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RL Circuits

Key idea: Solve RL transient problems using the time constant τ = L/R, current growth/decay equations, and exam-style worked examples.

  • GCE A-Level H3 Physics 2027
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Learning objectives

  • define self-inductance as the ratio of the e.m.f. induced in an electrical circuit / component to the rate of dI change of current causing it and use V = L to solve problems dt
  • show an understanding that mutual inductance is the tendency of an electrical circuit / component to oppose a change in the current in a nearby electrical circuit / component
  • show a qualitative understanding that dielectric materials enhance capacitance, and that dielectric breakdown can occur when the electric field is sufficiently strong (knowledge of the quantitative modification of electric fields in matter through the permittivity is not required)
  • show a qualitative understanding that ferromagnetic materials enhance inductance and that this enhancement is non-linear especially near saturation (knowledge of the quantitative modification of magnetic fields in matter through the permeability is not required)
  • Apply U = ½LI² to the energy stored in an inductor.
  • Derive the expression U = ½LI² for the potential energy stored in an inductor by considering the work done on charges.
  • solve problems using the formulae for the combined inductance of two or more inductors in series and in parallel
  • solve problems involving circuits with resistors, inductors, and sources of constant e.m.f. (includes solving first-order differential equations) [RL series circuits with constant e.m.f. source]
  • solve problems involving circuits with inductors and capacitors only (includes solving second-order differential equations) [LC series circuits without e.m.f. source]
  • solve problems involving circuits with resistors, inductors and capacitors only (candidates are not expected to solve the general second-order differential equations, though they can be asked to verify and use particular solutions). [RLC series circuits without e.m.f. source]

An RL series circuit has a resistor R and ideal inductor L connected to a constant e.m.f. source E. Take current around the loop as positive. When the switch changes, the inductor current cannot jump instantaneously; the result is a first-order transient with time constant: τ = L/R

1. Definitions (Must Know)

  • Inductor voltage: V_L = LdI/dt (sign depends on convention; it opposes the change).
  • Resistor voltage: V_R = IR
  • Time constant (RL): τ = L/R (s).
  • Steady state (long time after switch): dI/dt = 0 ⇒ V_L = 0 (ideal inductor behaves like a wire).
  • Symbols used in this lesson: E (V), R (Ω), L (H), I (A), t (s), τ (s).

2. Key Ideas (What Earns Marks)

  • Current continuity: in an ideal inductor, current cannot change instantaneously.
  • Use KVL with the inductor term to get the differential equation.
  • For a step to constant E (switch on): I(t) = (E/R)(1-e^(-t/τ))
  • For current decay after removing the source (switch off, RL loop): I(t) = I₀ e^(-t/τ)

RL transients (normalised)

Current grows toward its final value on switching on, and decays exponentially on switching off; both are governed by the same time constant τ = L/R.

Scroll across the graph to read all labels.

Current grows toward its final value on switching on, and decays exponentially on switching off; both are governed by the same time constant τ = L/R.Current grows toward its final value on switching on, and decays exponentially on switching off; both are governed by the same time constant τ = L/R.
At t = τ: current has risen to about 0.632 I∞ (or fallen to about 0.368 I0).
Open full-size graph
View figure data
Values for RL transients (normalised)
Time ratio (t/τ)Switch on: I/I∞ = 1 − e^(−t/τ)Switch off: I/I0 = e^(−t/τ)
001
0.50.3930.607
10.6320.368
1.50.7770.223
20.8650.135
30.950.05
40.9820.018

3. Detailed Explanations

A. Switching on (current growth): setting up the ODE

Apply Kirchhoff’s voltage law around the loop: E = IR + LdI/dt Rearrange: dI/dt + (R/L)I = E/L This is a first-order linear differential equation.

B. Solving (standard result)

The solution that satisfies I(0) = 0 is: I(t) = (E/R)(1-e^(-(R/L)t))

Using τ = L/R, I(t) = (E/R)(1-e^(-t/τ)). Key checkpoints:

  • t = 0 ⇒ I = 0
  • t → ∞ ⇒ I → E/R
  • initial slope: .dI/dt|ₜ₌₀ = E/L

C. Switching off (current decay)

If the source is removed but R and L remain in a closed loop, KVL gives: 0 = IR + LdI/dt So: dI/dt = -(R/L)I Solution: I(t) = I₀ e^(-t/τ)

4. Common Mistakes

  • Treating the inductor like a resistor: using V_L = IL (wrong).
  • Assuming current instantly becomes E/R at t = 0 (it starts at 0 and rises).
  • Forgetting steady-state behaviour: at long times, V_L = 0 for an ideal inductor.
  • Mixing up τ = L/R with RC time constants from capacitors.

5. Exam Tips

  • Write KVL first, then identify the form “L dI/dt + RI = E”.

  • If asked for time to reach a fraction of final current: I/I_∞ = 1-e^(-t/τ)

    Rearranging gives t = -τ ln(1-I/I_∞).

  • At t = τ, I reaches 1-e⁻¹ ≈ 0.632 of its final value.

