Lecture Notes on Inductance, RL, LC, and RLC Circuits

ILOS for Lecture 17

• Define self-inductance and its role in opposing changes in current, and identify the factors that affect its magnitude.

• Apply Faraday's Law to analyze RL circuits and determine the inductive time constant.

• Apply Faraday's Law to analyze LC circuits and compare their behavior with harmonic oscillators.

• Compute the energy stored in an inductor and derive expressions for magnetic energy density.

Inductance

• Self-Inductance and Inductor: Any circuit carrying a varying current has an induced EMF.

  • This effect is enhanced if the circuit includes a coil with $N$ turns of wire.

  • $\varepsilon = -N \frac{d\Phi_B}{dt} = -L \frac{dI}{dt}$

  • $L = \frac{N \Phi_B}{I}$ (self-inductance).

  • $L$ (inductance) measures the inductor's tendency to oppose any variations in current.

  • Inductance depends only on the geometry of the coil and the magnetic material.

  • $L \propto \frac{N^2}{l}$

• Units and Values

  • SI unit: Weber per Ampere (Wb/A) = Henry (H)

  • Typical values: μH ~ mH

  • Symbol: -₋₋₋-Mn-

• Example 30.3: Self-inductance of a toroidal solenoid.

  • Given: Cross-sectional area $A$, mean radius $r$, $N$ turns, nonmagnetic core.

  • Assume $B$ is uniform across the cross-section.

  • $N = 200$, $A = 5.0 cm^2$, $r = 0.10 m$

  • $L = \frac{N \Phi<em>B}{I} = \frac{N B A}{I} = \frac{\mu</em>0 N^2 A}{2 \pi r}$

  • B field inside $B = \frac{\mu_0Ni}{2 \pi r}$

  • $\Phi<em>B = BA = \frac{\mu</em>0 N i A}{2 \pi r}$

  • $L = \frac{\mu_0 N^2 A}{2 \pi r} = \frac{(4\pi \times 10^{-7}) \times 200^2 \times 5.0 \times 10^{-4}}{2 \pi \times 0.10} = 4 \times 10^{-5} H = 40 \mu H$

• Example 30.4: Induced EMF in the toroidal solenoid.

  • Current increases uniformly from 0 to 6.0 A in 3.0 ms.

  • $\varepsilon = -L \frac{dI}{dt} = -L \frac{\Delta I}{\Delta t} = -40 \times 10^{-6} \times \frac{6.0 - 0}{3.0 \times 10^{-3}} = -80 V$

  • Magnitude: $|\varepsilon| = 80 V$

  • Direction: Counterclockwise

• Example: Solenoid

  • $N$ turns, cross-sectional area $A$, length $l$

  • $B = \mu_0 n I$, where $n = \frac{N}{l}$

  • $L = \frac{N \Phi<em>B}{I} = \frac{N B A}{I} = \mu</em>0 n^2 A l$

  • Because increase uniformly instantaenous rate = avg rate.

RL Circuits

• Current Growth: Kirchhoff's loop rule no longer holds.

• Applying Faraday's Law:

  • $\varepsilon - iR - L \frac{di}{dt} = 0$

  • $\varepsilon = iR + L \frac{di}{dt}$

  • Solving the differential equation:

  • $I(t) = \frac{\varepsilon}{R} (1 - e^{-t/\tau_L})$

• Initial conditions:

  • At $t = 0$, $I(0) = 0$

  • For RL circuit: $I(t) = \frac{\varepsilon}{R} (1 - e^{-t/\tau_L})$

  • Inductive Time Constant: $\tau_L = \frac{L}{R}$

    • Units check: H/Ω = (V⋅s/A) / (V/A) = s

  • At $t = 0$, the inductor acts like a broken wire, opposing rapid changes in current.

  • As $t → ∞$, $I(t) → \frac{\varepsilon}{R}$, and the inductor acts like an ordinary connecting wire.

• An inductor in a circuit opposes rapid changes in current.

