physics2u
Depth
Derivation

The RC Circuit: Charging, Discharging and the Time Constant

Statement

For a series circuit consisting of a battery of constant emf \(\mathcal{E}\), a resistor of resistance \(R\), and a capacitor of capacitance \(C\), the charge on the capacitor as a function of time, starting from a discharged state at \(t=0\), is \(Q(t) = C\mathcal{E}\left(1 - e^{-t/\tau}\right)\) where \(\tau = RC\); the current is \(I(t) = \frac{\mathcal{E}}{R}e^{-t/\tau}\) and the capacitor voltage is \(V_C(t) = \mathcal{E}\left(1 - e^{-t/\tau}\right)\). For a discharging capacitor with initial charge \(Q_0\), the charge decays as \(Q(t) = Q_0 e^{-t/\tau}\).

Why it matters

The RC circuit is the simplest example of a first-order linear dynamical system. It introduces the concept of a time constant \(\tau\), which characterizes the exponential approach to equilibrium. This model underpins the design of timing circuits, filters, and signal processing components in electronics, and provides a template for understanding relaxation processes in physics and engineering.

It also illustrates the interplay between energy storage (capacitor) and energy dissipation (resistor), providing a clear framework for analyzing energy flow in electrical systems. The exponential behavior is universal in systems where the rate of change is proportional to the current state, making it a foundational concept in differential equations and system dynamics.

Assumptions
The battery emf \(\mathcal{E}\) is constant.If the battery voltage varies with time or load, the differential equation becomes non-homogeneous with a time-dependent forcing term, and the simple exponential solution no longer applies.
The capacitor is ideal — no leakage current.A real capacitor has finite insulation resistance, allowing charge to leak even when disconnected. This introduces an additional parallel conductance term, modifying the time constant and the final equilibrium state.
The resistor is ohmic — \(V = IR\) holds for all currents.If the resistance depends on voltage or current (e.g., a diode or a thermistor), the equation becomes nonlinear, and separation of variables may not yield a closed-form exponential solution.
Inductance is negligible.Real circuits have parasitic inductance from wires and components. If inductance \(L\) is non-negligible, the system becomes second-order (RLC circuit), and the response is oscillatory or damped oscillatory rather than purely exponential.
The circuit is lumped — propagation delays are negligible.At high frequencies or in long transmission lines, the finite speed of electromagnetic waves means the circuit cannot be treated as a single lumped element. Distributed parameter models (telegrapher's equations) are required.
Derivation
1
\[ \mathcal{E} - IR - V_C = 0 \]
Kirchhoff's voltage law (loop rule) for the series circuit: the sum of voltage drops equals the emf. A
2
\[ V_C = \frac{Q}{C} \]
Definition of capacitance: the voltage across a capacitor is proportional to the charge stored on it. A
3
\[ I = \frac{dQ}{dt} \]
Current is the rate of change of charge on the capacitor. A
4
\[ \mathcal{E} - R\frac{dQ}{dt} - \frac{Q}{C} = 0 \]
Substitute Steps 2 and 3 into Step 1 to obtain a first-order linear ODE in \(Q(t)\). B
5
\[ \frac{dQ}{dt} + \frac{1}{RC}Q = \frac{\mathcal{E}}{R} \]
Rearrange into standard linear form; the coefficient of \(Q\) is \(1/\tau\) where \(\tau = RC\). C
6
\[ \frac{dQ}{\mathcal{E}C - Q} = \frac{dt}{RC} \]
Separate variables: move all \(Q\) terms to the left and all \(t\) terms to the right. B
7
\[ \int_{0}^{Q(t)} \frac{dQ'}{\mathcal{E}C - Q'} = \int_{0}^{t} \frac{dt'}{RC} \]
Integrate both sides from the initial condition \(Q(0)=0\) to the general time \(t\). A
8
\[ -\ln(\mathcal{E}C - Q) + \ln(\mathcal{E}C) = \frac{t}{RC} \]
Evaluate the integral: \(\int \frac{dQ}{A-Q} = -\ln(A-Q)\). C
9
\[ \ln\left(\frac{\mathcal{E}C}{\mathcal{E}C - Q}\right) = \frac{t}{RC} \]
Combine logarithms using \(\ln a - \ln b = \ln(a/b)\). A
10
\[ \frac{\mathcal{E}C}{\mathcal{E}C - Q} = e^{t/RC} \]
Exponentiate both sides to eliminate the logarithm. A
11
\[ Q(t) = \mathcal{E}C\left(1 - e^{-t/RC}\right) \]
Solve for \(Q(t)\); this is the charging equation. B
12
\[ I(t) = \frac{dQ}{dt} = \frac{\mathcal{E}}{R}e^{-t/RC} \]
Differentiate \(Q(t)\) to find the current; it decays exponentially from \(\mathcal{E}/R\). A
13
\[ V_C(t) = \frac{Q(t)}{C} = \mathcal{E}\left(1 - e^{-t/RC}\right) \]
The capacitor voltage rises from 0 to \(\mathcal{E}\) with the same time constant. A
14
\[ Q_{\text{dis}}(t) = Q_0 e^{-t/RC} \]
For discharge, set \(\mathcal{E}=0\) in the ODE; the solution is a pure exponential decay. B
15
\[ W_R = \int_0^\infty I^2R\,dt = \frac{\mathcal{E}^2}{R}\int_0^\infty e^{-2t/RC}\,dt = \frac{\mathcal{E}^2}{R}\cdot\frac{RC}{2} = \tfrac12 C\mathcal{E}^2 \]
Energy bookkeeping for a full charge. The battery moves charge \(Q = C\mathcal{E}\) through its emf, supplying \(Q\mathcal{E} = C\mathcal{E}^2\); the capacitor ends with \(Q^2/2C = \tfrac12 C\mathcal{E}^2\); the integral of the Joule heating with Step 12 shows the resistor takes the other half, whatever the value of \(R\). B
Result
\[ Q(t) = C\mathcal{E}\left(1 - e^{-t/\tau}\right), \quad I(t) = \frac{\mathcal{E}}{R}e^{-t/\tau}, \quad V_C(t) = \mathcal{E}\left(1 - e^{-t/\tau}\right), \quad \tau = RC \]

