Difficulty: Intermediate | Prerequisites: Ohm's Law, Kirchhoff's Rules, basic calculus (separable differential equations).
RC circuits combine resistors and capacitors to create time-dependent behaviour. Unlike purely resistive circuits where everything reaches steady state instantly, RC circuits have voltages and currents that change exponentially over time. This is your first encounter with transient circuit analysis in the course, and the exponential charging/discharging patterns appear again in RL circuits, RLC circuits, and throughout electronics. The time constant τ = RC is the single most important quantity here.
When you close a switch to charge a capacitor through a resistor, the charge builds up exponentially toward its maximum value, and the current decays exponentially from its initial value. When discharging, both charge and current decay exponentially to zero. The time constant τ = RC sets the pace: larger τ means slower charging or discharging.
RC circuit
A circuit containing both a resistor (R) and a capacitor (C) in which current and voltage change with time rather than remaining constant.
Time constant (τ)
τ = RC. The time it takes for the charge (during discharge) to fall to about 37% of its initial value, or for the charge (during charging) to reach about 63% of its final value. Think of it as the circuit's characteristic "pace" for change.
Exponential decay
A function of the form f(t) = f₀ e^(−t/τ). The quantity drops by a factor of e (roughly 2.718) every time constant. This governs discharge curves.
Exponential growth to a limit
A function of the form f(t) = f_max(1 − e^(−t/τ)). The quantity rises quickly at first, then asymptotically approaches its maximum. This governs charging curves.
Separable differential equation
A first-order ODE that can be rearranged so all terms in one variable are on one side and all terms in the other variable are on the other side, then integrated. The RC circuit equations are solved this way.
You need to recognise and solve two forms:
Form 1 (exponential decay): dq/dt + (1/τ)q(t) = 0, with solution q(t) = q₀ e^(−t/τ)
Form 2 (sinusoidal oscillation): d²q/dt² + ω²q(t) = 0, with solution q(t) = q₀ sin(ωt + φ). This one is for LC circuits later, but the source material flags it for awareness.
Immediately after the switch closes: The charge on a capacitor is the same as it was immediately before. Capacitors cannot change their charge instantaneously.
After a long time (t → ∞): The current through the capacitor branch is zero. The capacitor is either fully charged (charging circuit) or fully discharged (discharging circuit).
At intermediate times: Use the exponential solutions.
A battery (V_battery), resistor (R), and capacitor (C) in series, with the switch just closed:
Apply KVR around the loop: −V_battery + q/C + IR = 0
At t = 0: The capacitor is uncharged (q₀ = 0), so q/C = 0.
Initial current: I₀ = V_battery / R (the capacitor acts like a short circuit at the very first instant).
At t → ∞: Current drops to zero (I_C = 0). The capacitor is fully charged.
Final charge: q_∞ = C · V_battery (the capacitor voltage equals the battery voltage).
At intermediate times: Since I = dq/dt, the KVR equation becomes a separable differential equation:
−V_battery + q/C + (dq/dt)R = 0
Solving with initial condition q(0) = 0:
Charge: q(t) = q_∞(1 − e^(−t/RC)) = CV_battery(1 − e^(−t/RC))
Current: I(t) = I₀ e^(−t/RC) = (V_battery/R) e^(−t/RC)
Behaviour: The charge rises from 0 toward CV_battery on an exponential curve. The current starts at V_battery/R and decays exponentially to zero.
Physically: At t = 0 the capacitor is uncharged, so it behaves like a wire and maximum current flows. As charge accumulates, the capacitor develops a voltage that opposes the battery, reducing the current until equilibrium is reached.
The capacitor starts fully charged (q₀ = CV_battery) and the switch moves to disconnect the battery, leaving the capacitor to discharge through R:
Apply KVR: q/C + IR = 0
At t = 0 (short term): q = q₀ = CV_battery.
Initial current: I₀ = −V_battery/R (negative because discharge current flows opposite to the charging direction).
At t → ∞ (long term): q_∞ = 0 and I = 0. Everything has discharged.
At intermediate times: q/C + (dq/dt)R = 0, which gives:
Charge: q(t) = q₀ e^(−t/RC)
Current: I(t) = I₀ e^(−t/RC)
Behaviour: Both charge and current decay exponentially to zero with time constant τ = RC.
τ has units of seconds (Ω × F = s).
A bigger τ means the circuit charges and discharges more slowly.
Increasing R: current flows more slowly, so charging takes longer.
Increasing C: more charge capacity to fill, so charging takes longer.
After 1τ: ~63% charged (or ~37% of initial charge remaining during discharge).
After 3τ: ~95% charged.
After 5τ: ~99% charged (often treated as "fully" charged in practice).
