that all initial capacitor energy is dissipated in R during source-free discharge.
Solutions
1. τ=0.2 s; vC=6e^(−t/0.2) V; iC(0+)=−6/20k=−0.3 mA; w0=½(10 µF)(36)=180 µJ.
2. 1=6e^(−t/0.2) ⇒ t=0.2 ln6≈0.358 s.
3. τ=(4 kΩ)(2 µF)=8 ms; vC=5−7e^(−t/8ms) V; iC=(7/4k)e^(−t/8ms)=1.75e^(−t/8ms) mA.
4. dv/dt|0+=(5−(−2))/8ms=875 V/s; i/C=1.75 mA/2 µF=875 V/s.
5. Rth=6k∥3k=2 kΩ, so τ=10 ms.
6. The time constant belongs to the actual post-switch topology; disconnected elements cannot carry post-switch current.
7. They are controlled by circuit variables and are part of the active linear network. Use a test source to compute port resistance.
8. ∫0∞(V0²/R)e^(−2t/RC)dt=½CV0².
Chapter 04 repeats the same discipline for RL networks. The state variable changes from capacitor voltage to inductor current and the time constant becomes L/Rth rather than RthC, but the first-order logic remains identical.