AcademyInductance
Academy
RL Transients
Level 1 - Physics topic page in Inductance.
Principle
An RL circuit changes current exponentially because an inductor resists current changes.
Notation
\(R\)
series resistance
\(\mathrm{\Omega}\)
\(L\)
inductance
\(\mathrm{H}\)
\(\tau\)
RL time constant
\(\mathrm{s}\)
\(I(t)\)
current as a function of time
\(\mathrm{A}\)
\(\mathcal E\)
source emf
\(\mathrm{V}\)
\(V_L\)
voltage across the inductor
\(\mathrm{V}\)
Method
Derivation 1: Current growth
After a battery is connected to a series \(RL\) circuit, Kirchhoff's loop law balances source, resistor drop, and inductor voltage.
Loop equation
\[\mathcal E-IR-L\frac{dI}{dt}=0\]
Time constant
\[\tau=\frac{L}{R}\]
Final current
\[I_f=\frac{\mathcal E}{R}\]
Growth solution
\[I(t)=I_f\left(1-e^{-t/\tau}\right)\]
Derivation 2: Current decay
When the source is removed and the inductor discharges through the resistor, the stored magnetic energy is dissipated in \(R\).
Decay equation
\[L\frac{dI}{dt}+IR=0\]
Decay solution
\[I(t)=I_0e^{-t/\tau}\]
Resistor power
\[P_R=I^2R\]
Rules
RL time constant
\[\displaystyle \tau=\frac{L}{R}\]
Current growth
\[\displaystyle I(t)=\frac{\mathcal E}{R}\left(1-e^{-tR/L}\right)\]
Current decay
\[I(t)=I_0e^{-tR/L}\]
Initial slope
\[\displaystyle \left.\frac{dI}{dt}\right|_{0}=\frac{\mathcal E}{L}\]
Examples
Question
A
\[0.50\,\mathrm{H}\]
inductor is in series with \[10\,\Omega\]
Find \(\tau\).Answer
\[\tau=\frac{L}{R}=\frac{0.50}{10}=5.0\times10^{-2}\,\mathrm{s}\]
Checks
- Inductor current cannot jump discontinuously in a finite-voltage circuit.
- At the first instant of switch-on, an ideal inductor behaves like an open branch.
- After a long time in DC, an ideal inductor behaves like a wire.