Charging:
q(t) curve: starts at 0, rises with decreasing slope, asymptotes to q_∞.
I(t) curve: starts at I₀, decays exponentially toward 0.
V_C(t) curve: mirrors q(t), rising from 0 toward V_battery.
Discharging:
q(t) curve: starts at q₀, decays exponentially toward 0.
I(t) curve: starts at I₀ (negative direction), decays exponentially toward 0.
V_C(t) curve: mirrors q(t), falling from V_battery toward 0.
Quantity | Charging | Discharging |
|---|---|---|
Charge q(t) | q_∞(1 − e^(−t/RC)) | q₀ e^(−t/RC) |
Current I(t) | I₀ e^(−t/RC) | I₀ e^(−t/RC) |
Initial current I₀ | V_battery / R | V_battery / R (opposite direction) |
Final charge q_∞ | C · V_battery | 0 |
Time constant | τ = RC | τ = RC |
RC circuits are the basis for timing circuits in electronics. The flash unit in a camera charges a capacitor through a resistor; the time constant determines how long you wait between flashes. RC filters in audio equipment separate high-frequency signals from low-frequency ones. Touchscreens on phones detect your finger by measuring changes in RC time constants.
Students often think the capacitor charges linearly. It does not. The rate of charging slows down as the capacitor voltage approaches the battery voltage, producing an exponential curve, not a straight line.
The current during charging is not constant. It is highest at t = 0 (when the capacitor is empty) and decreases as the capacitor fills.
"After a long time" does not mean "after exactly 1 time constant." It means t >> τ, practically 5τ or more, when the exponential terms are negligibly small.
When the switch position changes, the charge on the capacitor at that instant does not jump. The charge is continuous across the switching event. The current can change abruptly, but the charge cannot.
⚠️ Deriving the charging or discharging equations from KVR is a very common exam question. Practise the full derivation, not just the final formulas.
⚠️ You must know which expression to use (charging vs. discharging) based on the initial conditions.
⚠️ Expect graphical questions: given a q(t) or I(t) curve, identify whether the circuit is charging or discharging and estimate τ from the graph.
⚠️ Short-term and long-term behaviour shortcuts are heavily tested. "What is the current immediately after the switch closes?" and "What is the charge after a long time?" are bread-and-butter questions.
1. True or false: Immediately after closing the switch in a charging RC circuit, the capacitor behaves like a short circuit (wire).
A: True. It has no charge, so no voltage across it, and maximum current flows.
2. Fill in the blank: The time constant of an RC circuit with R = 5 kΩ and C = 2 μF is ______.
A: τ = RC = 5000 × 2×10⁻⁶ = 0.01 s = 10 ms.
3. True or false: During discharging, the current through the resistor increases over time.
A: False. It decreases exponentially.
4. Fill in the blank: After a very long time charging, the voltage across the capacitor equals ______.
A: The battery voltage (V_battery).
5. True or false: If you double both R and C, the time constant quadruples.
A: False. τ = RC, so doubling both gives τ_new = (2R)(2C) = 4RC. That is indeed four times larger. (The statement says "quadruples," which is correct. True.)
Q: A 10 V battery charges a 100 μF capacitor through a 50 kΩ resistor. What is the initial charging current, and what is the charge on the capacitor after one time constant?
A: I₀ = V/R = 10/50000 = 0.2 mA. τ = RC = 50000 × 100×10⁻⁶ = 5 s. After 1τ, q = q_∞(1 − e⁻¹) = CV(1 − 0.368) = (100×10⁻⁶)(10)(0.632) = 6.32 × 10⁻⁴ C = 0.632 mC.
Q: A fully charged capacitor (q₀ = 20 μC) discharges through a 1 kΩ resistor. The time constant is 2 ms. What is the charge after 4 ms?
A: 4 ms = 2τ. q(2τ) = 20 × e⁻² = 20 × 0.135 = 2.71 μC.
Q: Explain physically why the charging current decreases over time.
A: As charge builds on the capacitor, it develops a voltage that opposes the battery. The net voltage driving current through the resistor (V_battery − V_C) decreases, so by Ohm's law the current decreases.
Q: In a charging RC circuit, at what time is the current exactly half of its initial value?
A: I(t) = I₀ e^(−t/RC). Set I = I₀/2: e^(−t/RC) = 1/2, so t = RC ln 2 ≈ 0.693τ.
RC circuits rely on Kirchhoff's rules from the Simple Circuits topic. The exponential solutions here are direct analogues of what you will see in RL circuits (where τ = L/R instead of RC). The oscillatory differential equation (form 2) foreshadows LC and RLC circuit behaviour. Capacitor behaviour also connects back to the energy stored in electric fields from earlier in the course.
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