Understand RL transient behavior: Question bank — Hardware deep dive

Symbols used below (all built in the parent note): $i (t)$ = current through the series loop, $V$ = source voltage, $R$ = resistance, $L$ = inductance, $v_{L} = L d i / d t$ = inductor voltage, $v_{R} = i R$ = resistor voltage, $τ = L / R$ = time constant, $I_{0}$ = the current an inductor already carries at $t = 0$ .

True or false — justify

The inductor forbids a sudden jump in current, but a sudden jump in voltage across it is perfectly allowed.

True — only

i

must be continuous (else

v_{L} = L d i / d t

blows up);

v_{L}

itself leaps from

0

V

the instant the switch closes.

t = 0^{+}

in an energizing RL circuit, the inductor behaves like an open circuit.

True — current is still zero (it can't have jumped), so no current flows as if the branch were broken, and the full source voltage appears across

L

t \to \infty

(DC steady state) the inductor behaves like an open circuit.

False — the opposite:

d i / d t \to 0

v_{L} \to 0

, and the inductor becomes a plain wire (a short). See Steady-State Analysis.

Doubling

L

doubles the final steady current.

False — final current is

V / R

, which contains no

L

; doubling

L

only doubles

τ = L / R

, so the circuit reaches the same final value more slowly.

Doubling

R

makes the transient settle faster.

True —

τ = L / R

has

R

in the denominator, so larger

R

means smaller

τ

; the resistor dissipates the stored energy more quickly.

The energizing current

i (t) = \frac{V}{R} (1 - e^{- t / τ})

actually reaches

V / R

in finite time.

False —

e^{- t / τ}

never truly hits zero, so

i

only approaches

V / R

asymptotically; we call it "done" after

5 τ

because it's within

0.7%

During de-energizing, the current stays positive (same direction) the whole time.

True —

i (t) = I_{0} e^{- t / τ}

is a decaying positive number; it fades toward zero but never reverses sign in a simple RL discharge.

For RL circuits

τ = R L

, mirroring

τ = R C

for RC Transient Behavior.

False — RL uses a ratio

τ = L / R

, not a product; the unit check henry/ohm

=

seconds confirms it, while

R L

would give ohm·henry which is not seconds.

Spot the error

"When the switch closes,

i

instantly equals

V / R

because Ohm's law says so."

Ohm's law alone ignores the inductor; the KVL loop is

V = i R + L d i / d t

, and the

L d i / d t

term forces

i

to rise gradually from

i (0^{+}) = i (0^{-}) = 0

"The inductor voltage

v_{L}

grows over time as current builds up."

Backwards —

v_{L} = V e^{- t / τ}

decays; it is largest (

= V

) at switch-on when

d i / d t

is steepest, and shrinks as the current curve flattens.

"Bigger resistance stores more energy, so the transient lasts longer."

Energy is stored in the inductor (

\frac{1}{2} L I^{2}

), not the resistor; more

R

actually drains that energy faster, shrinking

τ

and shortening the transient.

"At

t = 0^{+}

, KVL is violated because

v_{R} = 0

but

v_{L} = V

, so they don't split the source."

They do add up:

v_{R} = i (0^{+}) R = 0 \cdot R = 0

and

v_{L} = V

, so

v_{R} + v_{L} = 0 + V = V

— Kirchhoff's Voltage Law holds exactly.

"To integrate the RL ODE you need a second boundary condition because it's second order."

It is first order (only

d i / d t

appears, no

d^{2} i / d t^{2}

); one condition,

i (0) = 0

i (0) = I_{0}

, fully pins the solution. See First-Order Differential Equations.

"Since

τ = L / R

has

R

, a superconducting loop (

R \to 0

) settles instantly."

Opposite — as

R \to 0

τ = L / R \to \infty

, so the current would take forever to change; with truly zero

R

an initial current persists undecayed.

"The 63.2% figure is the fraction remaining after one time constant."