  • $\frac{di}{dt} = \frac{\varepsilon}{L} - \frac{R}{L} i$

  • At $t = 0$, $\frac{di}{dt} = \frac{\varepsilon}{L}$

  • Steady state reached when $t → ∞$, $\frac{di}{dt} → 0$.

• Example 30.6: Sensitive electronic device connected to a source with an inductor in series.

  • Given: $R = 175 \Omega$, desired current $I = 36 mA$, maximum initial current rise $4.9 mA$ in $58 ms$.

  • (a) Required source EMF $\varepsilon$:

  • $\varepsilon = IR = 36 \times 10^{-3} \times 175 = 6.3 V$

  • (b) Required inductance $L$:

  • $I = \frac{\varepsilon}{R} (1 - e^{-t/\tau_L})$

  • $L = - \frac{tR}{\ln(1 - IR/\varepsilon)} = - \frac{58 \times 10^{-3} \times 175}{\ln(1 - \frac{175 \times 4.9 \times 10^{-3}}{6.3})} = 0.069 H$

  • (c) RL time constant $\tau_L$:

  • $\tau_L = \frac{L}{R} = \frac{0.069}{175} = 3.94 \times 10^{-4} s = 394 \mu s$

Energy Stored in an Inductor

• From the loop rule: $\varepsilon - iR - L \frac{di}{dt} = 0$

• Multiply by $i$:

  • $\varepsilon i = i^2 R + Li \frac{di}{dt}$

  • $P<em>{\varepsilon} = P</em>{\text{thermal}} + \frac{dU_B}{dt}$ (rate of energy stored in the inductor)

    • rate of energy stored in the inductor$\frac{dU_B}{dt} = L i \frac{di}{dt}$

• Integrating to find the total energy stored:

  • $U<em>B = \int</em>0^I L i \, di = \frac{1}{2} L I^2$

• Magnetic Energy in Inductor: $U<em>B = \frac{1}{2} L I^2$ (analogous to $U</em>E = \frac{1}{2} C V^2$)

• Example: Energy in a solenoid

  • $n = \frac{N}{l}$, $B = \mu_0 n I$

  • $U<em>B = \frac{1}{2} L I^2 = \frac{1}{2} \mu</em>0 n^2 l A I^2$

  • $\frac{U<em>B}{Al} = \frac{1}{2} \mu</em>0 n^2 I^2 = \frac{B^2}{2 \mu_0}$

  • Magnetic Energy Density: $u<em>B = \frac{B^2}{2 \mu</em>0}$

  • Electric Energy Density: $u<em>E = \frac{1}{2} \epsilon</em>0 E^2$

  • $U<em>B = \int u</em>B dV$, $U<em>E = \int u</em>E dV = \int \frac{1}{2} \epsilon_0 E^2 dV$

• Example 30.5: Storing electrical energy in a large inductor.

  • Given: Store $1.00 kWh$ of energy with a $200 A$ current.

  • $U_B = \frac{1}{2} L I^2$

  • $L = \frac{2 U_B}{I^2} = \frac{2 \times 1.00 \times 1000 \times 3600}{200^2} = 180 H$

  • A 180 H inductor using conventional wire would be very large (room-size) and have significant energy loss due to resistance ($P_{th} = I^2 R$).

  • $1 kWh = 1000 \times 3600 J$ (really big)

Current Decay

• Discharging a capacitor

  • RL Circuit: Switch is flipped to remove the EMF source.

  • Applying Kirchhoff's loop rule:

    • Loop constant is 0. Initially the value of battery (V) had a constannt value. However, after the current is off, then the value of battery should turn into $0$

  • $L \frac{di}{dt} + iR = 0$

• Initial condition:

  • $I(0) = I_0$

  • $i(t) = I<em>0 e^{-t/\tau</em>L}$

    • Eventually Current Tois will be 0.

• Example 30.7: Energy dissipation in a decaying R-L circuit.

