Inductance and RL Circuits

Self-Inductance

Part 1 of 7 — Definition and Calculation

What is Inductance?

When current flows through a coil, it creates a magnetic flux through itself. If the current changes, the flux changes, inducing an EMF that opposes the change (Lenz's law).

Self-inductance $L$ relates the flux linkage to the current:

$\Lambda = N\Phi_B = LI$

The induced EMF is:

$\mathcal{E} = -\frac{d\Lambda}{dt} = -L\frac{dI}{dt}$

Units

$[L] = \frac{\text{V} \cdot \text{s}}{\text{A}} = \text{Henry (H)}$

Quantity	Symbol	Unit
Inductance	$L$	H
Flux linkage	$\Lambda = N\Phi_B$	Wb (= V·s)
EMF	$\mathcal{E}$	V

Inductance of a Solenoid

A solenoid of length $\ell$, $N$ turns, cross-section $A$:

Magnetic field: $B = \mu_0 n I = \mu_0 (N/\ell) I$

Flux through one turn: $\Phi_B = BA = \mu_0 (N/\ell) I A$

Total flux linkage: $\Lambda = N\Phi_B = \mu_0 N^2 A I / \ell$

$\boxed{L_{\text{solenoid}} = \frac{\mu_0 N^2 A}{\ell} = \mu_0 n^2 A \ell}$

Key Dependencies

$L \propto N^2, \quad L \propto A, \quad L \propto 1/\ell$

Doubling the number of turns quadruples the inductance.

If a ferromagnetic core with permeability $\mu = \kappa_m \mu_0$ is inserted:

$L = \frac{\kappa_m \mu_0 N^2 A}{\ell}$

Inductance of a Toroid

A toroid with $N$ turns, inner radius $a$, outer radius $b$, height $h$:

$B = \frac{\mu_0 NI}{2\pi r} \quad (a < r < b)$

Flux through one turn:

$\Phi_B = \int_a^b B \cdot h\,dr = \frac{\mu_0 NIh}{2\pi}\int_a^b \frac{dr}{r} = \frac{\mu_0 NIh}{2\pi}\ln\frac{b}{a}$

$\boxed{L_{\text{toroid}} = \frac{\mu_0 N^2 h}{2\pi}\ln\frac{b}{a}}$

Inductance of a Coaxial Cable

Inner radius $a$, outer radius $b$, length $\ell$:

$L = \frac{\mu_0 \ell}{2\pi}\ln\frac{b}{a}$

(This is the same form as the toroid with $N = 1$.)

Inductance Per Unit Length

$\frac{L}{\ell} = \frac{\mu_0}{2\pi}\ln\frac{b}{a}$

Summary — Part 1

Geometry	Inductance
Solenoid	$L = \mu_0 N^2 A/\ell$
Toroid	$L = \mu_0 N^2 h \ln(b/a)/(2\pi)$
Coaxial cable	$L/\ell = \mu_0 \ln(b/a)/(2\pi)$
Induced EMF	$\mathcal{E} = -L\,dI/dt$

Next up: The RL circuit differential equation — Part 2.

RL Circuit Differential Equation

Part 2 of 7 — Setting Up and Solving

The RL Circuit

An inductor $L$ in series with a resistor $R$ and an EMF source $\mathcal{E}$.

Applying Kirchhoff's voltage law:

$\mathcal{E} - IR - L\frac{dI}{dt} = 0$

Rearranging:

$\boxed{L\frac{dI}{dt} + IR = \mathcal{E}}$

This is a first-order linear ODE with constant coefficients.

Standard Form

$\frac{dI}{dt} + \frac{R}{L}I = \frac{\mathcal{E}}{L}$

Comparing with $y' + Py = Q$:

• $P = R/L$
• $Q = \mathcal{E}/L$

Solving the ODE

Method 1: Integrating Factor

Multiply by $e^{Rt/L}$:

$\frac{d}{dt}\left[I \cdot e^{Rt/L}\right] = \frac{\mathcal{E}}{L}e^{Rt/L}$

Integrate both sides:

$I \cdot e^{Rt/L} = \frac{\mathcal{E}}{R}e^{Rt/L} + C$

$I(t) = \frac{\mathcal{E}}{R} + Ce^{-Rt/L}$

With $I(0) = 0$: $C = -\mathcal{E}/R$.

