physics2u
Depth
Derivation

The RL Circuit: Current Growth, Decay and the Time Constant L/R

Statement

For a series circuit consisting of a battery with constant EMF \(\mathcal{E}\), a resistor of resistance \(R\), and an inductor of inductance \(L\), the current \(I(t)\) satisfies the first-order linear differential equation \(L\frac{dI}{dt} + RI = \mathcal{E}\). Starting from zero current, the solution is \(I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right)\) where \(\tau = L/R\) is the time constant. If the battery is removed and the inductor is shorted through the resistor, the current decays as \(I(t) = I_0 e^{-t/\tau}\). The energy stored in the magnetic field is \(U = \frac{1}{2}LI^2\).

Why it matters

The RL circuit is the inductive twin of the RC circuit, and the electrical analogue of a mass set moving against linear drag: inductance plays the role of inertia, resistance the role of the drag coefficient, and the current relaxes exponentially towards its steady value. It is the fundamental building block for understanding how current behaves in circuits containing inductors, which are ubiquitous in power supplies, radio frequency circuits, and motor drives. The time constant \(\tau = L/R\) sets the timescale for transient behavior, determining how quickly a circuit reaches steady state or discharges.

Understanding RL dynamics is essential for designing circuits that filter signals, suppress voltage spikes, or store energy. The concept of inductive reactance and the opposition to change in current are central to AC circuit analysis, transformer operation, and the design of flyback converters. The energy storage capability of inductors is the basis for many power electronic applications, from switching power supplies to wireless charging.