Reading. The charge, current, and capacitor voltage all evolve exponentially with time constant \(\tau = RC\). The current starts at \(\mathcal{E}/R\) and decays to zero, while the capacitor voltage rises from zero to \(\mathcal{E}\). The time constant \(\tau\) is the time required for the charge to reach \(63.2\%\) of its final value.

Units check. \([RC] = \Omega \cdot \mathrm{F} = (\mathrm{V/A}) \cdot (\mathrm{C/V}) = (\mathrm{V \cdot C})/(\mathrm{A \cdot V}) = \mathrm{C/A} = \mathrm{s}\). Correct.

Limiting cases
  • \(t \ll \tau\): \(e^{-t/\tau} \approx 1 - t/\tau\), so \(Q(t) \approx C\mathcal{E}(t/\tau) = (\mathcal{E}/R)t\). The capacitor behaves like a short circuit; current is approximately constant \(\mathcal{E}/R\).
  • \(t \gg \tau\): \(e^{-t/\tau} \approx 0\), so \(Q(t) \approx C\mathcal{E}\), \(I(t) \approx 0\), \(V_C(t) \approx \mathcal{E}\). The capacitor is fully charged and acts like an open circuit.
  • \(R \to 0\): \(\tau \to 0\), so the capacitor charges instantaneously. In practice, inductance and resistance limit the current spike.
  • \(C \to 0\): \(\tau \to 0\), so the capacitor stores no charge and the circuit behaves like a pure resistor.
Breaks when
  • Inductance non-negligible. If \(L\) is significant, the circuit is RLC, and the response is oscillatory or damped oscillatory, not purely exponential. The time constant concept no longer applies in the same way.
  • Leaky capacitor. A real capacitor has finite insulation resistance \(R_{\text{leak}}\). The discharge equation becomes \(Q(t) = Q_0 e^{-t/(R_{\text{leak}}C)}\), and the charging equation is modified by a parallel conductance term.
  • Non-ideal battery. If the battery has internal resistance \(r\), the effective emf is reduced, and the time constant becomes \(\tau = (R+r)C\). The final voltage is also reduced if the battery sags under load.
  • Nonlinear components. If \(R\) or \(C\) depend on voltage or current, the ODE becomes nonlinear, and the exponential solution may not hold. Numerical methods may be required.
Failure modes
  • Confusing \(\tau\) with half-life. \(\tau = RC\) is the time to reach \(63.2\%\) of the final value, not \(50\%\). The half-life is \(t_{1/2} = \tau \ln 2 \approx 0.693\tau\).
  • Forgetting the initial condition. The general solution of the ODE includes an arbitrary constant. For charging, \(Q(0)=0\); for discharging, \(Q(0)=Q_0\). Using the wrong initial condition gives the wrong solution.
  • Assuming the current is constant. The current decays exponentially; it is only approximately constant for \(t \ll \tau\). Integrating \(I^2R\) over all time gives the total energy dissipated, not a constant power.
  • Misinterpreting the time constant. \(\tau\) is not the time to fully charge the capacitor; the capacitor is never fully charged in finite time. After \(5\tau\), the charge is \(99.3\%\) of the final value, which is often considered "fully charged" in practice.
Discussion

The RC circuit is a canonical example of a first-order linear system. The exponential behavior arises because the rate of change of charge is proportional to the difference between the current charge and the final equilibrium charge. This is a common pattern in physics: whenever a system relaxes toward equilibrium with a constant rate coefficient, the response is exponential.