$\boxed{I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-Rt/L}\right)}$

Method 2: Separation of Variables

$\frac{dI}{\mathcal{E}/L - (R/L)I} = dt$

$-\frac{L}{R}\ln\left(\frac{\mathcal{E}}{L} - \frac{R}{L}I\right) = t + C$

This yields the same result after applying $I(0) = 0$.

Voltage Across Each Element

Using $I(t) = \frac{\mathcal{E}}{R}(1 - e^{-Rt/L})$:

Across the resistor: $V_R(t) = IR = \mathcal{E}(1 - e^{-Rt/L})$

Across the inductor: $V_L(t) = L\frac{dI}{dt} = \mathcal{E}\,e^{-Rt/L}$

Verification: $V_R + V_L = \mathcal{E}(1 - e^{-Rt/L}) + \mathcal{E}\,e^{-Rt/L} = \mathcal{E}$ ✓

Behavior at Key Times

Time	$V_R$	$V_L$	$I$
$t = 0$	$0$	$\mathcal{E}$	$0$
$t = \tau$	$0.632\mathcal{E}$	$0.368\mathcal{E}$	$0.632(\mathcal{E}/R)$
$t \to \infty$	$\mathcal{E}$	$0$	$\mathcal{E}/R$

Summary — Part 2

Result	Expression
Differential equation	$L\,dI/dt + IR = \mathcal{E}$
Charging solution	$I(t) = (\mathcal{E}/R)(1-e^{-Rt/L})$
$V_R(t)$	$\mathcal{E}(1-e^{-Rt/L})$
$V_L(t)$	$\mathcal{E}\,e^{-Rt/L}$
Inductor at $t=0$	Open circuit
Inductor at $t=\infty$	Short circuit

Next up: RL charging and discharging curves — Part 3.

RL Charging and Discharging

Part 3 of 7 — Growth and Decay of Current

Charging (Growth)

Switch closes at $t = 0$, connecting $\mathcal{E}$, $R$, and $L$ in series:

$I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right), \quad \tau = \frac{L}{R}$

The current rises from 0 toward $\mathcal{E}/R$ exponentially.

Discharging (Decay)

If the EMF source is removed and the circuit is closed through $R$ only (initial current $I_0$):

$L\frac{dI}{dt} + IR = 0 \implies I(t) = I_0 e^{-t/\tau}$

The current decays exponentially from $I_0$ to 0.

Phase	Equation	$I(0)$	$I(\infty)$
Charging	$I = (\mathcal{E}/R)(1-e^{-t/\tau})$	0	$\mathcal{E}/R$
Discharging	$I = I_0 e^{-t/\tau}$	$I_0$	0

Derivation of the Decay Solution

Starting from $L\,dI/dt + IR = 0$:

$\frac{dI}{I} = -\frac{R}{L}dt$

$\int_{I_0}^{I} \frac{dI'}{I'} = -\frac{R}{L}\int_0^t dt'$

$\ln\frac{I}{I_0} = -\frac{Rt}{L}$

$I(t) = I_0 e^{-Rt/L} = I_0 e^{-t/\tau}$

Voltage During Decay

$V_R = IR = I_0 R e^{-t/\tau}$

$V_L = L\frac{dI}{dt} = L \cdot I_0 \left(-\frac{R}{L}\right)e^{-t/\tau} = -I_0 R e^{-t/\tau}$

Note: $V_L = -V_R$ (KVL with no EMF source). The inductor drives current through the resistor, acting as a temporary EMF source.

Progress at Multiple Time Constants

For charging ($I/I_{\max}$) and discharging ($I/I_0$):

$t/\tau$	Charging: $1-e^{-t/\tau}$	Discharging: $e^{-t/\tau}$
1	63.2%	36.8%
2	86.5%	13.5%
3	95.0%	5.0%
4	98.2%	1.8%
5	99.3%	0.7%

Rule of thumb: After $5\tau$, the transient is essentially complete (< 1% remaining).

Solving for Time

How long to reach a specific current $I_f$ during charging?

