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Switch questions with inductors look like they need differential equations. Most do not. If you know what an inductor looks like just after the switch moves and long after, the rest is Ohm's law. This post sets out the rules, works two circuits our LR growth and decay post does not cover, one with a capacitor too and one where the switch opens, and lists the traps.
Which exams test this?
The JEE (Advanced) 2026 Information Brochure lists "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 but does not name LR switching, and the NEET (UG) 2026 syllabus likewise lists self and mutual inductance. The NCERT Class 12 chapter Electromagnetic Induction (Reprint 2026-27) calls inductance "electrical inertia" that "opposes growth and decay of current", which is the idea behind every rule below. Check your own exam's current syllabus.
What are the rules?
Inductor current is continuous. The current just after the switch moves (t = 0⁺) equals the current just before (t = 0⁻).
At t = 0⁺, an inductor that had no current acts as an open gap. One that carried current I₀ acts as a source that keeps pushing I₀.
At t → ∞ (steady d.c.), an ideal inductor acts as a plain wire.
The voltage across an inductor can jump. It is L dI/dt. OpenStax University Physics, section 14.4 notes that the induced emf is greatest immediately after the switch is thrown and falls to zero as the current settles.
A capacitor is the mirror image: an uncharged capacitor acts as a wire at t = 0⁺ and as an open gap at t → ∞.
Worked example 1: an inductor and a capacitor together
A 24 V battery, a switch and R₁ = 4 Ω are in series with two parallel branches. Branch A is an inductor L = 0.5 H in series with R₂ = 8 Ω. Branch B is an uncharged capacitor C = 100 μF in series with R₃ = 4 Ω. The switch closes at t = 0.
Just after closing (t = 0⁺):
Inductor is a gap, capacitor is a wire. Current flows through R₁ and R₃ only: 24/(4 + 4) = 3 A, all through branch B.
Voltage across the branches: 24 − 3 × 4 = 12 V. No current in R₂, so all 12 V is across the inductor.
Initial rate of rise of inductor current: dI/dt = 12/0.5 = 24 A/s.
Long after (t → ∞):
Capacitor is a gap, inductor is a wire. Current: 24/(4 + 8) = 2 A, all through branch A.
Voltage across the branches: 2 × 8 = 16 V, so the capacitor ends at 16 V.
Stored energies: inductor ½ × 0.5 × 2² = 1.0 J; capacitor ½ × 100 × 10⁻⁶ × 16² = 0.0128 J.
The battery current falls from 3 A to 2 A, and the current shifts from one branch to the other.
Worked example 2: opening the switch
A 12 V battery and a switch feed a coil with L = 0.40 H and resistance r = 4 Ω. A 100 Ω resistor is connected directly across the coil. After a long time, the switch is opened.
Before opening: the coil is a 4 Ω wire, so it carries 12/4 = 3 A. The 100 Ω resistor carries 12/100 = 0.12 A.
Just after opening:
The coil still carries 3 A (rule 1). The only path is now through the 100 Ω resistor, so 3 A flows through it, in the opposite direction to its earlier 0.12 A.
Voltage across the 100 Ω resistor: 3 × 100 = 300 V, twenty-five times the battery's 12 V.
Time constant: τ = L/(R + r) = 0.40/104 = 3.8 ms.
Energy: the coil held ½ × 0.40 × 3² = 1.8 J. Shared in the ratio of resistances, 1.8 × 100/104 ≈ 1.73 J becomes heat in the 100 Ω resistor.
This is why opening an inductive circuit can give a large voltage: the inductor forces its old current through whatever path remains.
What mistakes do students make?
Letting inductor current jump when a switch opens. Find I at 0⁻ first, always.
Swapping the inductor and capacitor rules. Use the callout above.
Missing the direction change in a resistor that the inductor current is forced through.
Using the closing-circuit resistance for τ after opening. τ uses the resistance in the loop the inductor now sees.
Treating a real coil as ideal. At t → ∞ a coil is a wire of its own resistance, not zero.
A practice routine
Write the four rules and the capacitor mirror from memory.
Redo example 1 with R₃ = 12 Ω. (Check: 1.5 A at t = 0⁺, 2 A at t → ∞.)
Redo example 2 with a 1000 Ω resistor across the coil and find the voltage just after opening.
Solve three past JEE Advanced switch questions using only the rules before writing any exponential.
When the set-up of a switch problem is where you get stuck, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning behind each step.
Before your next mock, do practice step 3 and say out loud why the voltage can be larger than the battery's.