Assumptions
The inductance \(L\) is constant.If the core material saturates or the geometry changes with current, \(L\) becomes a function of \(I\), making the differential equation nonlinear. The linear solution \(I(t) = \frac{\mathcal{E}}{R}(1-e^{-t/\tau})\) no longer holds, and numerical methods are required.
The resistance \(R\) is constant and ohmic.For non-ohmic resistors (diodes, thermistors) or temperature-dependent resistance, \(R\) varies with \(I\) or time, complicating the differential equation. The simple exponential form is lost, and the time constant is no longer well-defined.
The inductor is ideal — no internal resistance or capacitance.Real inductors have winding resistance \(R_w\) and parasitic capacitance \(C_p\). At high frequencies, the parasitic capacitance creates a resonant circuit, and the simple RL model fails. The winding resistance adds to the total resistance, modifying the time constant to \(\tau = L/(R+R_w)\).
The circuit is lumped — dimensions are small compared to the wavelength of the signal.For high-frequency signals where the circuit dimensions are comparable to the wavelength, transmission line effects become important. The lumped element model breaks down, and distributed parameter models (telegrapher's equations) are required.
The magnetic field is confined to the inductor core.If significant magnetic flux links other parts of the circuit or external conductors, mutual inductance terms appear, coupling the equations. The single-inductor model is insufficient, and a system of coupled differential equations must be solved.
Derivation
1
\[ \mathcal{E} - IR - L\frac{dI}{dt} = 0 \]
Kirchhoff's voltage law (loop rule) for the series circuit: the sum of voltage drops equals the source EMF. The inductor voltage is \(V_L = L\frac{dI}{dt}\) by definition of inductance. A
2
\[ L\frac{dI}{dt} + RI = \mathcal{E} \]
Rearrange the loop equation to isolate the derivative term. This is a first-order linear ordinary differential equation with constant coefficients. A
3
\[ \frac{dI}{dt} + \frac{R}{L}I = \frac{\mathcal{E}}{L} \]
Divide through by \(L\) to put the equation in standard linear form. The coefficient of \(I\) is \(R/L\), which has units of \(\mathrm{s^{-1}}\). A
4
\[ \tau \equiv \frac{L}{R} \]
Define the time constant \(\tau\). It has units of \(\mathrm{H/\Omega = (V\cdot s/A)/(V/A) = s}\). This sets the timescale for the transient response. A
5
\[ \frac{dI}{dt} + \frac{1}{\tau}I = \frac{\mathcal{E}}{L} \]
Substitute \(\tau = L/R\) into the differential equation. The equation is now in a form where the time constant appears explicitly. A
6
\[ I(t) = I_{\text{part}}(t) + I_{\text{hom}}(t) \]
The general solution of a linear ODE is the sum of a particular solution (steady state) and the homogeneous solution (transient). This is the superposition principle for linear differential equations. C
7
\[ I_{\text{part}}(t) = \frac{\mathcal{E}}{R} \]
For the particular solution, assume \(I\) is constant (\(dI/dt = 0\)). Substituting into the ODE gives \(RI_{\text{part}} = \mathcal{E}\), so \(I_{\text{part}} = \mathcal{E}/R\). This is the steady-state current. B
8
\[ \frac{dI_{\text{hom}}}{dt} + \frac{1}{\tau}I_{\text{hom}} = 0 \]
The homogeneous equation is obtained by setting the source term to zero. This describes the natural response of the circuit. A
9
\[ I_{\text{hom}}(t) = A e^{-t/\tau} \]
Solve the homogeneous ODE by separation of variables: \(\frac{dI}{I} = -\frac{dt}{\tau}\), integrate both sides to get \(\ln I = -t/\tau + \ln A\), so \(I = A e^{-t/\tau}\). The constant \(A\) is determined by initial conditions. C
10
\[ I(t) = \frac{\mathcal{E}}{R} + A e^{-t/\tau} \]
Combine the particular and homogeneous solutions. The general solution has one arbitrary constant \(A\), as expected for a first-order ODE. A
11
\[ I(0) = 0 \quad \Longrightarrow \quad 0 = \frac{\mathcal{E}}{R} + A \quad \Longrightarrow \quad A = -\frac{\mathcal{E}}{R} \]
Apply the initial condition: at \(t=0\), the current is zero (inductor opposes sudden change). Solve for \(A\). B
12
\[ I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right) \]
Substitute \(A = -\mathcal{E}/R\) into the general solution. This is the current growth formula for an RL circuit starting from rest. A
13
\[ V_L(t) = L\frac{dI}{dt} = L\frac{d}{dt}\left[\frac{\mathcal{E}}{R}(1 - e^{-t/\tau})\right] = \mathcal{E} e^{-t/\tau} \]
Differentiate \(I(t)\) to find the inductor voltage. At \(t=0\), \(V_L = \mathcal{E}\); as \(t\to\infty\), \(V_L\to 0\). The inductor initially opposes the full source voltage. B
14
\[ V_R(t) = IR = \mathcal{E}(1 - e^{-t/\tau}) \]
The resistor voltage is \(IR\). At \(t=0\), \(V_R = 0\); as \(t\to\infty\), \(V_R\to \mathcal{E}\). The resistor voltage grows as the current increases. A
2
\[ L\frac{dI}{dt} + RI = 0,\quad I(0)=I_0 \;\Longrightarrow\; I(t) = I_0\,e^{-t/\tau} \]
Decay: remove the battery and close the loop through \(R\). The source term vanishes, so only the homogeneous solution remains, fixed by the current \(I_0\) flowing at the switching instant (the inductor current cannot jump). A
3
\[ P_L = V_L I = L I\frac{dI}{dt} = \frac{d}{dt}\left(\tfrac12 L I^2\right) \]
The power delivered to the inductor is voltage times current; for constant \(L\) it is an exact time derivative. B
4
\[ U = \int_0^{t} P_L\,dt\prime = \int_0^{I} L I\prime\,dI\prime = \tfrac12 L I^2 \]
Integrate from zero current: the stored magnetic energy depends only on the present current, not on how it was reached. B
5
\[ \int_0^\infty I_0^2 R\,e^{-2t/\tau}\,dt = I_0^2 R\,\frac{\tau}{2} = \tfrac12 L I_0^2 \]
During decay all the stored energy ends up as heat in the resistor, whatever the value of \(R\): a larger \(R\) dissipates faster but for a shorter time. B
6
\[ e^{-t_{1/2}/\tau} = \tfrac12 \;\Longrightarrow\; t_{1/2} = \tau\ln 2 = \frac{L}{R}\ln 2 \]
Half-life of the decaying current: the time for \(I\) to fall to half its value, the same from any starting point. A
Result
\[ I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right),\qquad I_{\text{decay}}(t) = I_0 e^{-t/\tau},\qquad \tau = \frac{L}{R},\qquad U = \tfrac12 L I^2 \]

Reading. The current starts at zero and asymptotically approaches the steady-state value \(\mathcal{E}/R\). The time constant \(\tau = L/R\) determines how quickly this approach occurs. After one time constant (\(t = \tau\)), the current reaches \(1 - e^{-1} \approx 63.2\%\) of its final value. After three time constants, it reaches \(1 - e^{-3} \approx 95.0\%\). After five time constants, it reaches \(1 - e^{-5} \approx 99.3\%\).