$I_f = \frac{\mathcal{E}}{R}(1-e^{-t/\tau}) \implies e^{-t/\tau} = 1 - \frac{I_f R}{\mathcal{E}}$

$t = -\tau \ln\left(1 - \frac{I_f R}{\mathcal{E}}\right)$

Summary — Part 3

Phase	Current	Key Feature
Charging	$(\mathcal{E}/R)(1-e^{-t/\tau})$	Approaches $\mathcal{E}/R$
Discharging	$I_0 e^{-t/\tau}$	Decays to 0
At $t = \tau$	63.2% of final (charging)	36.8% remaining (discharging)
At $t = 5\tau$	99.3% complete	< 1% remaining

Next up: The time constant $\tau = L/R$ in depth — Part 4.

Time Constant $\tau = L/R$

Part 4 of 7 — Physical Meaning and Applications

What Does $\tau = L/R$ Tell Us?

The time constant sets the timescale for the RL transient:

$\tau = \frac{L}{R}$

Physical interpretation:

• Large $L$: more energy stored per unit current → slower change
• Large $R$: more energy dissipated per unit current → faster decay
• $\tau$ is the time for the current to reach $1 - 1/e \approx 63.2\%$ of its final value (charging)
• $\tau$ is the time for the current to fall to $1/e \approx 36.8\%$ (discharging)

Units Check

$\frac{[L]}{[R]} = \frac{\text{H}}{\Omega} = \frac{\text{V}\cdot\text{s}/\text{A}}{\text{V}/\text{A}} = \text{s} \quad \checkmark$

The Initial Slope Interpretation

At $t = 0$ during charging:

$\frac{dI}{dt}\bigg|_{t=0} = \frac{\mathcal{E}}{L}$

If the current continued at this initial rate, it would reach $\mathcal{E}/R$ at time:

$t = \frac{\mathcal{E}/R}{\mathcal{E}/L} = \frac{L}{R} = \tau$

The time constant is the time the current would take to reach its final value if it maintained its initial rate of change.

Multiple Resistors

For complex circuits, the time constant uses the Thévenin resistance seen by the inductor:

$\tau = \frac{L}{R_{\text{Th}}}$

Example: If $L$ is in series with $R_1$ and both are in parallel with $R_2$:

When the source is removed (for decay), $R_{\text{Th}} = R_1 + R_2$ (series from L's perspective) → No! Actually from the inductor's terminals: $R_{\text{Th}} = R_1 + R_2$ if they're in series, or compute properly using Thévenin.

Comparison: RL vs. RC Time Constants

Circuit	Time Constant	Equation	Growing	Decaying
RC	$\tau = RC$	$V_C = \mathcal{E}(1-e^{-t/\tau})$	Voltage grows	Voltage decays
RL	$\tau = L/R$	$I = (\mathcal{E}/R)(1-e^{-t/\tau})$	Current grows	Current decays

Key Analogy

RC Quantity	RL Analog
Charge $Q$	Flux linkage $\Lambda = LI$
Voltage $V = Q/C$	Current $I = \Lambda/L$
$RC$	$L/R$
$\frac{1}{2}CV^2$	$\frac{1}{2}LI^2$

The mathematical structure is identical: $RC: \quad \frac{dQ}{dt} + \frac{Q}{RC} = \frac{\mathcal{E}}{R}$ $RL: \quad \frac{dI}{dt} + \frac{R}{L}I = \frac{\mathcal{E}}{L}$

Summary — Part 4

Concept	Detail
Time constant	$\tau = L/R$
Initial slope	$dI/dt
Thévenin approach	$\tau = L/R_{\text{Th}}$
RL ↔ RC analogy	$L/R \leftrightarrow RC$
After $5\tau$	Transient < 1%

Next up: Energy stored in an inductor — Part 5.

Energy Stored in an Inductor

Part 5 of 7 — $U = \frac{1}{2}LI^2$

Derivation from Power

The power delivered to an inductor is:

$P_L = V_L \cdot I = L\frac{dI}{dt} \cdot I$

The energy stored is the integral of power:

$U = \int_0^t P_L\,dt' = \int_0^I LI'\,dI' = \frac{1}{2}LI^2$

$\boxed{U = \frac{1}{2}LI^2}$

Comparison with Capacitor

Component	Energy	Field
Capacitor	$U = \frac{1}{2}CV^2$	Electric field
Inductor	$U = \frac{1}{2}LI^2$	Magnetic field

The energy is stored in the magnetic field created by the current flowing through the inductor.

