Inductance and RL Circuits

Self-inductance, mutual inductance, and RL circuit dynamics

Inductance and RL Circuits

Self-Inductance

Changing current in coil induces EMF in same coil:

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

where $L$ is inductance (or self-inductance).

Units: 1 henry (H) = 1 Wb/A = 1 V·s/A

Inductance of Solenoid

Long solenoid: $n$ turns/length, cross-sectional area $A$ , length $l$ :

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

Mutual Inductance

Current $I_1$ in coil 1 creates flux through coil 2:

$\mathcal{E}_2 = -M\frac{dI_1}{dt}$

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

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

Energy Stored in Inductor

$U_L = \frac{1}{2}LI^2$

Energy density in magnetic field: $u_B = \frac{B^2}{2\mu_0}$

For solenoid: $U = u_B \cdot \text{volume} = \frac{B^2}{2\mu_0}Al = \frac{1}{2}LI^2$

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

RL Circuit: Current Growth

Circuit: battery ( $\mathcal{E}$ ), resistor ( $R$ ), inductor ( $L$ ) in series.

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

Differential equation: $\frac{dI}{dt} = \frac{\mathcal{E}}{L} - \frac{R}{L}I$

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

Time constant: $\tau_L = \frac{L}{R}$

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

RL Circuit: Current Decay

Remove battery, current decays:

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

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

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

Energy Considerations

Current growth:

Energy from battery: $W = \int_0^\infty \mathcal{E}I \, dt = \frac{\mathcal{E}^2L}{R^2}$

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

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

(Equal amounts stored and dissipated)

LC Circuit

Inductor and capacitor (no resistance):

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

Oscillation: $Q(t) = Q_0\cos(\omega t + \phi)$

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

Energy oscillates between:

Electric: $U_E = Q^2/(2C)$
Magnetic: $U_B = LI^2/2$
Total: $U = U_E + U_B$ = constant

LRC Circuit

With resistance, oscillations damped:

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

Damped oscillation (for $R < 2\sqrt{L/C}$ ): $Q(t) = Q_0e^{-Rt/(2L)}\cos(\omega' t)$

where $\omega' = \sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}$

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

📚 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:

$\tau = \frac{L}{R} = \frac{0.4}{8.0}$

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

(b) Current at t = 0.08 s:

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

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

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

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

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

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

(c) Rate of energy storage:

Energy in inductor: $U_L = \frac{1}{2}LI^2$

$\frac{dU_L}{dt} = LI\frac{dI}{dt}$

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

$\frac{dI}{dt} = \frac{\varepsilon - IR}{L}$

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

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

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

Alternatively: $P_L = \varepsilon I - I^2 R = VI_L$ where $V_L = L\frac{dI}{dt}$

2Problem 2medium

❓ Question:

💡 Show Solution

Given:

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

(a) Time constant:

$\tau = \frac{L}{R} = \frac{0.4}{8.0}$

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

(b) Current at t = 0.08 s:

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

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

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

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

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

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

(c) Rate of energy storage:

Energy in inductor: $U_L = \frac{1}{2}LI^2$

$\frac{dU_L}{dt} = LI\frac{dI}{dt}$

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

$\frac{dI}{dt} = \frac{\varepsilon - IR}{L}$

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

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

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

Alternatively: $P_L = \varepsilon I - I^2 R = VI_L$ where $V_L = L\frac{dI}{dt}$

3Problem 3hard

❓ 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: $\varepsilon = 0$ , so: $-L\frac{dI}{dt} = IR$

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

Integrating: $\ln I = -\frac{R}{L}t + C$

At t = 0: I = I₀, so C = ln I₀

$\ln\left(\frac{I}{I_0}\right) = -\frac{R}{L}t$

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

(b) Time for I = 0.1I₀:

$0.1I_0 = I_0 e^{-Rt/L}$

$0.1 = e^{-Rt/L}$

$\ln(0.1) = -\frac{Rt}{L}$

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

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

This is $2.3\tau$ where $\tau = L/R = 0.025$ s

(c) Total energy dissipated:

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

$U_0 = 0.5 \text{ J}$

As t → ∞, all energy is dissipated: $E_{dissipated} = U_0 = \boxed{0.5 \text{ J}}$

Check: $\int_0^\infty I^2 R \, dt = \int_0^\infty I_0^2 e^{-2Rt/L} R \, dt = \frac{I_0^2 R L}{2R} = \frac{1}{2}LI_0^2$ ✓

4Problem 4hard

❓ Question:

💡 Show Solution

Given:

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

(a) Current vs. time:

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

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

Integrating: $\ln I = -\frac{R}{L}t + C$

At t = 0: I = I₀, so C = ln I₀

$\ln\left(\frac{I}{I_0}\right) = -\frac{R}{L}t$

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

(b) Time for I = 0.1I₀:

$0.1I_0 = I_0 e^{-Rt/L}$

$0.1 = e^{-Rt/L}$

$\ln(0.1) = -\frac{Rt}{L}$

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

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

This is $2.3\tau$ where $\tau = L/R = 0.025$ s

(c) Total energy dissipated:

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

$U_0 = 0.5 \text{ J}$

As t → ∞, all energy is dissipated: $E_{dissipated} = U_0 = \boxed{0.5 \text{ J}}$

Check: $\int_0^\infty I^2 R \, dt = \int_0^\infty I_0^2 e^{-2Rt/L} R \, dt = \frac{I_0^2 R L}{2R} = \frac{1}{2}LI_0^2$ ✓

5Problem 5hard

❓ 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: $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(0.1)(5.0 \times 10^{-5})}}$

$\omega = \frac{1}{\sqrt{5.0 \times 10^{-6}}} = \frac{1}{2.24 \times 10^{-3}}$

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

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

(b) Maximum current:

Energy conservation: $U_C = U_L$

$\frac{Q_0^2}{2C} = \frac{1}{2}LI_{max}^2$

$I_{max} = \frac{Q_0}{\sqrt{LC}} = Q_0 \omega$

$I_{max} = (1.0 \times 10^{-4})(447)$

$I_{max} = \boxed{0.0447 \text{ A} = 44.7 \text{ mA}}$

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

General solution: $Q(t) = Q_0 \cos(\omega t + \phi)$

At t = 0: Q = Q₀, so φ = 0

$\boxed{Q(t) = Q_0 \cos(\omega t) = (1.0 \times 10^{-4})\cos(447t) \text{ C}}$

Current: $I(t) = -\frac{dQ}{dt} = Q_0 \omega \sin(\omega t)$

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

Energy oscillates between capacitor and inductor:

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

6Problem 6hard

❓ Question:

💡 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: $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{(0.1)(5.0 \times 10^{-5})}}$

$\omega = \frac{1}{\sqrt{5.0 \times 10^{-6}}} = \frac{1}{2.24 \times 10^{-3}}$

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

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

(b) Maximum current:

Energy conservation: $U_C = U_L$

$\frac{Q_0^2}{2C} = \frac{1}{2}LI_{max}^2$

$I_{max} = \frac{Q_0}{\sqrt{LC}} = Q_0 \omega$

$I_{max} = (1.0 \times 10^{-4})(447)$

$I_{max} = \boxed{0.0447 \text{ A} = 44.7 \text{ mA}}$

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

General solution: $Q(t) = Q_0 \cos(\omega t + \phi)$

At t = 0: Q = Q₀, so φ = 0

$\boxed{Q(t) = Q_0 \cos(\omega t) = (1.0 \times 10^{-4})\cos(447t) \text{ C}}$

Current: $I(t) = -\frac{dQ}{dt} = Q_0 \omega \sin(\omega t)$

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

Energy oscillates between capacitor and inductor:

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

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