LR Circuits & Magnetic Energy

Inductors give circuits inertia and store energy.

10 minUpdated 22 October 2026LR circuittime constantmagnetic energy density

Lesson at a glanceShow the key points

Growth: i = (ℰ/R)(1 − e^(−t/τ)) (38.8); decay: i = i₀e^(−t/τ) (38.11); τ = L/R.
63 % after one τ (growth), 37 % left (decay).
U = ½Li² (38.12); u = B²/2μ₀ (38.13).

An inductor gives a circuit inertia. Current can’t jump on or off; it rises and falls exponentially, with a time constant L/R, and the inductor stores energy in its magnetic field.

Growth of current

An inductor L, a resistor R and a cell ℰ are joined in series and the switch is closed at t = 0. As the current rises, the inductor’s emf −L di/dt opposes it. The loop law gives

\mathcal E - L\frac{di}{dt} = Ri ;\Rightarrow; \int_0^i\frac{di}{\mathcal E - Ri} = \int_0^t\frac{dt}{L}

Growth(38.8–38.9)

i = i_0\left(1 - e^{-t/\tau}\right), \qquad i_0 = \frac{\mathcal E}{R}, \quad \tau = \frac LR

L/R has the dimensions of time: the time constant. After one time constant the current is i₀(1 − 1/e) = 63 % of its final value. In principle the final value is reached only as t → ∞; in practice a few time constants is enough. A small τ means a steep rise.

Interactive

Current in an LR circuit

Close the switch and the inductor's back-emf stops the current jumping up: it rises as i₀(1 − e^(−t/τ)) with τ = L/R. Remove the battery (keeping the loop closed) and the inductor keeps the current flowing for a while, decaying as i₀e^(−t/τ). The stored energy ½Li² is what keeps it going.

\tau = L/R

2.00ms

i

0.0mA

U = \tfrac12 Li^2

0.000mJ

t/\tau

0.00

Process

Inductance

L

50mH

Resistance

R

25Ω

Worked example

Switching on an LR circuit

A 50 mH inductor, a 25 Ω resistor and a 5.0 V cell are in series. Find the time constant, the final current, and the time to reach 90 % of it.

Decay of current

With a steady current i₀ flowing, the cell is replaced by a plain wire (a special switch closes the new path before opening the old one, so the circuit is never broken). Now the inductor’s emf is the only one: −L di/dt = Ri, so

Decay(38.10–38.11)

i = i_0\,e^{-t/\tau}

after one time constant 37 % remains

Worked example

Shorting the circuit

In the circuit above, after a long time the cell is shorted out and removed. Find the current 5.0 ms later.

Energy stored in an inductor

Multiply the growth equation by i dt and integrate from 0 to t:

\int_0^t\mathcal E i\,dt = \int_0^t i^2R\,dt + \tfrac12 Li^2

The work done by the cell equals the Joule heat in R plus a remaining term: the energy stored in the inductor’s magnetic field.

Energy in an inductor(38.12)

U = \tfrac12 Li^2

Worked example

Energy in a coil

How much energy is stored in an 80 mH inductor carrying 1.5 A?

Energy density of a magnetic field

For a long solenoid (radius r, length l, n turns per length, no magnetic material), L = μ₀n²πr²l and B = μ₀ni, so U = ½Li² = (B²/2μ₀)πr²l. Dividing by the volume πr²l:

Magnetic energy density(38.13)

u = \frac{B^2}{2\mu_0}

compare the electric field’s ½ε₀E²

Worked example

A strong field

What is the energy density of a 0.50 T field in air?

What to remember

1Growth: i = (ℰ/R)(1 − e^(−t/τ)) (38.8); decay: i = i₀e^(−t/τ) (38.11); τ = L/R.
2Energy stored: U = ½Li² (38.12).
3Magnetic energy density u = B²/2μ₀ (38.13).