Energy bookkeeping during charging reveals a surprising result: the battery supplies energy \(C\mathcal{E}^2\), the capacitor stores \(C\mathcal{E}^2/2\), and the resistor dissipates \(C\mathcal{E}^2/2\), independent of \(R\). This is because the current is higher for smaller \(R\), but the time is shorter, and the product \(I^2R \cdot t\) is invariant. This result is a direct consequence of the exponential decay of current.

The time constant \(\tau = RC\) sets the scale for all transient behavior. In practice, engineers use \(\tau\) to design timing circuits, filters, and signal processing components. The exponential response is also the basis for the concept of bandwidth in electronic circuits, where the cutoff frequency is \(f_c = 1/(2\pi RC)\).

At a deeper level, the lost half is the price of irreversibility. Charging the capacitor in one jump from \(0\) to \(\mathcal{E}\) always dissipates \(\tfrac12C\mathcal{E}^2\); charging it in \(N\) equal voltage steps of \(\mathcal{E}/N\), letting it settle after each, dissipates only \(N \times \tfrac12 C(\mathcal{E}/N)^2 = C\mathcal{E}^2/2N\). As \(N\to\infty\) the process becomes quasi-static and the loss vanishes, exactly as a gas compressed slowly enough produces no entropy. Adiabatic (energy-recovering) logic circuits exploit precisely this.

Common misconceptions. The capacitor does not "store" current; it stores charge. The current is the rate of flow of charge. The time constant is not a fixed delay; it is a scale parameter that characterizes the exponential decay. The energy dissipated in the resistor is not lost; it is converted to heat, which is a form of energy, but not recoverable as electrical energy.

Related pages. The Damped Harmonic Oscillator · AC Circuits: Phasors, Impedance and Series RLC Resonance · Ohm's Law, Resistance and Kirchhoff's Rules for DC Circuits · Electrostatic Energy and Field Energy Density.

Worked examples

Example 1. A 10 \(\mu\mathrm{F}\) capacitor is charged through a 10 k\(\Omega\) resistor by a 12 V battery. Find the time constant, the current at \(t = 0\), and the time required for the capacitor to reach 90% of its final charge.

1
\[ \tau = RC = (10 \times 10^3\ \Omega)(10 \times 10^{-6}\ \mathrm{F}) = 0.1\ \mathrm{s} \]
Calculate the time constant. A
2
\[ I(0) = \frac{\mathcal{E}}{R} = \frac{12\ \mathrm{V}}{10 \times 10^3\ \Omega} = 1.2\ \mathrm{mA} \]
The initial current is the emf divided by the resistance. A
3
\[ Q(t) = C\mathcal{E}(1 - e^{-t/\tau}) = 0.9 C\mathcal{E} \]
Set the charge to 90% of the final value. B
4
\[ 1 - e^{-t/\tau} = 0.9 \Rightarrow e^{-t/\tau} = 0.1 \Rightarrow t = -\tau \ln(0.1) = \tau \ln(10) \]
Solve for \(t\) using logarithms. A
5
\[ t = 0.1\ \mathrm{s} \times 2.3026 \approx 0.23\ \mathrm{s} \]
Calculate the numerical value. A
\[ \tau = 0.1\ \mathrm{s}, \quad I(0) = 1.2\ \mathrm{mA}, \quad t_{90\%} \approx 0.23\ \mathrm{s} \]

Reading. The time constant is 0.1 s, the initial current is 1.2 mA, and it takes about 0.23 s to reach 90% of the final charge.

Units check. \([RC] = \Omega \cdot \mathrm{F} = \mathrm{s}\). \([\mathcal{E}/R] = \mathrm{V}/\Omega = \mathrm{A}\). Correct.

Example 2. A 100 \(\mu\mathrm{F}\) capacitor is initially charged to 50 V and then discharged through a 2 k\(\Omega\) resistor. Find the time constant, the initial current, and the time required for the voltage to drop to 10 V.