It's the fraction of the journey completed (

1 - e^{- 1} = 0.632

) when rising; the fraction remaining is

e^{- 1} = 36.8%

— don't swap them.

Why questions

Why is

R

in the denominator of

τ = L / R

instead of the numerator?

Because a larger

R

pulls more current-driving voltage away from the inductor and dissipates energy faster, so the current changes more quickly — smaller

τ

Why does the inductor take the full source voltage at

t = 0^{+}

Current is momentarily still

0

, so

v_{R} = i R = 0

; KVL then dumps the entire

V

onto

L

, which is exactly what forces

d i / d t

to be steepest right at the start.

Why is the shape a single exponential rather than, say, a straight line?

The rate of current change

d i / d t = (V - i R) / L

depends on how far

i

still is from

V / R

; that "rate proportional to remaining gap" is the defining signature of exponential approach.

Why do we say the transient is "essentially over" at

5 τ

and not require infinity?

After

5 τ

e^{- 5} \approx 0.0067

, so less than

0.7%

of the change remains — below the precision of any real measurement, so practically it's finished.

Why can an inductor store energy while a resistor cannot?

The inductor holds energy in its magnetic field (

\frac{1}{2} L I^{2}

) and can return it; the resistor only converts energy to heat, which is irreversible. See Inductor Fundamentals.

Why does the current in de-energizing decay to zero and not to some nonzero value?

With the source removed (

V = 0

), KVL becomes

i R + L d i / d t = 0

, whose only steady state (

d i / d t = 0

) is

i = 0

; there is nothing left to sustain any current.

Edge cases

What is

τ

L = 0

(ideal wire, no inductance)?

τ = 0/ R = 0

, so the transient is instantaneous and current jumps straight to

V / R

— the pure-resistive Ohm's law case with no delay.

What happens to

v_{L}

at the exact instant of switch-on if we insist

i

jumps?

d i / d t

would be infinite, demanding

v_{L} = L d i / d t = \infty

— physically impossible, which is precisely why

i

must stay continuous instead.

In de-energizing, what does

v_{L}

do at

t = 0^{+}

compared to

v_{R}

They are equal and opposite:

v_{L} = - i (0^{+}) R = - I_{0} R

, because with

V = 0

the inductor must supply the exact voltage the resistor drops to keep current flowing.

R

is enormous and the source is suddenly disconnected, why can a spark appear across the switch?

A huge

R

(or an open switch,

R \to \infty

) forces

d i / d t

to be very large to maintain continuity, producing a big

v_{L} = L d i / d t

voltage spike that can arc across the gap.

What is the steady-state current when the source

V = 0

but the inductor started with

I_{0}

It decays to

0

i (t) = I_{0} e^{- t / τ}

; the stored magnetic energy is entirely dissipated as heat in

R

, leaving no current.

If two RL circuits share the same

τ

but one has double

L

and double

R

, do they rise identically?

The shape

i / I_{final}

t

is identical (same

τ

), but their final currents differ since

I_{final} = V / R

depends on

R

separately.

Recall One-line self-test before you leave

Cover everything: state (1) why current can't jump, (2) why $v_{L}$ is largest at $t = 0^{+}$ , (3) why bigger $R$ settles faster. If all three come with a because, you've mastered the traps. Answer ::: (1) infinite $v_{L} = L d i / d t$ is impossible so $i$ is continuous; (2) $i = 0$ so full $V$ lands on $L$ ; (3) $R$ in denominator of $τ = L / R$ shrinks the time constant.

Connections

Parent: RL Transient Behavior — the full derivation these traps probe.
Time Constant — the $τ = L / R$ many items hinge on.
Inductor Fundamentals — source of $v_{L} = L d i / d t$ and $\frac{1}{2} L I^{2}$ .
Kirchhoff's Voltage Law — the loop equation used in the "spot the error" set.
First-Order Differential Equations — why it's first order, one boundary condition.
Steady-State Analysis — the "inductor is a short at DC" edge case.
RC Transient Behavior — the $τ = R C$ contrast trap.