Check yourself

Right after the switch closes in an LR circuit, the current is…

Doubling L in an LR circuit (same R) changes the time constant by…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadRL circuitsOpenStax University Physics Vol. 2 §14.4 ReadEnergy in a magnetic fieldOpenStax University Physics Vol. 2 §14.3 ReadRL circuitWikipedia

PreviousSelf-Induction Next Mutual Induction & the Induction Coil

LR Circuits & Magnetic Energy

Inductors give circuits inertia and store energy.

10 minUpdated 22 October 2026LR circuittime constantmagnetic energy density

Lesson at a glanceShow the key points

Growth: i = (ℰ/R)(1 − e^(−t/τ)) (38.8); decay: i = i₀e^(−t/τ) (38.11); τ = L/R.
63 % after one τ (growth), 37 % left (decay).
U = ½Li² (38.12); u = B²/2μ₀ (38.13).

An inductor gives a circuit inertia. Current can’t jump on or off; it rises and falls exponentially, with a time constant L/R, and the inductor stores energy in its magnetic field.

Growth of current

An inductor L, a resistor R and a cell ℰ are joined in series and the switch is closed at t = 0. As the current rises, the inductor’s emf −L di/dt opposes it. The loop law gives

\mathcal E - L\frac{di}{dt} = Ri ;\Rightarrow; \int_0^i\frac{di}{\mathcal E - Ri} = \int_0^t\frac{dt}{L}

Growth(38.8–38.9)

i = i_0\left(1 - e^{-t/\tau}\right), \qquad i_0 = \frac{\mathcal E}{R}, \quad \tau = \frac LR

Interactive

Current in an LR circuit

\tau = L/R

2.00ms

i

0.0mA

U = \tfrac12 Li^2

0.000mJ

t/\tau

0.00

Process

Inductance

L

50mH

Resistance

R

25Ω

Worked example

Switching on an LR circuit

A 50 mH inductor, a 25 Ω resistor and a 5.0 V cell are in series. Find the time constant, the final current, and the time to reach 90 % of it.

Decay of current

Decay(38.10–38.11)

i = i_0\,e^{-t/\tau}

after one time constant 37 % remains

Worked example

Shorting the circuit

In the circuit above, after a long time the cell is shorted out and removed. Find the current 5.0 ms later.

Energy stored in an inductor

Multiply the growth equation by i dt and integrate from 0 to t:

\int_0^t\mathcal E i\,dt = \int_0^t i^2R\,dt + \tfrac12 Li^2

The work done by the cell equals the Joule heat in R plus a remaining term: the energy stored in the inductor’s magnetic field.

Energy in an inductor(38.12)

U = \tfrac12 Li^2

Worked example

Energy in a coil

How much energy is stored in an 80 mH inductor carrying 1.5 A?

Energy density of a magnetic field

For a long solenoid (radius r, length l, n turns per length, no magnetic material), L = μ₀n²πr²l and B = μ₀ni, so U = ½Li² = (B²/2μ₀)πr²l. Dividing by the volume πr²l:

Magnetic energy density(38.13)

u = \frac{B^2}{2\mu_0}

compare the electric field’s ½ε₀E²

Worked example

A strong field

What is the energy density of a 0.50 T field in air?

What to remember

1Growth: i = (ℰ/R)(1 − e^(−t/τ)) (38.8); decay: i = i₀e^(−t/τ) (38.11); τ = L/R.
2Energy stored: U = ½Li² (38.12).
3Magnetic energy density u = B²/2μ₀ (38.13).

Check yourself

Right after the switch closes in an LR circuit, the current is…

Doubling L in an LR circuit (same R) changes the time constant by…

Keep exploring

Hand-picked videos, simulations and readings that explain this idea from a different angle.

ReadRL circuitsOpenStax University Physics Vol. 2 §14.4 ReadEnergy in a magnetic fieldOpenStax University Physics Vol. 2 §14.3 ReadRL circuitWikipedia

PreviousSelf-Induction Next Mutual Induction & the Induction Coil