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The LR Circuit Transient

An inductor resists changes in the current through it. Close a switch onto a battery and the current does not jump to its final value $V/R$; it climbs there smoothly, $I(t) = (V/R)(1 - e^{-t/\tau})$, with the time constant $\tau = L/R$. The reason is the inductor's back-EMF, $V_L = L\,dI/dt$, which starts equal to the full battery voltage and opposes the rising current, then fades as the current levels off; at every instant $V_R + V_L = V$. While the current builds, so does the inductor's magnetic field and the energy stored in it, $U = \tfrac12 L I^2$. Open the switch and the field cannot vanish instantly either: the current decays as $I_0 e^{-t/\tau}$ and the stored energy drains out as heat in the resistor. Flip the switch and watch the current ramp up and down, the field lines grow and collapse, and the back-EMF reverse to keep the current flowing.

Figure 1. The LR circuit transient. Top: the battery, switch, resistor, and inductor; the current (green) flows with the inductor's magnetic field (blue) and stored energy building, and the back-EMF (orange) opposing the change. Bottom: the current rising exponentially to V/R with time constant tau = L/R, and the back-EMF, with the decay when the switch opens. Method: backward-Euler integration of L dI/dt + R I = V. Source: Halliday, Resnick and Walker, Fundamentals of Physics, Ch. 30.
V4.0 V
L2.0 H
R2.0 ohm
on (charging)

WHAT TO TRY

  • Flip the switch off and on: the current ramps down and up exponentially, and the back-EMF flips sign to keep the current flowing as the field collapses.
  • Raise the inductance or lower the resistance: the time constant $\tau = L/R$ grows and the current takes longer to settle.
  • Watch the field lines through the coil grow as the current rises and the stored energy $U = \tfrac12 L I^2$ climb in the readout.
  • At one time constant the current has reached about 63 percent of its final value $V/R$ (the green dashed line).