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LR circuit questions look like calculus problems, but most of them can be solved with three rules and one exponential. This post explains how current grows and decays in an LR circuit, how to handle the switch problems that JEE likes, works through examples with real numbers, and ends with a practice routine. It builds on self-inductance and the energy ½LI², so revise those first if they are shaky.
Where is this in the syllabus?
The JEE (Advanced) 2026 Information Brochure lists "Self and mutual inductance; RC, LR, LC and LCR (in series) circuits with d.c. and a.c. sources". The NTA Syllabus for JEE (Main) 2026 lists self and mutual inductance and the LCR series circuit, but does not name LR transients. The NCERT Class 12 chapter on electromagnetic induction (Reprint 2026-27) calls inductance "electrical inertia" that "opposes growth and decay of current", and derives the stored energy ½LI². Check your own exam year's syllabus. Our post comparing JEE Main and JEE Advanced explains how the two papers differ.
How does current grow when you close the switch?
Take a battery of emf E, a resistor R and an inductor L in series. Kirchhoff's loop rule gives E − IR − L(dI/dt) = 0. Solving this differential equation, as in MIT's 8.02 course notes on inductance, gives
I(t) = (E/R)(1 − e^(−t/τ)), with τ = L/R.
The inductor's emf is largest at t = 0 and dies away, so after a long time the inductor behaves like a plain wire and the current is E/R. The MIT notes also show the energy balance: the battery's power IE equals the heat I²R plus the rate of energy stored in the inductor.
How does it decay when the battery is removed?
If the current has reached I0 and the battery is taken out while the inductor and resistor stay in a closed loop, the current falls as
I(t) = I0 e^(−t/τ), with the same τ = L/R.
The inductor drives the current for a while using its stored energy. If you simply break the circuit with no closed path, the current tries to drop to zero instantly, L(dI/dt) becomes very large, and you get a spark across the switch.
Worked example 1: growth with numbers
E = 20 V, R = 10 Ω, L = 2 H.
Time constant: τ = L/R = 2 ÷ 10 = 0.2 s.
Final current: I0 = E/R = 20 ÷ 10 = 2 A.
Current at t = τ = 0.2 s: I = 2(1 − e^(−1)) = 2(1 − 0.368) = 2 × 0.632 ≈ 1.26 A.
Time to reach half the final current: 1 − e^(−t/τ) = 0.5, so t = τ ln 2 = 0.2 × 0.693 ≈ 0.139 s.
Initial rate of rise: at t = 0, I = 0, so the whole 20 V is across the inductor and dI/dt = E/L = 20 ÷ 2 = 10 A/s.
Energy stored at steady state: ½LI0² = ½ × 2 × 2² = 4 J.
Worked example 2: decay and the heat produced
Now remove the battery in example 1, leaving L and R in a loop. I = 2e^(−t/0.2).
After 0.4 s (2τ): I = 2e^(−2) = 2 × 0.135 ≈ 0.27 A.
After 5τ = 1 s: e^(−5) ≈ 0.0067, so I ≈ 0.013 A, under 1% of the start.
Total heat in R during the decay: ∫I²R dt = I0²R × τ/2 = 4 × 10 × 0.1 = 4 J. All the stored energy ends up as heat, as it should.
Worked example 3: finding τ when there are two resistors
A 12 V battery is connected in series with R1 = 4 Ω. After R1, the circuit splits into two parallel branches: R2 = 4 Ω in one, and an ideal inductor L = 0.5 H in the other. The switch closes at t = 0.
Just after closing: the inductor current is still 0, so it acts like a break. Battery current = 12 ÷ (4 + 4) = 1.5 A.
Long after: the inductor acts as a wire and shorts out R2. Battery current = 12 ÷ 4 = 3 A, all of it through L.
Time constant: replace the battery by a wire and find the resistance the inductor sees. R1 and R2 are then in parallel: 4 × 4 ÷ (4 + 4) = 2 Ω. So τ = 0.5 ÷ 2 = 0.25 s, not L/R1 or L/R2.
Inductor current: I_L = 3(1 − e^(−t/0.25)) A.
What mistakes do students make with LR circuits?
Using the wrong resistance in τ. Always use the resistance seen by the inductor with the sources switched off.
Letting the inductor current jump. Voltages across an inductor can jump; its current cannot.
Mixing up RC and LR. In RC, τ = RC and the capacitor acts as a wire at first; in LR, τ = L/R and the inductor acts as a break at first.
Forgetting the decay loop. Decay needs a closed path; without one the problem is about sparks, not exponentials.
Reading graphs wrongly. At t = τ, a growing current has reached about 63% of its final value; a decaying one has fallen to about 37%.
A practice routine
Derive the growth equation yourself from the loop rule. Our post on differential equations for JEE covers the method.
Redo example 3 with R2 = 12 Ω. Check: τ = 0.5 ÷ 3 ≈ 0.167 s.
Sketch I against t and the inductor voltage against t for growth and decay on one page.
Solve five past JEE Advanced LR questions, writing the three switch rules beside each.
When the set-up of a switch problem is the part that stops you, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning behind each step.
Today, redo worked example 3 without looking, then change one resistor and predict τ before calculating it.
