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6.5 - Circuits with Resistors and Inductors (LR Circuits)

Introduction to LR Circuits

An LR circuit has a resistor and an inductor wired in series. The resistor burns off energy while the inductor stores it in a magnetic field.
The key behavior: the current changes exponentially over time, and a time constant sets the pace.

Kirchhoff’s Loop Rule for LR Circuits

Apply Kirchhoff’s Loop Rule to an LR circuit with a battery:
E=IR+LdIdt\mathcal{E} = IR + L\frac{dI}{dt}
where: - E\mathcal{E} = EMF of the battery (V). - II = current through the circuit (A). - RR = resistance (Ω). - LL = inductance (H). - dIdt\frac{dI}{dt} = rate of change of the current (A/s).
What each term is: - IRIR is the voltage drop across the resistor, dissipated as heat. - LdIdtL\frac{dI}{dt} is the self-induced EMF in the inductor, fighting the change in current.
This is a differential equation, because that dIdt\frac{dI}{dt} is sitting right in it.

Time Constant for LR Circuits

The time constant tells you how fast the current changes in an LR circuit.
τ=LR\tau = \frac{L}{R}
where: - τ\tau (tau) = time constant (s). - LL = inductance (H). - RR = resistance (Ω).
What it means: - Bigger τ\tau: the current changes slowly (big inductance or small resistance). - Smaller τ\tau: the current changes fast (small inductance or big resistance).
After one time constant (t=τt = \tau): - Charging: the current has climbed to about 63% of its max. - Discharging: the current has dropped to about 37% of where it started.
After 5τ5\tau: the circuit is about 99% of the way there, so essentially done.

Another Way to Read τ

Right after the switch closes, the current is zero, so the loop rule gives E=LdIdt\mathcal{E} = L\frac{dI}{dt}. The current starts out rising at dIdt=EL\frac{dI}{dt} = \frac{\mathcal{E}}{L}. If it kept rising at that starting rate, it would reach its final value ER\frac{\mathcal{E}}{R} after:
t=E/RE/L=LR=τt = \frac{\mathcal{E}/R}{\mathcal{E}/L} = \frac{L}{R} = \tau
So the time constant is the time the circuit would take to reach steady state if it kept changing at its starting rate. It doesn’t, because the rate keeps dropping, so it only gets to 63% by then. The same idea works for a current dying out: the starting slope is −I0τ-\frac{I_0}{\tau}, and that line hits zero at t=τt = \tau, when the real current is at 37%.
6.5 - Circuits with Resistors and Inductors (LR Circuits)

Circuit Schematic and Behavior

The LR Circuit Setup

6.5 - Circuits with Resistors and Inductors (LR Circuits)
The charging phase: the switch connects the battery, and current builds up in the inductor.
The discharging phase: the switch cuts the battery out, and the inductor dumps its stored energy through resistor R2R_2.

Charging and Discharging Graphs

Charging: the Current Builds

When the switch closes and hooks up the battery, the current grows exponentially from zero up to its maximum.
6.5 - Circuits with Resistors and Inductors (LR Circuits)
The charging equation:
I(t)=I0(1−e−t/τ)I(t) = I_0\left(1 - e^{-t/\tau}\right)
where I0=ER1I_0 = \frac{\mathcal{E}}{R_1} is the steady-state current.

Discharging: the Current Decays

When the switch opens and cuts out the battery, the current falls exponentially from its max down to zero.
6.5 - Circuits with Resistors and Inductors (LR Circuits)
The discharging equation:
I(t)=I0e−t/τI(t) = I_0 e^{-t/\tau}
where I0I_0 is whatever the current was when the discharge started.

Example Problem: Deriving the Discharge Equation

Problem: An LR circuit has been hooked up to a battery long enough to reach a steady-state current I0I_0. Then the switch opens, cutting the battery out and letting the inductor discharge through resistor R2R_2.
Derive the equation showing that the current is:
I(t)=I0e−t/τI(t) = I_0 e^{-t/\tau}
where τ=LR2\tau = \frac{L}{R_2}.
6.5 - Circuits with Resistors and Inductors (LR Circuits)