1
\[ \tau = RC = (2 \times 10^3\ \Omega)(100 \times 10^{-6}\ \mathrm{F}) = 0.2\ \mathrm{s} \]
Calculate the time constant. A
2
\[ I(0) = \frac{V_0}{R} = \frac{50\ \mathrm{V}}{2 \times 10^3\ \Omega} = 25\ \mathrm{mA} \]
The initial current is the initial voltage divided by the resistance. A
3
\[ V(t) = V_0 e^{-t/\tau} = 10\ \mathrm{V} \]
Set the voltage to 10 V. B
4
\[ e^{-t/\tau} = \frac{10}{50} = 0.2 \Rightarrow t = -\tau \ln(0.2) = \tau \ln(5) \]
Solve for \(t\) using logarithms. A
5
\[ t = 0.2\ \mathrm{s} \times 1.6094 \approx 0.32\ \mathrm{s} \]
Calculate the numerical value. A
\[ \tau = 0.2\ \mathrm{s}, \quad I(0) = 25\ \mathrm{mA}, \quad t_{10\mathrm{V}} \approx 0.32\ \mathrm{s} \]

Reading. The time constant is 0.2 s, the initial current is 25 mA, and it takes about 0.32 s for the voltage to drop to 10 V.

Units check. \([RC] = \Omega \cdot \mathrm{F} = \mathrm{s}\). \([V/R] = \mathrm{V}/\Omega = \mathrm{A}\). Correct.

Problems
  1. A 5 \(\mu\mathrm{F}\) capacitor is charged through a 1 k\(\Omega\) resistor by a 9 V battery. Find the time constant and the initial current.
    Solution \(\tau = RC = (1 \times 10^3\ \Omega)(5 \times 10^{-6}\ \mathrm{F}) = 5 \times 10^{-3}\ \mathrm{s} = 5\ \mathrm{ms}\). \(I(0) = \mathcal{E}/R = 9\ \mathrm{V}/1000\ \Omega = 9\ \mathrm{mA}\).
  2. A 10 \(\mu\mathrm{F}\) capacitor is initially charged to 20 V and discharged through a 5 k\(\Omega\) resistor. Find the time required for the voltage to drop to 5 V.
    Solution \(\tau = RC = (5 \times 10^3\ \Omega)(10 \times 10^{-6}\ \mathrm{F}) = 0.05\ \mathrm{s}\). \(V(t) = V_0 e^{-t/\tau} \Rightarrow 5 = 20 e^{-t/0.05} \Rightarrow e^{-t/0.05} = 0.25 \Rightarrow t = -0.05 \ln(0.25) = 0.05 \ln(4) \approx 0.05 \times 1.386 = 0.0693\ \mathrm{s}\).
  3. A 20 \(\mu\mathrm{F}\) capacitor is charged through a 2 k\(\Omega\) resistor by a 15 V battery. Find the time required for the current to drop to 10% of its initial value.
    Solution \(\tau = RC = (2 \times 10^3\ \Omega)(20 \times 10^{-6}\ \mathrm{F}) = 0.04\ \mathrm{s}\). \(I(t) = I_0 e^{-t/\tau} \Rightarrow 0.1 I_0 = I_0 e^{-t/0.04} \Rightarrow e^{-t/0.04} = 0.1 \Rightarrow t = -0.04 \ln(0.1) = 0.04 \ln(10) \approx 0.04 \times 2.3026 = 0.0921\ \mathrm{s}\).
  4. A 50 \(\mu\mathrm{F}\) capacitor is initially charged to 30 V and discharged through a 10 k\(\Omega\) resistor. Find the energy dissipated in the resistor during the entire discharge process.
    Solution The initial energy stored in the capacitor is \(U_0 = \frac{1}{2}CV_0^2 = \frac{1}{2}(50 \times 10^{-6}\ \mathrm{F})(30\ \mathrm{V})^2 = \frac{1}{2}(50 \times 10^{-6})(900) = 0.0225\ \mathrm{J}\). All of this energy is dissipated in the resistor, so \(E_{\text{diss}} = 0.0225\ \mathrm{J}\).
  5. A 100 \(\mu\mathrm{F}\) capacitor is charged through a 1 k\(\Omega\) resistor by a 12 V battery. Find the total energy supplied by the battery, the energy stored in the capacitor, and the energy dissipated in the resistor.
    Solution The final charge is \(Q_f = C\mathcal{E} = (100 \times 10^{-6}\ \mathrm{F})(12\ \mathrm{V}) = 1.2 \times 10^{-3}\ \mathrm{C}\). The energy supplied by the battery is \(E_{\text{bat}} = Q_f \mathcal{E} = (1.2 \times 10^{-3}\ \mathrm{C})(12\ \mathrm{V}) = 0.0144\ \mathrm{J}\). The energy stored in the capacitor is \(U_C = \frac{1}{2}C\mathcal{E}^2 = \frac{1}{2}(100 \times 10^{-6}\ \mathrm{F})(12\ \mathrm{V})^2 = \frac{1}{2}(100 \times 10^{-6})(144) = 0.0072\ \mathrm{J}\). The energy dissipated in the resistor is \(E_{\text{diss}} = E_{\text{bat}} - U_C = 0.0144 - 0.0072 = 0.0072\ \mathrm{J}\).