Skip to main content

Growth and decay of current in LR circuit

Short answer

For growth, i = (ε/R)(1 - e^(-t/τ)) with τ = L/R. For decay, i = i0 e^(-t/τ). The time constant τ is the time for current to reach 63.2% of final value in growth or fall to 36.8% in decay; after ~5τ it's within 1% of final.

Derive the relation for rise (growth) and fall (decay) of current in an LR circuit. Plot graph between current and time for both cases and explain the figure and the inductive time constant.

Circuit: battery emf ε, resistor R, inductor L in series with a switch. The inductive time constant is τ = L/R.

        +----- R -----+
        |             |
      ε =|            L (coil)
        |             |
        +---- S ------+

Growth of current (switch to the battery)

  1. Kirchhoff's loop rule: ε − iR − L di/dt = 0
  2. Rearrange: di/(ε/R − i) = (R/L) dt
  3. Integrate from i = 0 at t = 0: −ln[(ε/R − i)/(ε/R)] = Rt/L
  4. i = (ε/R)(1 − e^(−t/τ)) = i₀(1 − e^(−t/τ)), where i₀ = ε/R

Decay of current (battery removed, circuit closed through R)

  1. Loop rule with ε = 0: iR + L di/dt = 0
  2. di/i = −(R/L) dt. Integrate from i = i₀ at t = 0:
  3. i = i₀ e^(−t/τ)

Graph

  i      Growth                    i      Decay
 i0 |- - - - - - - _____         i0 |.
    |         . '                   | `.
    |      .'                       |   `.
    |    .'                         |     ` .
    |  .'                           |         ` .  .  .
    |.'_____________________ t      |_____________________ t
    0   tau   2tau   3tau           0   tau   2tau   3tau

The growth curve rises towards i₀ and the decay curve falls towards zero; the two are mirror images.

Explanation

  • Time constant τ = L/R (seconds): the time in which the current grows to 63.2% of i₀ (1 − e⁻¹), or decays to 36.8% (e⁻¹) of i₀.
  • At t = 0 the inductor opposes any change: growth starts with i = 0 and slope ε/L.
  • After about 5τ the current is within 1% of its final value.
  • A large L or small R gives a larger τ, so the current changes more slowly.

✓ Answer checked by Claude Opus 5.5 (high effort)

All Engineering Physics chapter 8 questions →