  • What fraction of the original energy is dissipated after $2.3 \tau_L$?

  • $i(t) = I<em>0 e^{-t/\tau</em>L}$

  • $U<em>B(t) = \frac{1}{2} L i(t)^2 = \frac{1}{2} L I</em>0^2 e^{-2t/\tau_L}$

• $U<em>B(2.3 \tau</em>L) = \frac{1}{2} L I<em>0^2 e^{-2(2.3)} = \frac{1}{2} L I</em>0^2 e^{-4.6} = 0.01 \frac{1}{2} L I_0^2$

• $U<em>B (2.3 \tau</em>L) = \frac{1}{2} L I_0^2 e^{-4.6}$

  • Fraction remaining: $e^{-4.6} = 0.01$

  • Therefore, 99% of the energy has been dissipated.

LC Circuits (LC Oscillations)

• Method 1: Faraday's Law

  • $L \frac{di}{dt} + \frac{q}{C} = 0$

  • $\frac{d^2 q}{dt^2} + \frac{1}{LC} q = 0$

• Method 2: Conservation of Energy

  • $U<em>{tot} = U</em>B + U_E = \frac{1}{2} L I^2 + \frac{q^2}{2C} = \text{constant}$

  • $\frac{dU_{tot}}{dt} = 0$

  • $L I \frac{dI}{dt} + \frac{q}{C} \frac{dq}{dt} = 0$ (since $I = \frac{dq}{dt}$)

  • $\frac{d^2 q}{dt^2} + \frac{1}{LC} q = 0$

• Differential equation for simple harmonic motion:
  * block-spring system with the angular frequency being $w= \sqrt{\frac{k}{m}}$.

  • Solution: $q(t) = Q \cos(\omega t + \phi)$

    • $Q$: Maximum charge stored in the capacitor.

    • $\omega$: Angular frequency of the electromagnetic oscillation, $\omega = \frac{1}{\sqrt{LC}}$ (check units).

    • $\phi$: Phase constant.

    • Checking the SI unit of $w$

      • $\sqrt{\frac{1}{LC}} = \sqrt{\frac{1}{H \times F}} = \sqrt{\frac{A}{Vs} \frac{V}{C}} = \sqrt{\frac{A}{s \times A \times sV}} = \sqrt{\frac{C V}{s^2 VC}} = \frac{1}{s}$

        • I = A = C/s

• Current:

  • $I(t) = \frac{dq}{dt} = -\omega Q \sin(\omega t + \phi)$

  • Amplitude of current: $\omega Q$

  • $\begin{aligned}<br>&amp;t=0 \quad \phi = \frac{\pi}{2} \quad I=0<br>\&amp; t=\frac{T}{4} \quad \phi =0 \quad I=\pm wQ<br>\end{aligned}$

• Energy in the Capacitor and Inductor:

  • $U_E(t) = \frac{q^2}{2C} = \frac{Q^2}{2C} \cos^2(\omega t + \phi)$

  • $U_B(t) = \frac{1}{2} L I^2 = \frac{1}{2} L \omega^2 Q^2 \sin^2(\omega t + \phi) = \frac{Q^2}{2C} \sin^2(\omega t + \phi)$

• Total Electromagnetic Energy:

  • $U<em>{tot} = U</em>B(t) + U<em>E(t) = \frac{Q^2}{2C} = \frac{1}{2} L I</em>{max}^2 = \text{constant}$

  • Maximum $U<em>E$ = Maximum $U</em>B$

• Example 30.8: LC circuit with a charged capacitor and an inductor.