Magnetic Energy Density

For a solenoid: $B = \mu_0 nI$ and $L = \mu_0 n^2 A\ell$.

$U = \frac{1}{2}LI^2 = \frac{1}{2}(\mu_0 n^2 A\ell)I^2 = \frac{(\mu_0 nI)^2}{2\mu_0}(A\ell) = \frac{B^2}{2\mu_0} \cdot (\text{volume})$

The magnetic energy density is:

$\boxed{u_B = \frac{B^2}{2\mu_0}}$

This is a general result valid for any magnetic field, not just solenoids.

Example Calculation

A 1 T field stores:

$u_B = \frac{(1)^2}{2(4\pi \times 10^{-7})} = \frac{1}{8\pi \times 10^{-7}} \approx 4 \times 10^5 \text{ J/m}^3$

For comparison, the electric energy density $u_E = \frac{1}{2}\epsilon_0 E^2$: a field of $3 \times 10^6$ V/m (near breakdown) stores only $\sim 40$ $J/m^{3}$.

Energy Budget During RL Charging

As current grows from 0 to $I_f = \mathcal{E}/R$:

Energy from the source:

$U_{\text{source}} = \int_0^\infty \mathcal{E} I\,dt = \mathcal{E}\int_0^\infty \frac{\mathcal{E}}{R}(1-e^{-t/\tau})\,dt$

$= \frac{\mathcal{E}^2}{R}\left[t + \tau e^{-t/\tau}\right]_0^\infty$

This integral diverges! But during the transient (finite time), we can compute:

$U_{\text{source}}(\text{transient only}) = \frac{\mathcal{E}^2\tau}{R} = \frac{L\mathcal{E}^2}{R^2}$

Actually, the steady-state power $\mathcal{E}^2/R$ continues forever. The finite transient energy is:

$U_{\text{stored}} = \frac{1}{2}LI_f^2 = \frac{L\mathcal{E}^2}{2R^2}$

$U_{\text{dissipated in transient}} = \frac{L\mathcal{E}^2}{2R^2}$

Like RC circuits: during the transient, equal energy is stored and dissipated.

Summary — Part 5

Formula	Expression
Stored energy	$U = \frac{1}{2}LI^2$
Energy density	$u_B = B^2/(2\mu_0)$
Rate of energy storage	$dU/dt = LI\,dI/dt$
RL charging energy split	$U_R = U_L = L\mathcal{E}^2/(2R^2)$

Next up: Problem-solving workshop — Part 6.

Problem-Solving Workshop

Part 6 of 7 — AP Physics C: E&M Style Problems

Strategy for RL Circuit Problems

1. Identify the circuit phase: charging or discharging.
2. Determine the time constant $\tau = L/R_{\text{Th}}$.
3. Find initial and final conditions:
   • Charging: $I(0) = 0$, $I(\infty) = \mathcal{E}/R$
   • Discharging: $I(0) = I_0$, $I(\infty) = 0$
4. Write the solution: $I(t) = I(\infty) + [I(0) - I(\infty)]e^{-t/\tau}$
5. Compute voltages, power, or energy as needed.

This general formula works for any RL transient: $I(t) = I_f + (I_i - I_f)e^{-t/\tau}$.

Worked Example: Two-Resistor RL Circuit

A circuit has $\mathcal{E} = 30$ V, $R_1 = 10\,\Omega$ (in series with $L = 0.1$ H), and $R_2 = 15\,\Omega$ in parallel with the $R_1$-$L$ branch.

Steady State ($t \to \infty$):

Inductor acts as short circuit: $V_L = 0$, so $V_{R_1} = IR_1$. $I_L = \frac{\mathcal{E}}{R_1} \cdot \frac{R_2}{R_1 + R_2}$

Wait — let's be careful. At steady state, $L$ is a wire. The parallel combination is $R_1 \| R_2 = 10 \times 15/25 = 6\,\Omega$... but this depends on the exact topology.

