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Inductance and RL Circuits

Self-inductance, mutual inductance, and RL circuit dynamics

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Inductance and RL Circuits

Self-Inductance

Changing current in coil induces EMF in same coil:

E=−LdIdt\mathcal{E} = -L\frac{dI}{dt}

where LL is inductance (or self-inductance).

Units: 1 henry (H) = 1 Wb/A = 1 V·s/A

Inductance of Solenoid

Long solenoid: nn turns/length, cross-sectional area AA, length ll:

L=μ0n2Al=μ0N2lAL = \mu_0 n^2 Al = \mu_0 \frac{N^2}{l}A

Mutual Inductance

Current I1I_1 in coil 1 creates flux through coil 2:

E2=−MdI1dt\mathcal{E}_2 = -M\frac{dI_1}{dt}

Mutual inductance: M=Φ21I1=Φ12I2M = \frac{\Phi_{21}}{I_1} = \frac{\Phi_{12}}{I_2}

(Same value both ways: M12=M21M_{12} = M_{21})

Energy Stored in Inductor

UL=12LI2U_L = \frac{1}{2}LI^2

Energy density in magnetic field: uB=B22μ0u_B = \frac{B^2}{2\mu_0}

For solenoid: U=uB⋅volume=B22μ0Al=12LI2U = u_B \cdot \text{volume} = \frac{B^2}{2\mu_0}Al = \frac{1}{2}LI^2

(Using B=μ0nIB = \mu_0 nI and L=μ0n2AlL = \mu_0 n^2 Al)

RL Circuit: Current Growth

Circuit: battery (E\mathcal{E}), resistor (RR), inductor (LL) in series.

Kirchhoff's loop rule: E−IR−LdIdt=0\mathcal{E} - IR - L\frac{dI}{dt} = 0

Differential equation: dIdt=EL−RLI\frac{dI}{dt} = \frac{\mathcal{E}}{L} - \frac{R}{L}I

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

Time constant: τL=LR\tau_L = \frac{L}{R}

After time τL\tau_L: current reaches (1−1/e)≈63%(1 - 1/e) \approx 63\% of final value.

RL Circuit: Current Decay

Remove battery, current decays:

LdIdt+IR=0L\frac{dI}{dt} + IR = 0

Solution: I(t)=I0e−Rt/LI(t) = I_0 e^{-Rt/L}

Current decays exponentially with time constant τL=L/R\tau_L = L/R.

Energy Considerations

Current growth:

Energy from battery: W=∫0∞EI dt=E2LR2W = \int_0^\infty \mathcal{E}I \, dt = \frac{\mathcal{E}^2L}{R^2}

Energy stored in inductor: UL=12L(ER)2U_L = \frac{1}{2}L\left(\frac{\mathcal{E}}{R}\right)^2

Energy dissipated in resistor: UR=12L(ER)2U_R = \frac{1}{2}L\left(\frac{\mathcal{E}}{R}\right)^2

(Equal amounts stored and dissipated)

LC Circuit

Inductor and capacitor (no resistance):

Ld2Qdt2+QC=0L\frac{d^2Q}{dt^2} + \frac{Q}{C} = 0

Oscillation: Q(t)=Q0cos⁡(ωt+ϕ)Q(t) = Q_0\cos(\omega t + \phi)

where ω=1/LC\omega = 1/\sqrt{LC} (angular frequency).

Energy oscillates between:

  • Electric: UE=Q2/(2C)U_E = Q^2/(2C)
  • Magnetic: UB=LI2/2U_B = LI^2/2
  • Total: U=UE+UBU = U_E + U_B = constant

LRC Circuit

With resistance, oscillations damped:

Ld2Qdt2+RdQdt+QC=0L\frac{d^2Q}{dt^2} + R\frac{dQ}{dt} + \frac{Q}{C} = 0

Damped oscillation (for R<2L/CR < 2\sqrt{L/C}): Q(t)=Q0e−Rt/(2L)cos⁡(ω′t)Q(t) = Q_0e^{-Rt/(2L)}\cos(\omega' t)

where ω′=1LC−R24L2\omega' = \sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}

Quality factor: Q=ω0LR=1RLCQ = \frac{\omega_0 L}{R} = \frac{1}{R}\sqrt{\frac{L}{C}}

📚 Practice Problems

1Problem 1medium

❓ Question:

An RL circuit consists of R = 8.0 Ω, L = 0.4 H, and a battery ε = 12 V. The switch is closed at t = 0. Find: (a) the time constant, (b) the current at t = 0.08 s, and (c) the rate of energy storage in the inductor at this time.