  • Given: $C = 25 \mu F$, $L = 10 mH$, $V = 300 V$

  • (a) Frequency and period of oscillation:

  • $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{1.0 \times 10^{-3} \times 25 \times 10^{-6}}} = 2000 \text{ rad/s}$

  • $f = \frac{\omega}{2 \pi} = \frac{2000}{2 \pi} = 318 Hz$

  • $T = \frac{1}{f} = \frac{1}{318} = 3.1 \times 10^{-3} s = 3.1 ms$

  • (b) Capacitor charge and circuit current at $t = 1.2 ms$:

  • $Q = CV = 25 \times 10^{-6} \times 300 = 7.5 \times 10^{-3} C$

  • $q(t) = Q \cos(\omega t) = 7.5 \times 10^{-3} \cos(2000 \times 1.2 \times 10^{-3}) = -5.5 \times 10^{-3} C$

  • $I(t) = -\omega Q \sin(\omega t) = -2000 \times 7.5 \times 10^{-3} \sin(2000 \times 1.2 \times 10^{-3}) = -10 A$
    $1.2 m s = 0.39 T$ ( betwen $\frac{T}{4}$ and $\frac{T}{2}$ discharging )

• Example 30.9: Energy in the LC circuit.

  • Given: $C = 25 \mu F$, $L = 10 mH$, $V = 300 V$

  • (a) At $t = 0$:

  • $U_B = 0$

  • $U_E = \frac{Q^2}{2C} = \frac{(7.5 \times 10^{-3})^2}{2 \times 25 \times 10^{-6}} = 1.1 J$

  • (b) At $t = 1.2 ms$:

  • $U_B = \frac{1}{2} L I^2 = \frac{1}{2} \times 10 \times 10^{-3} \times (-10)^2 = 0.5 J$

  • $U_E = \frac{q^2}{2C} = \frac{(-5.5 \times 10^{-3})^2}{2 \times 25 \times 10^{-6}} = 0.6 J$

ILOS for Lecture 18

• Apply Faraday's Law to analyze RLC circuits and compare their behavior with damped harmonic oscillators.

• Calculate mutual inductance based on its definition, and apply the reciprocity theorem when necessary.

• Apply Faraday's Law to analyze forced RLC circuits and compare their behavior with forced harmonic oscillators.
  \begin{array}{llll}\\\text { inductor } & U{B}=L I & +q(t) & \frac{d}{L} I {L} & U{p}=i X \\text { m } w{0}= & \frac{I} & I & d i \ \hline & L & I[T + & & + E=0 \\\\\ q(t)=Q{m} \cos (w+) & \\\\text { r } \text { circuit } & 1 & \
  i(t) m=-Q{m} w \sin (w t) & R & - & \frac{q}{D} \end{array} \begin{array}{lc}4 & -Q{m} w \sin (w t)\2 &(t)=I_{0} e \int f(t)=Q \cos (w+ & b) \\end{array}

RLC Circuit - Damped Oscillation

• Method 1: Faraday's Law
  * Loop constant is 0. Initially $\frac{d E}{d t}=L \frac{d i}{d t}$ $L \frac{d^{2} i}{d t}+R i+ \frac{9}{0}$

• Similar to damped mechanical oscillator.

  • $\sum F<em>x = m a</em>x = m \frac{d^2 x}{dt^2} + b \frac{dx}{dt} + kx = 0$

  • Correspondence:

  • $m \leftrightarrow L$

  • $b \leftrightarrow R$

  • $k \leftrightarrow \frac{1}{C}$

  • $\omega \leftrightarrow \omega$

Method 2

• Conservation of Energy

  • $U<em>{tot} = U</em>B + U_E = \frac{1}{2} L I^2 + \frac{q^2}{2C}$

  • $\frac{dU_{tot}}{dt} = \frac{d}{dt}(\frac{1}{2} L I^2 + \frac{q^2}{2C})$

• $\frac{d^{2} q}{d t^{2}}+ \frac{R}{L} \frac{dq}{dt}+ \frac{\partial}{L o} =0$

• Damped Oscillation and solution:
  $q (t) Q e^{-}$ \cos ( w't + ) Damped EM Oscillation

• ${\text {Angular Frequency }} W' = \sqrt{\omega_{0} - \frac{R^{2}}{2 L} } = \sqrt{\omega - {R2}}{2}$

• Damped
  $$f(t) = Ae **