Units check. \([L/R] = \mathrm{H/\Omega = (V\cdot s/A)/(V/A) = s}\). The exponent \(-t/\tau\) is dimensionless, as required. The current \(\mathcal{E}/R\) has units \(\mathrm{V/\Omega = A}\). Consistent.

Limiting cases
  • \(t \to 0^+\): \(I(0^+) = 0\), \(V_L(0^+) = \mathcal{E}\), \(V_R(0^+) = 0\). The inductor behaves as an open circuit, opposing the sudden change in current.
  • \(t \to \infty\): \(I(\infty) = \mathcal{E}/R\), \(V_L(\infty) = 0\), \(V_R(\infty) = \mathcal{E}\). The inductor behaves as a short circuit, and the circuit reaches steady state.
  • \(t = \tau\): \(I(\tau) = \frac{\mathcal{E}}{R}(1 - e^{-1}) \approx 0.632\frac{\mathcal{E}}{R}\). The current has reached 63.2% of its final value.
  • \(t = 3\tau\): \(I(3\tau) = \frac{\mathcal{E}}{R}(1 - e^{-3}) \approx 0.950\frac{\mathcal{E}}{R}\). The current has reached 95.0% of its final value, often considered "settled" in engineering practice.
  • \(t = 5\tau\): \(I(5\tau) = \frac{\mathcal{E}}{R}(1 - e^{-5}) \approx 0.993\frac{\mathcal{E}}{R}\). The current has reached 99.3% of its final value, considered fully settled in most applications.
Breaks when
  • Core saturation. If the magnetic core saturates, the inductance \(L\) decreases with increasing current, making the differential equation nonlinear. The exponential solution no longer holds, and the current rises faster than predicted by the linear model.
  • High-frequency operation. At frequencies where the wavelength is comparable to the circuit dimensions, parasitic capacitance and radiation effects become important. The lumped RL model fails, and transmission line theory is required.
  • Non-ideal switch. If the switch has finite resistance or inductance, or if it opens abruptly, large voltage spikes can occur due to the inductor's opposition to current change. The simple model assumes an ideal switch with zero resistance and no inductance.
  • Temperature-dependent resistance. If the resistor's temperature changes significantly with current, \(R\) becomes a function of \(I\), making the differential equation nonlinear. The time constant is no longer constant, and the exponential form is lost.
Failure modes
  • Assuming current changes instantaneously. The inductor opposes changes in current, so \(I(0^+) = 0\) for a circuit starting from rest. Students often incorrectly assume the current jumps to \(\mathcal{E}/R\) at \(t=0\).
  • Confusing voltage and current waveforms. The inductor voltage \(V_L(t) = \mathcal{E}e^{-t/\tau}\) decays exponentially, while the current \(I(t) = \frac{\mathcal{E}}{R}(1 - e^{-t/\tau})\) grows exponentially. Students often mix up which quantity decays and which grows.
  • Forgetting the initial condition. The constant \(A\) in the general solution is determined by the initial current. If the circuit starts with a non-zero current, the solution is \(I(t) = \frac{\mathcal{E}}{R} + (I_0 - \frac{\mathcal{E}}{R})e^{-t/\tau}\). Students often forget to apply the initial condition correctly.
  • Misinterpreting the time constant. The time constant \(\tau = L/R\) is not the time to reach steady state; it is the time to reach 63.2% of the final value. Students often think \(\tau\) is the time to reach 100% of the final value, which is never reached in finite time.
Discussion

The RL circuit is a first-order system, meaning its response is determined by a single time constant \(\tau = L/R\). This is in contrast to RLC circuits, which are second-order systems with two time constants (or a natural frequency and damping ratio). The first-order nature of the RL circuit makes its analysis straightforward: the solution is a single exponential, and the transient behavior is completely characterized by \(\tau\).