Solution

Step 1: apply Kirchhoff’s Loop Rule with no battery
With the battery gone, the loop equation is:
0=−IR2−LdIdt0 = -IR_2 - L\frac{dI}{dt}
Where the signs come from (going around the loop in the direction of the current): - −IR2-IR_2: moving with the current through a resistor, the potential drops by IR2IR_2. - −LdIdt-L\frac{dI}{dt}: this is the potential change across the inductor when you move with the current. The minus sign is part of that rule for every inductor. It isn’t there because the current is falling. In fact, since the current is falling, dIdt<0\frac{dI}{dt} < 0, so this term is positive: the inductor acts like a small battery pushing the current forward, which is how it keeps the current going.
Step 2: rearrange the differential equation
−LdIdt=IR2-L\frac{dI}{dt} = IR_2LdIdt=−IR2L\frac{dI}{dt} = -IR_2dIdt=−R2LI\frac{dI}{dt} = -\frac{R_2}{L}I
Step 3: separate the variables
dII=−R2Ldt\frac{dI}{I} = -\frac{R_2}{L}dt
Step 4: integrate both sides
∫I0I(t)dII=−R2L∫0tdt\int_{I_0}^{I(t)} \frac{dI}{I} = -\frac{R_2}{L}\int_0^t dt
The left side:
[ln⁡I]I0I(t)=ln⁡I(t)−ln⁡I0=ln⁡(I(t)I0)[\ln I]_{I_0}^{I(t)} = \ln I(t) - \ln I_0 = \ln\left(\frac{I(t)}{I_0}\right)
The right side:
−R2L[t]0t=−R2Lt-\frac{R_2}{L}[t]_0^t = -\frac{R_2}{L}t
Step 5: put them together
ln⁡(I(t)I0)=−R2Lt\ln\left(\frac{I(t)}{I_0}\right) = -\frac{R_2}{L}t
Step 6: exponentiate both sides
eln⁡(I(t)/I0)=e−R2t/Le^{\ln(I(t)/I_0)} = e^{-R_2 t/L}I(t)I0=e−R2t/L\frac{I(t)}{I_0} = e^{-R_2 t/L}
Step 7: spot the time constant
Since τ=LR2\tau = \frac{L}{R_2}, we have:
R2L=1τ\frac{R_2}{L} = \frac{1}{\tau}
So:
I(t)I0=e−t/τ\frac{I(t)}{I_0} = e^{-t/\tau}
Step 8: the final result
I(t)=I0e−t/τ\boxed{I(t) = I_0 e^{-t/\tau}}
Which confirms the exponential decay of the current during discharge.

Example Problem: Deriving the Charging Equation

Problem: A battery of EMF E\mathcal{E}, a resistor RR, and an inductor LL are in series. The switch closes at t=0t = 0, when I=0I = 0. Solve the loop equation for I(t)I(t).
Step 1: write the loop rule
E−IR−LdIdt=0\mathcal{E} - IR - L\frac{dI}{dt} = 0
Step 2: separate the variables
dIER−I=RL dt\frac{dI}{\frac{\mathcal{E}}{R} - I} = \frac{R}{L}\,dt
Step 3: integrate, starting from zero current
∫0IdI′ER−I′=RL∫0tdt′⇒−ln⁡(ER−IER)=RLt\int_0^{I}\frac{dI'}{\frac{\mathcal{E}}{R} - I'} = \frac{R}{L}\int_0^t dt' \quad\Rightarrow\quad -\ln\left(\frac{\frac{\mathcal{E}}{R} - I}{\frac{\mathcal{E}}{R}}\right) = \frac{R}{L}t
Step 4: solve for the current
1−IRE=e−Rt/L⇒I(t)=ER(1−e−t/τ),τ=LR1 - \frac{IR}{\mathcal{E}} = e^{-Rt/L} \quad\Rightarrow\quad \boxed{I(t) = \frac{\mathcal{E}}{R}\left(1 - e^{-t/\tau}\right), \quad \tau = \frac{L}{R}}
Check it: at t=0t = 0, I=0I = 0. As t→∞t \to \infty, I→ERI \to \frac{\mathcal{E}}{R}, as if the inductor were a plain wire. And the starting slope is ER⋅1τ=EL\frac{\mathcal{E}}{R}\cdot\frac{1}{\tau} = \frac{\mathcal{E}}{L}, the same as plugging I=0I = 0 into the loop rule.

Right After a Switch Is Closed or Opened

The current through an inductor can’t jump, so right after a switch moves, the inductor’s current is whatever it was right before. Its EMF does whatever it takes to make that true.
Switch just closed, no current yet: for that instant the inductor carries zero current, like a break in its branch. Its induced EMF is equal in size and opposite in direction to the potential difference applied across its branch, which is exactly what keeps the current at zero. In a single series loop, that means ∣EL∣=E|\mathcal{E}_L| = \mathcal{E} and VR=0V_R = 0.
Switch just opened (or the source cut out) while current flows: the inductor keeps the same current going, now through whatever path is left. Its EMF becomes whatever that path needs, ∣EL∣=I0Rpath|\mathcal{E}_L| = I_0R_{path}. If RpathR_{path} is large, that can be much bigger than the battery’s EMF, which is why opening a switch on an inductor can make a spark.
After that, everything in the circuit (the current, the inductor’s voltage, the stored energy) changes exponentially until it settles at its new steady state.

