LR Circuits
A short lesson with diagrams, a worked example, a problem to try, and a quick check. Then quiz yourself on 6 practice questions.
In a circuit with a resistor and inductor, the current can’t jump. The loop rule gives ε − IR − L dI/dt = 0, so the current rises exponentially toward ε/R: I = (ε/R)(1 − e−t/τ), with time constant τ = L/R.
1. Growth and decay
2. The two extremes
Just after closing the switch: I = 0, so the resistor has no voltage and the inductor has all of ε. The current starts rising at dI/dt = ε/L.
The inductor acts like a break in the wire at first.
A long time later: I is steady, so dI/dt = 0 and the inductor has no voltage. I = ε/R, as if the inductor were a plain wire.
Opening the switch suddenly forces dI/dt to be huge, so the inductor can make a very large voltage: that’s the spark at a switch.
3. Example, then try it
A 6.0 V battery is connected at t = 0 to a 2.0 Ω resistor in series with a 0.50 H inductor. Find τ, the initial rate of current increase, the final current, and the final stored energy.
τ = L/R = 0.25 s; dI/dt at t = 0 is ε/L = 12 A/s.
Ifinal = ε/R = 3.0 A; U = ½(0.50)(3.0)² ≈ 2.3 J.
Derive I(t) for a battery ε switched on at t = 0 in series with R and L. Then, for ε = 9.0 V, R = 3.0 Ω and L = 0.60 H, find how long the current takes to reach 90% of its final value.
Separate: L dI/dt = ε − IR → dI/(ε/R − I) = (R/L) dt.
Integrate from 0: −ln[(ε/R − I)/(ε/R)] = Rt/L, so I = (ε/R)(1 − e−Rt/L).
Numbers: τ = 0.60/3.0 = 0.20 s and Ifinal = 3.0 A.
90%: e−t/τ = 0.10 → t = τ ln 10 = (0.20)(2.30) ≈ 0.46 s.
Answer: I = (ε/R)(1 − e−t/τ); it reaches 90% (2.7 A) after about 0.46 s.
4. Common mistakes
In an LR circuit, the inductance is doubled. What happens?
τ = L/R doubles. The final current ε/R doesn’t depend on L.
Ready to practice?
6 questions on LR Circuits, each with a hint and an explanation for every wrong answer.