If $R_1$ and $L$ are in the same branch (series), and that branch is in parallel with $R_2$:

At steady state: voltage across each branch $= \mathcal{E} = 30$ V.

• $I_{\text{branch 1}} = 30/R_1 = 30/10 = 3$ A (since $L$ is a short)
• $I_{\text{branch 2}} = 30/R_2 = 30/15 = 2$ A
• $I_{\text{total}} = 5$ A

Time constant: $\tau = L/R_1 = 0.1/10 = 0.01$ s = 10 ms (the resistance in the inductor's branch).

Energy Dissipation During Decay

During RL decay with $I(t) = I_0 e^{-t/\tau}$:

$P_R(t) = I^2 R = I_0^2 R e^{-2t/\tau}$

Total energy dissipated:

$U_R = \int_0^\infty I_0^2 R e^{-2t/\tau}\,dt = I_0^2 R \cdot \frac{\tau}{2} = I_0^2 R \cdot \frac{L}{2R} = \frac{1}{2}LI_0^2$

This equals the initial magnetic energy stored — energy is conserved.

Workshop Summary

Problem Type	Key Formula
General RL transient	$I(t) = I_f + (I_i - I_f)e^{-t/\tau}$
Multiple resistors	$\tau = L/R_{\text{Th}}$
Time to reach target	$t = -\tau\ln[(I_f - I_{\text{target}})/(I_f - I_i)]$
Energy verification	$\int P_R\,dt = \frac{1}{2}LI_0^2$

Next up: Review and applications — Part 7.

Review & Applications

Part 7 of 7 — Comprehensive Assessment

Formula Reference

Concept	Formula
Self-inductance	$\Lambda = LI$, $\mathcal{E} = -L\,dI/dt$
Solenoid	$L = \mu_0 N^2 A/\ell$
Toroid	$L = \mu_0 N^2 h \ln(b/a)/(2\pi)$
RL charging	$I = (\mathcal{E}/R)(1-e^{-t/\tau})$
RL discharging	$I = I_0 e^{-t/\tau}$
Time constant	$\tau = L/R$
Stored energy	$U = \frac{1}{2}LI^2$
Energy density	$u_B = B^2/(2\mu_0)$
General transient	$I(t) = I_f + (I_i - I_f)e^{-t/\tau}$

Real-World Applications

1. Ignition Coils

Car ignition systems use RL decay to generate high voltages. When current through the inductor is interrupted:

$\mathcal{E}_{\text{induced}} = -L\frac{dI}{dt}$

A rapid $dI/dt$ (fast switch-off) produces thousands of volts to create a spark.

2. Electromagnetic Relays

An inductor creates a magnetic field to pull a switch contact. The RL time constant determines how quickly the relay engages.

Flyback protection: When the relay opens, collapsing $B$ induces large $\mathcal{E}$. A diode across the inductor provides a current path, preventing voltage spikes.

3. Energy Storage (SMES)

Superconducting Magnetic Energy Storage uses $R \approx 0$ coils:

• $\tau = L/R \to \infty$ (current persists indefinitely)
• Energy stored as $\frac{1}{2}LI^2$ with no resistive losses

4. Transformers

Mutual inductance $M$ couples two coils: $\mathcal{E}_2 = -M\frac{dI_1}{dt}, \quad M = k\sqrt{L_1 L_2}$ where $k$ is the coupling coefficient ($0 \leq k \leq 1$).

🎉 Topic Complete!

You've mastered Inductance & RL Circuits for AP Physics C: E&M:

Part	Topic	Status
1	Self-inductance definition and calculation	✅
2	RL circuit differential equation	✅
3	RL charging and discharging	✅
4	Time constant $\tau = L/R$	✅
5	Energy stored in inductor	✅
6	Problem-solving workshop	✅
7	Review & applications	✅

Key takeaway: RL circuits are governed by the same first-order ODE structure as RC circuits, with the duality $L \leftrightarrow C$, $I \leftrightarrow V$. Master the general transient formula $I(t) = I_f + (I_i - I_f)e^{-t/\tau}$, the energy relation $U = \frac{1}{2}LI^2$, and the critical behavior of inductors as open circuits at $t = 0$ and short circuits at $t = \infty$.