6. Worked Examples

Modelled example 1

Current growth at a given time

Core

Problem

An RL circuit has R = 4.0 Ω, L = 2.0 H and E = 12 V applied at t = 0. Find I at 0.50 s.
Study the worked solution

  1. Find the circuit scales

    Method

    τ = 0.50 s and I_∞ = 3.0 A.

    Reason

    τ = L/R sets the growth rate and E/R is the steady current.

    Working

    τ = 2.0/4.0 = 0.50 s, I_∞ = 12/4.0 = 3.0 A

  2. Evaluate the growth law

    Method

    I = 1.9 A.

    Reason

    At t = τ, the growth factor is 1-e⁻¹.

    Working

    I = 3.0(1-e⁻¹) = 1.9 A

Guided practice 2

Time to reach 90% of final current

About 5 min

Problem

For the same circuit, find the time to reach 0.90I_∞.

Try this before viewing the solution

Hints

Hint 1: isolate the exponential
Set 0.90 = 1-e^(-t/τ).
View solution step by step

  1. Isolate the decay factor

    Method

    e^(-t/τ) = 0.10.

    Reason

    The remaining gap to the final current is ten percent.

    Working

    0.90 = 1-e^(-t/τ)

  2. Take logarithms

    Method

    t = 1.2 s.

    Reason

    t = τ ln 10 with τ = 0.50 s.

    Working

    t = (0.50)(2.303) = 1.2 s

Common misconception 3

Current decay

Find and correct the mistake

Learner claim

A 1.5 H inductor discharges through 3.0 Ω from I₀ = 2.0 A. A learner uses the growth expression I₀(1-e^(-t/τ)). Explain why that expression is wrong here and find I at 1.0 s.

Try this before viewing the solution

Correct transient form

View solution step by step

  1. Find the time constant

    Method

    τ = 0.50 s.

    Reason

    τ = L/R.

    Working

    τ = 1.5/3.0 = 0.50 s

  2. Use decay from the initial value

    Method

    I = 0.27 A.

    Reason

    The source-free current approaches zero.

    Working

    I = 2.0e^(-1.0/0.50) = 0.27 A

Examiner practice 4

Inductor voltage during current growth

4 marks

Examination question

For the R = 4.0 Ω, L = 2.0 H, E = 12 V circuit, find |V_L| at 0.50 s. [4 marks]

Try this before viewing the solution

View solution step by step

  1. State current growth

    1 mark

    Method

    I = (E/R)(1-e^(-t/τ)).

    Reason

    This is the switch-on solution.

    Working

    τ = L/R

  2. Differentiate

    1 mark

    Method

    dI/dt = (E/L)e^(-t/τ).

    Reason

    Differentiating converts Rτ to L.

    Working

    dI/dt = (E/L)e^(-t/τ)

  3. Relate voltage

    1 mark

    Method

    |V_L| = E e^(-t/τ).

    Reason

    |V_L| = L|dI/dt|.

    Working

    |V_L| = L(E/L)e^(-t/τ)

  4. Evaluate

    1 mark

    Method

    |V_L| = 4.4 V.

    Reason

    t = τ = 0.50 s.

    Working

    |V_L| = 12e⁻¹ = 4.4 V

Challenge 5

Energy stored at a given time (link to inductor energy)

Minimal support

Independent transfer

For the circuit in Example A, find the inductor energy at 0.50 s.

Try this before viewing the solution

Hints

Hint 1: retain the unrounded current
Use I = 3.0(1-e⁻¹) A before squaring.
View solution step by step

  1. Use instantaneous current

    Method

    I = 3.0(1-e⁻¹) A.

    Reason

    Stored energy depends on the current at that instant.

    Working

    I = I_∞(1-e^(-t/τ))

  2. Calculate energy

    Method

    U = 3.6 J.

    Reason

    Substitute the unrounded current into U = (1/2)LI².

    Working

    U = (1/2)(2.0)[3.0(1-e⁻¹)]² = 3.6 J

7. Mind Stretchers

Mind stretcher 1: Why the inductor voltage can exceed the source voltageExtension

When switching off, why can the induced voltage across the inductor become very large?

Answer

Because V_L = L dI/dt. If the circuit tries to force the current to drop very quickly (large negative dI/dt), the inductor generates a large induced e.m.f. to oppose that rapid change. This is why sparks can occur at switches.

Mind stretcher 2: Which is continuous: current or voltage?Extension

At the instant a switch is toggled in an ideal RL circuit, which quantity is continuous and which can change abruptly: the current through the inductor or the voltage across it? Explain briefly.

Answer

The inductor current is continuous (it cannot change instantaneously). The inductor voltage can change abruptly because V_L = L dI/dt and the derivative can change at the switching instant depending on the circuit conditions.

8. Optional/Enrichment: Non-ideal Inductors

Real inductors have internal resistance and can saturate (with ferromagnetic cores). These effects modify the transient behaviour, but H3 RL questions typically model inductors as ideal unless stated otherwise.

Next step

Continue to LC Circuits, where a capacitor and inductor exchange energy in a second-order oscillation.

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H3 Physics
Edition
GCE A-Level H3 Physics 2027