RL circuit — growth and decay of current: Question bank

Before we start, a one-line reminder of every symbol so nothing is assumed:

$ε$ = battery EMF (the "push", in volts).
$R$ = resistance (ohms, $Ω$ ); $L$ = inductance (henry, H).
$I$ = current (amps); $\frac{d I}{d t}$ = how fast current is changing.
$τ = L / R$ = the time constant (seconds) — the natural clock of the circuit.
$V_{L} = L d I / d t$ = voltage across the inductor.

True or false — justify

An inductor at the very instant the switch closes behaves like a plain wire

False. At

t = 0

it maximally opposes change, so no current flows through it — it acts like an open circuit. It becomes a plain wire only as

t \to \infty

During growth the inductor voltage

V_{L}

is largest at

t = 0

True.

V_{L} = ε e^{- t / τ}

starts at the full

ε

(inductor absorbs the whole battery push) and decays to

0

as the resistor takes over.

t = τ

the current in a growing RL circuit is exactly half its final value

False. It reaches

1 - e^{- 1} \approx 63.2%

, not

50%

. Halfway happens later, at

t = τ ln 2 \approx 0.69 τ

Doubling

R

(keeping

L

ε

fixed) makes the circuit settle faster

True for the timescale —

τ = L / R

halves, so it settles quicker. But the final current

ε / R

also halves, so it climbs to a lower ceiling faster.

The decay current reaches exactly zero after

5 τ

False.

e^{- t / τ}

never truly hits zero; at

5 τ

it is

\approx 0.7%

of the start — "practically over" but mathematically nonzero forever.

Growth and decay both use the 63.2% figure at

t = τ

False. Growth rises to 63.2% of max; decay falls to 36.8% of its start. They are complementary:

63.2% + 36.8% = 100%

If you double

L

, the final steady current changes

False. The final current is

ε / R

— no

L

in it. Doubling

L

only makes the approach slower (bigger

τ

); the destination is unchanged.

The energy

\frac{1}{2} L I_{0}^{2}

stored in the inductor is dissipated instantly when the battery is removed

False. It drains gradually through

R

over the decay, heating the resistor; the timescale of that dumping is again

τ = L / R

Why questions

Why does the inductor voltage

V_{L}

fall during growth while the resistor voltage rises?

They must sum to

ε

(KVL). Early on the inductor hogs the push (

V_{L} \approx ε

, current tiny); as current climbs,

I R

grows and

V_{L} = ε e^{- t / τ}

shrinks to keep the total at

ε

Why is the RL story called "an exponential approach to equilibrium"?

The rate of change is proportional to how far you are from the target (

d I / d t \propto ε - I R

). Whenever change-rate is proportional to remaining distance, the solution is an exponential — like RC charging.

Why does a large

L

make the circuit "sluggish"?

Large

L

means a large back-EMF for a given

d I / d t

, so the same current change is fought harder. It is electromagnetic inertia — the current version of a heavy mass resisting acceleration.

Why can current keep flowing after the battery is removed?

The inductor stored energy

\frac{1}{2} L I_{0}^{2}

in its magnetic field. By Faraday+Lenz, as that field collapses it drives an EMF that maintains the current, which then decays through

R

Why do we check units to remember

τ = L / R

and not

R / L

τ

must be seconds.

[L] / [R] = H /Ω = (V\cdots/A) / (V/A) = s

✓, while

R / L

gives

1/ s

— a frequency, not a time. Units settle the ambiguity instantly.

Why is the inductor the "opposite" of the capacitor at

t = 0

A capacitor starts as a short (allows a current surge, blocks voltage jump), an inductor starts as an open (blocks current surge, allows a voltage jump). They swap roles because inductors resist current change while capacitors resist voltage change.

Edge cases

What happens to

τ

and the current if

L \to 0

(ideal wire, no inductor)?

τ = L / R \to 0

: the smoothing vanishes and current jumps instantly to

ε / R

. The circuit reduces to a pure resistor obeying plain Ohm's law.

What happens if

R \to 0

during decay (a superconducting-like short)?

τ = L / R \to \infty

: the decay never dies — with no resistor to dissipate energy, current

I_{0}

circulates essentially forever. This is the idealized "persistent current" limit.

ε = 0

from the start (no battery ever connected), what is the current?

Zero for all time. The growth formula gives

I = (0/ R) (1 - e^{- t / τ}) = 0

; with no push and no stored energy there is nothing to drive current.

At exactly

t = 0

in growth, what are

I

\frac{d I}{d t}

, and

V_{L}

I = 0

(no jump),

\frac{d I}{d t} = ε / L

(steepest here), and

V_{L} = ε

(inductor takes the whole EMF since

I R = 0

t \to \infty

in growth, what happens to

\frac{d I}{d t}

and

V_{L}

Both

\to 0

: current levels off at

ε / R

, so it stops changing, and the inductor voltage

L d I / d t

collapses to zero — the inductor now behaves like a plain wire.

If you connect two identical inductors in series, how does

τ

change?

Series inductances add (

L \to 2 L

), so

τ = 2 L / R

doubles — the circuit gets twice as sluggish, though the final current

ε / R

is unchanged.

Recall One-line self-test

Growth rises to 63% at $τ$ ; decay falls to 37% at $τ$ ; inductor = Open at start, wire at end; $τ = L / R$ (units force it). ::: If you can say all four from memory with reasons, you've cleared the trap set.

Connections

Inductance and self-induction — why the $L d I / d t$ opposition exists.
Lenz's law — the sign of the back-EMF in every "why" above.
Energy stored in a magnetic field — the $\frac{1}{2} L I^{2}$ that powers decay.
Kirchhoff's voltage law — the loop rule behind the sign-error traps.
RC circuit — charging and discharging — the dual whose confusion causes the $τ$ mix-ups.

Question bank — RL circuit — growth and decay of current