Engineering · Electrical engineering

Current rise in an RL circuit

After a DC source is switched on, current in a series RL circuit rises exponentially toward its steady value.

AdvancedUniversity/collegeBachelor foundationsSubject specialization

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Formula

i (t) = \frac{U _{0}}{R} (1 - e^{- tR / L}), τ = \frac{L}{R}

i in A; U₀ in V; R in Ω; L in H; t and τ in s

Variables & units

Variables & units
Symbol	Meaning	Unit
$i(t)$	Current at time t after switch-on	A
$U₀$	Constant DC source voltage	V
$R$	Total series resistance	Ω
$L$	Inductance	H
$τ$	Inductive time constant L/R	s

LaTeX

i(t)=\frac{U_0}{R}(1-e^{-tR/L}),\qquad \tau=\frac{L}{R}

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Derivation & background

The step response assumes a constant DC source, linear components, initially zero inductor current and a series resistance R. The inductor opposes rapid current change; at long times current approaches U₀/R. When switched off, current can decay exponentially through a closed discharge path. Include winding resistance in R.

Exam blueprint

Validity range

Derivation steps

After a DC source is switched on, current in a series RL circuit rises exponentially toward its steady value.

1State the model assumptions and list quantities with units.
2Substitute into i(t)=(U₀/R)(1−e⁻ᵗᴿ/ᴸ); τ=L/R; convert quantities to SI first.
3For U₀=12 V, R=6 Ω and L=3 H, τ=0.5 s and i(τ)=2(1−e⁻¹)≈1.26 A.

Rearrangements

Base equation for the target quantity

i(t)=\frac{U_0}{R}(1-e^{-tR/L}),\qquad \tau=\frac{L}{R}

Preserve signs, boundary conditions and units when rearranging.

Task variant

Work through the numerical example for “Current rise in an RL circuit” and check its unit and magnitude.

For U₀=12 V, R=6 Ω and L=3 H, τ=0.5 s and i(τ)=2(1−e⁻¹)≈1.26 A.

Common mistakes

Substituting a related quantity that shares the same symbol.

Use variable definitions and check: i in A; U₀ in V; R in Ω; L in H; t and τ in s.

Exam context

Analyze switching transients in coils, relays and simple actuators.

Formula cluster

Electrical engineering

After a DC source is switched on, current in a series RL circuit rises exponentially toward its steady value.

Σ I hinein = Σ I hinaus; Σ ΔV = 0

Kirchhoff’s junction and loop rules

ℰ = −N dΦB/dt

Faraday’s law of induction

C(s) = KP + KI/s + KDs

Ideal PID controller transfer function

Worked example

For U₀=12 V, R=6 Ω and L=3 H, τ=0.5 s and i(τ)=2(1−e⁻¹)≈1.26 A.

Applications

Analyze switching transients in coils, relays and simple actuators.

How do you calculate with Current rise in an RL circuit?

Here is how to work through a typical Current rise in an RL circuit (i(t)=(U₀/R)(1−e⁻ᵗᴿ/ᴸ); τ=L/R) task step by step:

1
Task
Work through the numerical example for “Current rise in an RL circuit” and check its unit and magnitude.
Solution path
For U₀=12 V, R=6 Ω and L=3 H, τ=0.5 s and i(τ)=2(1−e⁻¹)≈1.26 A.

Frequently asked questions about Current rise in an RL circuit

Why does current not instantly reach U₀/R?+

An inductor generates a back emf proportional to current change. At switch-on, the requested change is initially large, so the inductor slows the rise. As current grows, the rate falls and current approaches U₀/R asymptotically. An ideal inductor’s current cannot jump. This step response assumes an idealized linear inductor and a defined current path. Total R includes winding, source and other series resistance. Ideal inductor current cannot jump instantly; after switch-off, a real circuit needs a safe path for stored magnetic energy.

What happens after one time constant?+

At switch-on, after τ=L/R, current reaches about 63.2% of steady value. Each further time constant removes the same fraction of remaining error. During switch-off through a resistive path, current falls to about 36.8% of its previous value. This applies to a first-order circuit with constant parameters. This step response assumes an idealized linear inductor and a defined current path. Total R includes winding, source and other series resistance. Ideal inductor current cannot jump instantly; after switch-off, a real circuit needs a safe path for stored magnetic energy.