The inductor's opposition to change in current is a direct consequence of Faraday's law of induction. When the current changes, the magnetic flux through the inductor changes, inducing an EMF that opposes the change (Lenz's law). This induced EMF is proportional to the rate of change of current, \(V_L = L\frac{dI}{dt}\), and is the key term that makes the RL circuit dynamic rather than algebraic.

The energy stored in the inductor's magnetic field is \(U = \frac{1}{2}LI^2\). This can be derived by integrating the power delivered to the inductor: \(P = V_L I = L\frac{dI}{dt}I\), so \(U = \int_0^I L I' dI' = \frac{1}{2}LI^2\). During the current growth phase, the battery supplies energy to both the resistor (dissipated as heat) and the inductor (stored in the magnetic field). During the decay phase, the inductor releases its stored energy, which is dissipated in the resistor.

The decay result explains the spark at a switch. If a current \(I_0\) is interrupted by opening the circuit, the effective resistance jumps from \(R\) to a very large value, so \(\tau = L/R\) collapses and \(|V_L| = L\,|dI/dt|\) becomes enormous: the stored energy \(\tfrac12 L I_0^2\) must go somewhere, and it goes into an arc across the opening contacts. This is why relay coils and motor windings are fitted with a flyback diode, which gives the current a low-resistance path to decay through, turning the destructive spike into a gentle exponential decay with time constant \(L/R_{\text{loop}}\).

Common misconceptions. The inductor does not "store current"; it stores energy in its magnetic field. The current is a flow of charge, not a substance that can be stored. The inductor opposes changes in current, not current itself. In steady state, the inductor behaves as a short circuit, and the current is determined solely by the resistor and source. Another misconception is that the inductor voltage is always equal to the source voltage; in fact, \(V_L\) and \(V_R\) sum to the source voltage, and their relative magnitudes change with time.

Worked examples

Example 1. A 12 V battery is connected to a series circuit with a 100 Ω resistor and a 0.5 H inductor. Find the time constant, the current at \(t = 0.01\) s, and the inductor voltage at \(t = 0.01\) s.

1
\[ \tau = \frac{L}{R} = \frac{0.5\ \mathrm{H}}{100\ \mathrm{\Omega}} = 0.005\ \mathrm{s} = 5\ \mathrm{ms} \]
Calculate the time constant. A
2
\[ I(t) = \frac{\mathcal{E}}{R}(1 - e^{-t/\tau}) = \frac{12}{100}(1 - e^{-0.01/0.005}) = 0.12(1 - e^{-2}) \]
Substitute the given values into the current formula. A
3
\[ I(0.01) = 0.12(1 - 0.1353) = 0.12(0.8647) = 0.1038\ \mathrm{A} \]
Evaluate the exponential and compute the current. A
4
\[ V_L(t) = \mathcal{E}e^{-t/\tau} = 12 e^{-0.01/0.005} = 12 e^{-2} = 12(0.1353) = 1.62\ \mathrm{V} \]
Calculate the inductor voltage at \(t = 0.01\) s. A
\[ \tau = 5\ \mathrm{ms}, \quad I(0.01\ \mathrm{s}) = 0.104\ \mathrm{A}, \quad V_L(0.01\ \mathrm{s}) = 1.62\ \mathrm{V} \]

Reading. After 2 time constants (\(t = 0.01\) s = \(2\tau\)), the current has reached 86.5% of its final value (0.12 A), and the inductor voltage has decayed to 13.5% of its initial value (12 V).

Units check. \([L/R] = \mathrm{H/\Omega = s}\). The exponent \(-t/\tau\) is dimensionless. Current in amperes, voltage in volts. Consistent.

Example 2. A 0.2 H inductor with a steady current of 3 A is disconnected from the battery and connected to a 50 Ω resistor. Find the time constant, the current at \(t = 0.005\) s, and the total energy dissipated in the resistor.

