Inductance in DC and AC Circuits
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Direct answer
An inductor opposes change, not current: through the self-induced emf e = −L dI/dt it delays both growth and decay of current in an LR circuit, with every stage set by the time constant τ = L/R. Closing the switch, current climbs as I = I0(1 − e^−t/τ) toward I0 = V/R; opening it, current falls as I0 e^−t/τ — after one time constant the circuit is 63% of the way up (or down to 37%). The coil banks the delay as magnetic energy U = ½LI², which is why interrupting an inductive circuit throws a spark: the stored energy must go somewhere, fast. In AC service the same instinct appears as inductive reactance XL = 2πfL, rising with frequency, with the current lagging the voltage by a quarter cycle.
What you must remember
- Growth and decay: I = I0(1 − e^−t/τ) on switching in, I = I0 e^−t/τ on switching out, with I0 = V/R — the final value is set by R alone.
- Time constant: τ = L/R seconds; larger L slows the circuit, larger R hastens it; at t = τ growth reaches 63.2%, decay leaves 36.8%.
- Energy storage: U = ½LI² in the magnetic field; energy density of a magnetic field u = B²/2μ0 (NCERT's result); solenoid inductance L = μ0n²Al.
- AC behaviour: XL = 2πfL; I = V/XL; current lags voltage by 90° in a pure inductor; at f = 0, XL = 0 — the inductor passes DC and blocks fast AC.
- The switching-off spike: with I collapsing quickly, e = −L dI/dt becomes large — roughly a kilovolt when 5 A dies in 10 ms through 2 H — the ignition-coil principle.
- Numerical anchors: L = 2 H with R = 4 Ω gives τ = 0.5 s; at V = 20 V, I0 = 5 A and stored energy = 25 J.
- Sign reading: the minus sign in e = −L dI/dt is Lenz's law — the induced emf opposes the change, not the current itself.
Watching current grow and collapse
Close a 20 V battery on 4 Ω and 2 H in series. The destination is I0 = V/R = 5 A; the pace is τ = L/R = 0.5 s. At t = 0.5 s, I = 5 × 0.632 ≈ 3.16 A; at t = 1 s, about 4.3 A; only near 2.5 s (five time constants) is the current effectively 5 A, at which point the coil stores U = ½ × 2 × 25 = 25 J. Now break the circuit: suppose the current dies in 10 ms — then e = −L dI/dt ≈ 2 × (5/0.01) = 1000 V appears across the gap, a kilovolt from a 20 V battery, which is exactly how ignition systems make sparks. Finally swap the battery for AC: at 50 Hz, XL = 2π × 50 × 2 = 628 Ω; at 5 kHz it is a hundredfold higher — the same coil becomes a roadblock for high frequencies.
Where NEET sets the trap
The division of labour is the concept most often asserted: the final current depends on R only (the inductor is a plain wire in steady DC), while L decides only how fast that value is approached — an assertion-reason favourite both directions. The 63/37 numbers get swapped in options; remember that 63% belongs to growth, 37% to what remains after decay. Energy items punish linearity: U ∝ I², so doubling current quadruples the stash. In AC questions, students quote XL = ωL correctly and then forget the phase (current lags by π/2) when asked for power: the pure inductor consumes no average power. The ignition-spike logic is asked qualitatively — why a spark, why the ammeter reading must not be broken abruptly — and the answer is always ½LI² demanding a path.
Frequently asked questions
What is the time constant of an LR circuit?
τ = L/R — the time in which growing current reaches 63.2% of its final value, or decaying current falls to 36.8%.
What is the steady current in a 20 V circuit with 4 Ω and 2 H in series?
5 A: for steady DC the inductor offers no opposition (its reactance at f = 0 is zero), so I0 = V/R; the 2 H only sets the arrival time.
How much energy is stored in a 2 H inductor carrying 5 A?
U = ½LI² = 25 J, held in the magnetic field and released as a spark or arc if the circuit is broken.
How does an inductor treat DC versus high-frequency AC?
As a plain conductor for steady DC (XL = 0 at f = 0) and increasingly as a block as frequency rises, since XL = 2πfL.
Why does a spark jump when an inductive circuit is opened?
The collapsing field drives a large induced emf (e = −L dI/dt) across the break to dissipate the stored ½LI² — the working principle of ignition coils.