Why does a coil need a discharge path?+

The coil stores U=½Li² and tends to maintain current after supply disconnect. Without a path, voltage can rise sharply until an unwanted arc or component breakdown occurs. A flyback diode or other clamp limits voltage but also changes decay time. This step response assumes an idealized linear inductor and a defined current path. Total R includes winding, source and other series resistance. Ideal inductor current cannot jump instantly; after switch-off, a real circuit needs a safe path for stored magnetic energy.

How do I check the result and units for Current rise in an RL circuit?+

Write the quantity and unit from the variable list next to every value: i in A; U₀ in V; R in Ω; L in H; t and τ in s. Substitute only after converting to a consistent unit system, and state sign or reference direction explicitly. The example For U₀=12 V, R=6 Ω and L=3 H, τ=0.5 s and i(τ)=2(1−e⁻¹)≈1.26 A. is a calculation check: result, unit and order of magnitude must fit the described situation. Round after intermediate calculations so small contributions are not lost prematurely. This step response assumes an idealized linear inductor and a defined current path. Total R includes winding, source and other series resistance. Ideal inductor current cannot jump instantly; after switch-off, a real circuit needs a safe path for stored magnetic energy.

Which model limit matters for Current rise in an RL circuit?+

The background states this starting condition: The step response assumes a constant DC source, linear components, initially zero inductor current and a series resistance. That limits when the result is a valid model prediction. Compare the problem’s boundary conditions with this assumption before substituting values; real components, material properties or loads can introduce additional effects. The cited reference supports the relation used here but does not remove the need to check whether the specific case fits the model. This step response assumes an idealized linear inductor and a defined current path. Total R includes winding, source and other series resistance. Ideal inductor current cannot jump instantly; after switch-off, a real circuit needs a safe path for stored magnetic energy.

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Subject references and further reading

This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.

[1]OpenStax, University Physics Volume 2, §14.4 “RL Circuits”.

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Current rise in an RL circuit

After a DC source is switched on, current in a series RL circuit rises exponentially toward its steady value.

AdvancedUniversity/collegeBachelor foundationsSubject specialization

Free to browse · flashcards or a multiple-choice quiz · import from Essential

Symbol

Meaning

Unit

i(t)

Current at time t after switch-on

U₀

Constant DC source voltage

R

Total series resistance

L

Inductance

τ

Inductive time constant L/R

Derivation & background

Exam blueprint

Validity range

Derivation steps

After a DC source is switched on, current in a series RL circuit rises exponentially toward its steady value.

1State the model assumptions and list quantities with units.
2Substitute into i(t)=(U₀/R)(1−e⁻ᵗᴿ/ᴸ); τ=L/R; convert quantities to SI first.
3For U₀=12 V, R=6 Ω and L=3 H, τ=0.5 s and i(τ)=2(1−e⁻¹)≈1.26 A.

How do you calculate with Current rise in an RL circuit?

Here is how to work through a typical Current rise in an RL circuit (i(t)=(U₀/R)(1−e⁻ᵗᴿ/ᴸ); τ=L/R) task step by step:

Task

Work through the numerical example for “Current rise in an RL circuit” and check its unit and magnitude.

Solution path

For U₀=12 V, R=6 Ω and L=3 H, τ=0.5 s and i(τ)=2(1−e⁻¹)≈1.26 A.

Frequently asked questions about Current rise in an RL circuit

Why does current not instantly reach U₀/R?+

What happens after one time constant?+

Why does a coil need a discharge path?+

How do I check the result and units for Current rise in an RL circuit?+

Which model limit matters for Current rise in an RL circuit?+

Retain Current rise in an RL circuit for exams

Create a curated FSRS exam set for i(t)=(U₀/R)(1−e⁻ᵗᴿ/ᴸ); τ=L/R: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.

Curated formula set · LaTeX · FSRS spaced repetition · import from Essential