Example Problem: Numbers for a Series LR Circuit

Problem: A 9.0 V battery, a 3.0 Ω resistor, and a 0.60 H inductor are in series. The switch closes at t=0t = 0. Find (a) the time constant, (b) how fast the current increases right after the switch closes, (c) the current at t=0.40t = 0.40 s, (d) when the current reaches 2.0 A, and (e) the energy stored in the inductor once the current is steady.
(a) Time constant
τ=LR=0.60 H3.0 Ω=0.20 s\tau = \frac{L}{R} = \frac{0.60 \text{ H}}{3.0\ \Omega} = 0.20 \text{ s}
(b) Starting rate: at t=0t = 0, I=0I = 0, so the whole 9.0 V is across the inductor:
dIdt=EL=9.0 V0.60 H=15 A/s\frac{dI}{dt} = \frac{\mathcal{E}}{L} = \frac{9.0 \text{ V}}{0.60 \text{ H}} = 15 \text{ A/s}
(c) Current at 0.40 s: the final current is I0=9.0 V3.0 Ω=3.0I_0 = \frac{9.0 \text{ V}}{3.0\ \Omega} = 3.0 A, so:
I=(3.0 A)(1−e−(0.40 s)/(0.20 s))=(3.0 A)(1−e−2)=2.6 AI = (3.0 \text{ A})\left(1 - e^{-(0.40 \text{ s})/(0.20 \text{ s})}\right) = (3.0 \text{ A})(1 - e^{-2}) = 2.6 \text{ A}
(d) Time to reach 2.0 A: set I=2.0I = 2.0 A and solve:
1−e−t/τ=2.03.0⇒e−t/τ=13⇒t=τln⁡3=(0.20 s)(1.10)=0.22 s1 - e^{-t/\tau} = \frac{2.0}{3.0} \quad\Rightarrow\quad e^{-t/\tau} = \frac{1}{3} \quad\Rightarrow\quad t = \tau\ln 3 = (0.20 \text{ s})(1.10) = 0.22 \text{ s}
(e) Stored energy at steady state
UL=12LI02=12(0.60 H)(3.0 A)2=2.7 JU_L = \frac{1}{2}LI_0^2 = \frac{1}{2}(0.60 \text{ H})(3.0 \text{ A})^2 = 2.7 \text{ J}

Steady-State Behavior

After a long time (many time constants), the LR circuit settles into steady state.

Charging Steady State

Long after the switch closes (t≫τt \gg \tau): - The current maxes out: I=I0=ERI = I_0 = \frac{\mathcal{E}}{R}. - The current stops changing: dIdt=0\frac{dI}{dt} = 0. - The inductor voltage is VL=LdIdt=0V_L = L\frac{dI}{dt} = 0, so it acts like a wire. - All the voltage drops across the resistor: VR=EV_R = \mathcal{E}.
Why: once the current stops changing, the inductor has no induced EMF left. It’s just a short circuit at that point.

Discharging Steady State

Long after the switch opens (t≫τt \gg \tau): - The current hits zero: I=0I = 0. - Every bit of energy that was in the inductor has been burned as heat in the resistor. - The circuit is done.

Energy Considerations

The energy the inductor starts with:
U0=12LI02U_0 = \frac{1}{2}LI_0^2
During the discharge, all of it gets dissipated as heat in the resistor:
Edissipated=∫0∞I2(t)R dtE_{dissipated} = \int_0^\infty I^2(t) R \, dt
Substitute I(t)=I0e−t/τI(t) = I_0 e^{-t/\tau}:
Edissipated=∫0∞(I0e−t/τ)2R dtE_{dissipated} = \int_0^\infty (I_0 e^{-t/\tau})^2 R \, dtEdissipated=I02R∫0∞e−2t/τ dtE_{dissipated} = I_0^2 R \int_0^\infty e^{-2t/\tau} \, dtEdissipated=I02R⋅τ2=I02R⋅L2RE_{dissipated} = I_0^2 R \cdot \frac{\tau}{2} = I_0^2 R \cdot \frac{L}{2R}Edissipated=12LI02E_{dissipated} = \frac{1}{2}LI_0^2
Which is exactly the energy it started with. ✓ Energy is conserved.