1
\[ \tau = \frac{L}{R} = \frac{0.2\ \mathrm{H}}{50\ \mathrm{\Omega}} = 0.004\ \mathrm{s} = 4\ \mathrm{ms} \]
Calculate the time constant for the decay circuit. A
2
\[ I(t) = I_0 e^{-t/\tau} = 3 e^{-0.005/0.004} = 3 e^{-1.25} \]
Use the decay formula with initial current \(I_0 = 3\) A. A
3
\[ I(0.005) = 3(0.2865) = 0.860\ \mathrm{A} \]
Evaluate the exponential and compute the current. A
4
\[ U = \frac{1}{2}LI_0^2 = \frac{1}{2}(0.2)(3)^2 = 0.1 \times 9 = 0.9\ \mathrm{J} \]
Calculate the initial energy stored in the inductor. A
5
\[ P_R(t) = I^2R = (I_0 e^{-t/\tau})^2 R = I_0^2 R e^{-2t/\tau} \]
The power dissipated in the resistor at time \(t\). B
6
\[ E_{\text{diss}} = \int_0^\infty P_R(t) dt = \int_0^\infty I_0^2 R e^{-2t/\tau} dt = I_0^2 R \left[\frac{-\tau}{2}e^{-2t/\tau}\right]_0^\infty = I_0^2 R \frac{\tau}{2} \]
Integrate the power over all time to find the total energy dissipated. B
7
\[ E_{\text{diss}} = (3)^2(50)\frac{0.004}{2} = 9 \times 50 \times 0.002 = 0.9\ \mathrm{J} \]
Substitute the values. The total energy dissipated equals the initial energy stored, as expected. A
\[ \tau = 4\ \mathrm{ms}, \quad I(0.005\ \mathrm{s}) = 0.860\ \mathrm{A}, \quad E_{\text{diss}} = 0.9\ \mathrm{J} \]

Reading. After 1.25 time constants, the current has decayed to 28.7% of its initial value. The total energy dissipated in the resistor (0.9 J) equals the initial energy stored in the inductor, confirming energy conservation.

Units check. \([L/R] = \mathrm{s}\). \([LI^2] = \mathrm{H\cdot A^2 = (V\cdot s/A)A^2 = V\cdot A\cdot s = J}\). Power in watts, energy in joules. Consistent.

Problems
  1. A 24 V battery is connected to a series circuit with a 200 Ω resistor and a 1 H inductor. Find the time constant and the current at \(t = 0.01\) s.
    Solution \(\tau = L/R = 1/200 = 0.005\) s. \(I(t) = \frac{24}{200}(1 - e^{-0.01/0.005}) = 0.12(1 - e^{-2}) = 0.12(1 - 0.1353) = 0.12(0.8647) = 0.104\) A.
  2. An inductor with \(L = 0.5\) H and a steady current of 4 A is disconnected and connected to a 100 Ω resistor. Find the time constant and the current at \(t = 0.005\) s.
    Solution \(\tau = L/R = 0.5/100 = 0.005\) s. \(I(t) = 4 e^{-0.005/0.005} = 4 e^{-1} = 4(0.3679) = 1.47\) A.
  3. A 12 V battery is connected to a series circuit with a 50 Ω resistor and a 0.2 H inductor. Find the inductor voltage at \(t = 0.004\) s.
    Solution \(\tau = L/R = 0.2/50 = 0.004\) s. \(V_L(t) = \mathcal{E}e^{-t/\tau} = 12 e^{-0.004/0.004} = 12 e^{-1} = 12(0.3679) = 4.41\) V.
  4. An inductor with \(L = 1\) H and a steady current of 2 A is disconnected and connected to a 25 Ω resistor. Find the total energy dissipated in the resistor.
    Solution \(U = \frac{1}{2}LI_0^2 = \frac{1}{2}(1)(2)^2 = 2\) J. The total energy dissipated equals the initial energy stored, so \(E_{\text{diss}} = 2\) J.
  5. A 36 V battery is connected to a series circuit with a 100 Ω resistor and a 0.5 H inductor. Find the time at which the current reaches 90% of its final value.
    Solution \(I(t) = \frac{\mathcal{E}}{R}(1 - e^{-t/\tau})\). Set \(I(t) = 0.9 \frac{\mathcal{E}}{R}\): \(0.9 = 1 - e^{-t/\tau}\), so \(e^{-t/\tau} = 0.1\), \(-t/\tau = \ln(0.1) = -2.3026\), \(t = 2.3026\tau\). \(\tau = L/R = 0.5/100 = 0.005\) s. \(t = 2.3026(0.005) = 0.0115\) s.