What an inductor does
An inductor is a coil of wire. Current in it makes a magnetic field. If the current changes, the field changes, and by Faraday's law this induces an emf in the coil itself. By Lenz's law this emf opposes the change.
εL = −L dI/dt
L is the inductance, measured in henry (H): 1 H gives 1 V of back emf when the current changes by 1 A every second.
- Steady current: dI/dt = 0, so no back emf. The inductor acts like a plain wire.
- Fast change: big back emf. The inductor fights it.
Think of it as a heavy flywheel for charge: hard to start, hard to stop.
Current growth when the switch closes
Kirchhoff's loop rule round the circuit:
ε − IR − L dI/dt = 0
At t = 0 the current is 0, so the whole battery voltage is across the inductor: dI/dt = ε/L (fastest rise). The inductor acts like an open switch for that first instant.
Solving gives
I(t) = (ε/R)(1 − e−t/τ), with τ = L/R (seconds).
- t = τ: I = 63% of ε/R
- t = 2τ: 86%; t = 3τ: 95%; t = 5τ: over 99%
- Long time: I = ε/R, just as if the coil were a wire.
The voltage across the inductor falls the opposite way: VL = ε e−t/τ.
Current decay and stored energy
If the battery is removed and the coil is left connected to R, the current does not stop at once:
I(t) = I₀ e−t/τ
After τ, 37% is left; after 5τ, less than 1%.
Where does the energy come from? Building up the current stored energy in the coil's magnetic field:
U = ½ L I²
During decay this energy turns into heat in the resistor. If the circuit is simply broken (no path), the current tries to drop to zero almost instantly, dI/dt is huge, and the back emf can make a spark.
Comparing τ with the RC circuit
Both give exponential curves with a time constant, but:
- RC: τ = RC. Bigger R makes it slower.
- LR: τ = L/R. Bigger R makes it faster (and the final current smaller).
Uses of LR behaviour: chokes and filters that block sudden spikes, ignition coils, relays, and the soft start of motors.
Try it: in free play, set L = 2 H, R = 4 Ω (τ = 0.5 s). Predict where the green τ line will fall, then change R to 8 Ω and predict again.
Key formulas and definitions
- ε_L = −L dI/dt
- Loop rule: ε − IR − L dI/dt = 0
- τ = L / R
- Growth: I = (ε/R)(1 − e^(−t/τ))
- Decay: I = I₀ e^(−t/τ)
- V_L = ε e^(−t/τ) (growth)
- U = ½ L I²
- At t = τ: 63% (growth), 37% (decay)
Worked examples
1. An inductor of 2 H and a 4 Ω resistor are in series with a 12 V battery. Find τ and the final current.
τ = L/R = 2/4 = 0.5 s. Final current = ε/R = 12/4 = 3 A.
2. In the same circuit, find the current 0.5 s after the switch closes.
t = τ, so I = 3 × (1 − e⁻¹) = 3 × 0.632 ≈ 1.90 A.
3. Find the rate of rise of current just after the switch closes.
At t = 0, I = 0, so L dI/dt = ε → dI/dt = 12/2 = 6 A/s.
4. How much energy is stored in the inductor when the current is steady?
U = ½ L I² = ½ × 2 × 3² = 9 J.
5. The battery is removed and the coil discharges through the 4 Ω resistor. Find the current after 1.0 s.
t = 2τ: I = 3 e⁻² = 3 × 0.135 ≈ 0.41 A.
6. How long does it take the growing current to reach half its final value?
½ = 1 − e^(−t/τ) → e^(−t/τ) = ½ → t = τ ln 2 = 0.5 × 0.693 ≈ 0.35 s.
Common mistakes
- Thinking the inductor blocks current forever. It only slows changes; at steady state it acts like a wire.
- Writing τ = LR or τ = R/L. The time constant is L divided by R (check: henry/ohm = second).
- Using 37% for growth. In growth the current reaches 63% at τ; 37% is what is left in decay.
- Forgetting the minus sign in ε_L = −L dI/dt: the back emf always opposes the change.