Voltage and Energy Over Time

The current isn’t the only thing that changes exponentially. In a series LR circuit, the inductor’s voltage and stored energy follow from I(t)I(t), with RR and τ\tau for the loop the current is in:
Switch closed (current building up): - VL=LdIdt=Ee−t/τV_L = L\frac{dI}{dt} = \mathcal{E}e^{-t/\tau}, starting at E\mathcal{E} and decaying to 0. - UL=12LI2=12LI02(1−e−t/τ)2U_L = \frac{1}{2}LI^2 = \frac{1}{2}LI_0^2\left(1 - e^{-t/\tau}\right)^2, starting at 0 and rising to 12LI02\frac{1}{2}LI_0^2.
Switch opened (current dying out): - ∣VL∣=I0Re−t/τ|V_L| = I_0Re^{-t/\tau}, starting at I0RI_0R and decaying to 0. - UL=12LI02e−2t/τU_L = \frac{1}{2}LI_0^2e^{-2t/\tau}, starting at 12LI02\frac{1}{2}LI_0^2 and decaying to 0.
6.5 - Circuits with Resistors and Inductors (LR Circuits)
Each graph levels off at a value set by the circuit’s starting conditions. Notice that the energy lags behind the current. At t=τt = \tau the building current is at 63%, but the energy is only at (0.63)2≈0.40(0.63)^2 \approx 0.40 of its final value. When the current dies out, the energy falls twice as fast as the current (time constant τ2\frac{\tau}{2}), so at t=τt = \tau only e−2≈0.14e^{-2} \approx 0.14 of it is left.

Example Problem: An Inductor with Several Resistors

Problem: In the circuit below, the switch SS has been open for a long time. It closes at t=0t = 0.
6.5 - Circuits with Resistors and Inductors (LR Circuits)
  1. Find I1I_1, I2I_2, and ILI_L right after the switch closes, and the rate dILdt\frac{dI_L}{dt} at that moment.
  1. Find all three currents and the energy stored in the inductor after the switch has been closed a long time.
  1. Then the switch is opened again. Find the current in each resistor and the size of the inductor’s EMF right after it opens, the time constant of the decay, and the total thermal energy produced afterward.

Solution

(a) Right after closing
The inductor’s current was zero, and it can’t jump, so IL=0I_L = 0. For this instant, the inductor’s branch carries no current, and the circuit acts like just R1R_1 and R2R_2 in series:
I1=I2=12 V4.0 Ω+12 Ω=0.75 AI_1 = I_2 = \frac{12 \text{ V}}{4.0\ \Omega + 12\ \Omega} = 0.75 \text{ A}
The potential difference across R2R_2 is (0.75 A)(12 Ω)=9.0(0.75 \text{ A})(12\ \Omega) = 9.0 V. The inductor’s branch is in parallel with R2R_2, so it has 9.0 V across it too. With no current, R3R_3 has no voltage across it, so all 9.0 V is across the inductor:
dILdt=VLL=9.0 V0.50 H=18 A/s\frac{dI_L}{dt} = \frac{V_L}{L} = \frac{9.0 \text{ V}}{0.50 \text{ H}} = 18 \text{ A/s}
(b) Long after closing
Now the inductor acts like a plain wire, so R2R_2 (12 Ω) is in parallel with R3R_3 (6.0 Ω):
R23=(12 Ω)(6.0 Ω)12 Ω+6.0 Ω=4.0 Ω,I1=12 V4.0 Ω+4.0 Ω=1.5 AR_{23} = \frac{(12\ \Omega)(6.0\ \Omega)}{12\ \Omega + 6.0\ \Omega} = 4.0\ \Omega, \qquad I_1 = \frac{12 \text{ V}}{4.0\ \Omega + 4.0\ \Omega} = 1.5 \text{ A}
The parallel part has (1.5 A)(4.0 Ω)=6.0(1.5 \text{ A})(4.0\ \Omega) = 6.0 V across it, so I2=6.0 V12 Ω=0.50I_2 = \frac{6.0 \text{ V}}{12\ \Omega} = 0.50 A and IL=6.0 V6.0 Ω=1.0I_L = \frac{6.0 \text{ V}}{6.0\ \Omega} = 1.0 A. The energy stored in the inductor is:
UL=12LIL2=12(0.50 H)(1.0 A)2=0.25 JU_L = \frac{1}{2}LI_L^2 = \frac{1}{2}(0.50 \text{ H})(1.0 \text{ A})^2 = 0.25 \text{ J}
(While the current builds, the time constant is LReq\frac{L}{R_{eq}}, where ReqR_{eq} is the resistance the inductor sees with the battery replaced by a wire: R3R_3 in series with R1R_1 and R2R_2 in parallel, Req=6.0 Ω+3.0 Ω=9.0 ΩR_{eq} = 6.0\ \Omega + 3.0\ \Omega = 9.0\ \Omega, so τ=0.056\tau = 0.056 s.)
(c) Right after opening
The battery and R1R_1 are cut out, so I1=0I_1 = 0. The inductor still carries 1.0 A, and the only path left is the loop through R3R_3 and R2R_2. So R3R_3 carries 1.0 A, and R2R_2 carries 1.0 A too, now flowing up (the opposite way from before). The inductor has to push 1.0 A through 18 Ω:
∣EL∣=(1.0 A)(6.0 Ω+12 Ω)=18 V|\mathcal{E}_L| = (1.0 \text{ A})(6.0\ \Omega + 12\ \Omega) = 18 \text{ V}
That’s more than the 12 V the battery supplied. The current then dies out with:
τ=LR2+R3=0.50 H18 Ω=0.028 s\tau = \frac{L}{R_2 + R_3} = \frac{0.50 \text{ H}}{18\ \Omega} = 0.028 \text{ s}
All 0.25 J that was stored in the inductor ends up as thermal energy in R2R_2 and R3R_3.
What it tells you: right after a switch closes, an inductor with no current acts like a break in its branch. A long time later, it acts like a plain wire. Right after a switch opens, it acts like a current source that keeps its current the same for that instant.