💡 Show Solution

Given:

  • R = 8.0 Ω
  • L = 0.4 H
  • ε = 12 V
  • t = 0.08 s

(a) Time constant:

τ=LR=0.48.0\tau = \frac{L}{R} = \frac{0.4}{8.0}

τ=0.05 s=50 ms\tau = \boxed{0.05 \text{ s} = 50 \text{ ms}}

(b) Current at t = 0.08 s:

For RL circuit with battery: I(t)=Imax(1−e−t/τ)I(t) = I_{max}(1 - e^{-t/\tau})

where Imax=ε/R=12/8.0=1.5I_{max} = \varepsilon/R = 12/8.0 = 1.5 A

I(0.08)=1.5(1−e−0.08/0.05)I(0.08) = 1.5(1 - e^{-0.08/0.05})

I(0.08)=1.5(1−e−1.6)I(0.08) = 1.5(1 - e^{-1.6})

I(0.08)=1.5(1−0.2019)I(0.08) = 1.5(1 - 0.2019)

I(0.08)=1.5(0.7981)=1.20 AI(0.08) = 1.5(0.7981) = \boxed{1.20 \text{ A}}

(c) Rate of energy storage:

Energy in inductor: UL=12LI2U_L = \frac{1}{2}LI^2

dULdt=LIdIdt\frac{dU_L}{dt} = LI\frac{dI}{dt}

From differential equation: ε=IR+LdIdt\varepsilon = IR + L\frac{dI}{dt}

dIdt=ε−IRL\frac{dI}{dt} = \frac{\varepsilon - IR}{L}

At t = 0.08 s: dIdt=12−(1.20)(8.0)0.4=12−9.60.4=6.0 A/s\frac{dI}{dt} = \frac{12 - (1.20)(8.0)}{0.4} = \frac{12 - 9.6}{0.4} = 6.0 \text{ A/s}

dULdt=LIdIdt=(0.4)(1.20)(6.0)\frac{dU_L}{dt} = LI\frac{dI}{dt} = (0.4)(1.20)(6.0)

dULdt=2.88 W\frac{dU_L}{dt} = \boxed{2.88 \text{ W}}

Alternatively: PL=εI−I2R=VILP_L = \varepsilon I - I^2 R = VI_L where VL=LdIdtV_L = L\frac{dI}{dt}

2Problem 2hard

❓ Question:

An inductor L = 0.25 H carrying current I₀ = 2.0 A is suddenly connected to a resistor R = 10 Ω (battery removed). Find: (a) the current as a function of time, (b) the time for the current to decrease to 10% of its initial value, and (c) the total energy dissipated in the resistor.

💡 Show Solution

Given:

  • L = 0.25 H
  • I₀ = 2.0 A
  • R = 10 Ω
  • Discharging RL circuit

(a) Current vs. time:

For discharging: ε=0\varepsilon = 0, so: −LdIdt=IR-L\frac{dI}{dt} = IR

dII=−RLdt\frac{dI}{I} = -\frac{R}{L}dt

Integrating: ln⁡I=−RLt+C\ln I = -\frac{R}{L}t + C

At t = 0: I = I₀, so C = ln I₀

ln⁡(II0)=−RLt\ln\left(\frac{I}{I_0}\right) = -\frac{R}{L}t

I(t)=I0e−Rt/L=2.0e−10t/0.25=2.0e−40t A\boxed{I(t) = I_0 e^{-Rt/L} = 2.0 e^{-10t/0.25} = 2.0e^{-40t} \text{ A}}

(b) Time for I = 0.1I₀:

0.1I0=I0e−Rt/L0.1I_0 = I_0 e^{-Rt/L}

0.1=e−Rt/L0.1 = e^{-Rt/L}

ln⁡(0.1)=−RtL\ln(0.1) = -\frac{Rt}{L}

t=−Lln⁡(0.1)R=−(0.25)(−2.303)10t = -\frac{L\ln(0.1)}{R} = -\frac{(0.25)(-2.303)}{10}

t=0.0576 s=57.6 mst = \boxed{0.0576 \text{ s} = 57.6 \text{ ms}}

This is 2.3τ2.3\tau where τ=L/R=0.025\tau = L/R = 0.025 s

(c) Total energy dissipated:

Initial energy stored in inductor: U0=12LI02=12(0.25)(2.0)2U_0 = \frac{1}{2}LI_0^2 = \frac{1}{2}(0.25)(2.0)^2

U0=0.5 JU_0 = 0.5 \text{ J}

As t → ∞, all energy is dissipated: Edissipated=U0=0.5 JE_{dissipated} = U_0 = \boxed{0.5 \text{ J}}

Check: ∫0∞I2R dt=∫0∞I02e−2Rt/LR dt=I02RL2R=12LI02\int_0^\infty I^2 R \, dt = \int_0^\infty I_0^2 e^{-2Rt/L} R \, dt = \frac{I_0^2 R L}{2R} = \frac{1}{2}LI_0^2 ✓

3Problem 3hard

❓ Question:

An LC circuit consists of L = 0.1 H and C = 50 μF. The capacitor is initially charged to Q₀ = 1.0 × 10⁻⁴ C, then connected to the inductor at t = 0. Find: (a) the angular frequency of oscillation, (b) the maximum current, and (c) expressions for Q(t) and I(t).

💡 Show Solution

Given:

  • L = 0.1 H
  • C = 50 μF = 5.0 × 10⁻⁵ F
  • Q₀ = 1.0 × 10⁻⁴ C
  • I₀ = 0 (initially)

(a) Angular frequency:

For LC circuit: ω=1LC=1(0.1)(5.0×10−5)\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(0.1)(5.0 \times 10^{-5})}}

ω=15.0×10−6=12.24×10−3\omega = \frac{1}{\sqrt{5.0 \times 10^{-6}}} = \frac{1}{2.24 \times 10^{-3}}

ω=447 rad/s\omega = \boxed{447 \text{ rad/s}}

Frequency: f=ω/(2π)=71.2f = \omega/(2\pi) = 71.2 Hz

(b) Maximum current:

Energy conservation: UC=ULU_C = U_L

Q022C=12LImax2\frac{Q_0^2}{2C} = \frac{1}{2}LI_{max}^2

Imax=Q0LC=Q0ωI_{max} = \frac{Q_0}{\sqrt{LC}} = Q_0 \omega

Imax=(1.0×10−4)(447)I_{max} = (1.0 \times 10^{-4})(447)

Imax=0.0447 A=44.7 mAI_{max} = \boxed{0.0447 \text{ A} = 44.7 \text{ mA}}

(c) Q(t) and I(t):

General solution: Q(t)=Q0cos⁡(ωt+ϕ)Q(t) = Q_0 \cos(\omega t + \phi)

At t = 0: Q = Q₀, so φ = 0

Q(t)=Q0cos⁡(ωt)=(1.0×10−4)cos⁡(447t) C\boxed{Q(t) = Q_0 \cos(\omega t) = (1.0 \times 10^{-4})\cos(447t) \text{ C}}

Current: I(t)=−dQdt=Q0ωsin⁡(ωt)I(t) = -\frac{dQ}{dt} = Q_0 \omega \sin(\omega t)

I(t)=Imaxsin⁡(ωt)=0.0447sin⁡(447t) A\boxed{I(t) = I_{max} \sin(\omega t) = 0.0447\sin(447t) \text{ A}}

Energy oscillates between capacitor and inductor:

  • At ωt=0,π,2π\omega t = 0, \pi, 2\pi: all energy in C
  • At ωt=π/2,3π/2\omega t = \pi/2, 3\pi/2: all energy in L
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