Theory
Time constant and physical meaning
A series LR circuit responds exponentially because the inductor resists changes in current.
The characteristic timescale is the time constant
\[
\tau=\frac{L}{R},
\]
where \(L\) is inductance (H) and \(R\) is resistance (Ω).
After one time constant:
- growth: \(I(\tau)\approx 0.632\,I_\infty\)
- decay: \(I(\tau)\approx 0.368\,I_0\)
Current growth (step voltage applied)
For a DC source \(\varepsilon\) applied at \(t=0\), Kirchhoff’s loop rule gives
\[
L\frac{dI}{dt}+RI=\varepsilon.
\]
With \(I(0)=0\), the solution is
\[
I(t)=\frac{\varepsilon}{R}\left(1-e^{-t/\tau}\right),
\qquad
I_\infty=\frac{\varepsilon}{R}.
\]
The resistor and inductor voltages (growth) are
\[
V_R(t)=I(t)R,
\qquad
V_L(t)=\varepsilon - V_R(t)=L\frac{dI}{dt}.
\]
Initially \(V_L(0)=\varepsilon\) and \(V_R(0)=0\); as \(t\to\infty\), \(V_L\to 0\) and \(V_R\to\varepsilon\).
Current decay (source removed)
If the source is removed and the current starts at \(I(0)=I_0\), the loop equation becomes
\[
L\frac{dI}{dt}+RI=0.
\]
The solution is the exponential decay
\[
I(t)=I_0 e^{-t/\tau}.
\]
During decay, the inductor supplies energy back to the resistor. The voltages satisfy
\[
V_L(t)+V_R(t)=0,
\qquad
V_R(t)=I(t)R,
\qquad
V_L(t)=-V_R(t).
\]
Graph interpretation
The calculator plots \(I(t)\) and draws a dashed marker at \(t=\tau\). This marker is useful because \(\tau\) is
the point where the exponential has a standard fraction of its initial/steady value:
- growth: \(I(\tau)=I_\infty(1-e^{-1})\approx 0.632\,I_\infty\)
- decay: \(I(\tau)=I_0 e^{-1}\approx 0.368\,I_0\)