Summary of Key Formulas

Time Constant

τ=LR\tau = \frac{L}{R}

Charging Current

I(t)=I0(1−e−t/τ)I(t) = I_0\left(1 - e^{-t/\tau}\right)
where I0=ERI_0 = \frac{\mathcal{E}}{R}

Discharging Current

I(t)=I0e−t/τI(t) = I_0 e^{-t/\tau}

Voltage Across the Inductor

VL(t)=LdIdtV_L(t) = L\frac{dI}{dt}
  • Charging: VL=Ee−t/τV_L = \mathcal{E}e^{-t/\tau}, decaying from E\mathcal{E} down to 0.
  • Discharging: VL=−I0Re−t/τV_L = -I_0 R e^{-t/\tau}, starting at max and decaying to 0.

Comparison: RC vs. LR Circuits

Property
RC Circuit
LR Circuit
Time constant
τ=RC\tau = RC
τ=LR\tau = \frac{L}{R}
Stores
Electric field energy
Magnetic field energy
What changes
Voltage (on the capacitor)
Current (through the inductor)
Can’t change instantly
Voltage across C
Current through L
Charging equation
V(t)=V0(1−e−t/τ)V(t) = V_0(1 - e^{-t/\tau})
I(t)=I0(1−e−t/τ)I(t) = I_0(1 - e^{-t/\tau})
Discharging equation
V(t)=V0e−t/τV(t) = V_0 e^{-t/\tau}
I(t)=I0e−t/τI(t) = I_0 e^{-t/\tau}
Steady state (DC)
Open circuit (no current)
Short circuit (no voltage)
Same skeleton: both give you exponential behavior with basically identical math.

Common Mistakes

  1. Using the wrong resistance in the time constant. When the circuit has multiple resistors (R1R_1 for charging, R2R_2 for discharging), make sure you grab the right one. For discharge, τ=L/R2\tau = L/R_2, where R2R_2 is the resistance in the discharge path, not the original charging resistor.
  1. Dropping the negative sign in the differential equation. When you derive the discharge equation, it’s dIdt=−RLI\frac{dI}{dt} = -\frac{R}{L}I, negative because the current is falling. Don’t lose that sign.
  1. Mixing up the charging and discharging equations. Charging looks like I(t)=I0(1−e−t/τ)I(t) = I_0(1 - e^{-t/\tau}), starting at zero and growing. Discharging looks like I(t)=I0e−t/τI(t) = I_0 e^{-t/\tau}, starting at max and decaying. That (1−e−t/τ)(1 - e^{-t/\tau}) piece belongs to charging only.
  1. Wrong integration limits. When you solve the discharge differential equation, integrate from I0I_0 to I(t)I(t), not from 0, since you start at the maximum current I0I_0 when the discharge kicks off.
  1. Thinking the current jumps when the switch opens. Current through an inductor has to be continuous. The instant the switch opens, II is the same value just before and just after, even though the voltage across the inductor can spike dramatically.

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