196 lines
8.7 KiB
TeX
196 lines
8.7 KiB
TeX
\subsection{LR Circuits and Transients}
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This subsection introduces the series LR circuit, derives the time-dependent current during both the growth and decay phases, and discusses the energy stored in the inductor's magnetic field. The inductor opposes changes in current through a self-induced back EMF, producing an exponential transient with characteristic time constant $\tau = L/R$.
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\dfn{LR circuit and self-induced EMF}{Consider a series circuit consisting of a battery with constant EMF $\mathcal{E}$, a resistor of resistance $R$, an inductor of inductance $L$, and a switch, all connected in a single closed loop. When current $I$ flows through the inductor, any change in current induces a back EMF across the inductor given by
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\[
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\mathcal{E}_L = -L\,\frac{dI}{dt}.
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\]
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By Kirchhoff's loop rule, the sum of potential differences around the loop is zero. Traversing the loop in the direction of the current:
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\[
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\mathcal{E} - IR - L\,\frac{dI}{dt} = 0.
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\]
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This first-order differential equation governs the time evolution of the current $I(t)$. The inductor's back EMF opposes any change in current, analogous to how inertia opposes changes in velocity in mechanics.}
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\nt{The inductor acts as a ``magnetic inertia'' element: just as a mass cannot change its velocity instantaneously, current through an inductor cannot change instantaneously. At the instant the switch is closed, the current is zero and the full battery EMF appears across the inductor. After a long time, the current reaches a steady value and the inductor behaves as a short circuit (zero voltage drop).}
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\mprop{Time constant of an LR circuit}{The \emph{time constant} $\tau$ characterizes how quickly the current changes in an LR circuit:
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\[
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\tau = \frac{L}{R}.
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\]
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The SI unit of inductance is the henry (H) and the SI unit of resistance is the ohm ($\Omega$). Since $1\,\mathrm{H} = 1\,\mathrm{V\!\cdot\!s/A}$ and $1\,\Omega = 1\,\mathrm{V/A}$, the ratio $L/R$ has units of seconds, confirming that $\tau$ is a characteristic time. At $t = \tau$, the current during growth has reached $(1 - e^{-1}) \approx 63.2\%$ of its maximum value.}
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\thm{Current growth in a series LR circuit}{Consider a series LR circuit with battery EMF $\mathcal{E}$, resistance $R$, and inductance $L$. The switch is closed at $t = 0$, with the initial current $I(0) = 0$. The current as a function of time is
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\[
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I(t) = \frac{\mathcal{E}}{R}\,\bigl(1 - e^{-t/\tau}\bigr),
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\]
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where $\tau = L/R$. The maximum (steady-state) current is
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\[
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I_{\text{max}} = \frac{\mathcal{E}}{R}.
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\]}
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\pf{Current growth derivation}{We solve the differential equation obtained from Kirchhoff's loop rule:
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\[
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\mathcal{E} - IR - L\,\frac{dI}{dt} = 0.
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\]
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Rearrange to isolate the time derivative:
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\[
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L\,\frac{dI}{dt} = \mathcal{E} - IR.
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\]
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Separate variables, bringing all $I$ terms to one side:
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\[
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\frac{dI}{\mathcal{E} - IR} = \frac{dt}{L}.
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\]
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Integrate both sides. On the left, substitute $u = \mathcal{E} - IR$, so $du = -R\,dI$:
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\[
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\int_{0}^{I(t)} \frac{dI}{\mathcal{E} - IR} = \int_{0}^{t} \frac{dt}{L}.
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\]
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The left-hand integral is
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\[
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-\frac{1}{R}\,\ln\Bigl(\mathcal{E} - IR\Bigr)\,\bigg|_{0}^{\,I(t)}
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= -\frac{1}{R}\,\ln\!\left(\frac{\mathcal{E} - IR(t)}{\mathcal{E}}\right).
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\]
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The right-hand integral is $t/L$. Thus,
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\[
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-\frac{1}{R}\,\ln\!\left(\frac{\mathcal{E} - IR(t)}{\mathcal{E}}\right) = \frac{t}{L}.
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\]
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Multiply by $-R$:
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\[
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\ln\!\left(\frac{\mathcal{E} - IR(t)}{\mathcal{E}}\right) = -\frac{R}{L}\,t.
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\]
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Exponentiate both sides:
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\[
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\frac{\mathcal{E} - IR(t)}{\mathcal{E}} = e^{-Rt/L}.
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\]
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Solve for $I(t)$:
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\[
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I(t) = \frac{\mathcal{E}}{R}\,\bigl(1 - e^{-Rt/L}\bigr).
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\]
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Since $\tau = L/R$, this is equivalent to $I(t) = (\mathcal{E}/R)\,(1 - e^{-t/\tau})$.
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In the limit $t \to \infty$, the exponential vanishes and $I \to \mathcal{E}/R$. \Qed}
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\thm{Current decay in a series LR circuit}{Consider a series LR circuit carrying an initial current $I_0$ (established in steady state with a battery that is then disconnected, leaving only $R$ and $L$ in a closed loop). With $I(0) = I_0$, the current as a function of time is
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\[
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I(t) = I_0\,e^{-t/\tau},
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\]
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where $\tau = L/R$.}
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\pf{Current decay derivation}{With the battery removed, Kirchhoff's loop rule gives
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\[
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IR + L\,\frac{dI}{dt} = 0.
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\]
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Separate variables:
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\[
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\frac{dI}{I} = -\frac{R}{L}\,dt.
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\]
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Integrate both sides from $t = 0$ to $t$:
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\[
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\int_{I_0}^{I(t)} \frac{dI}{I} = -\frac{R}{L}\,\int_{0}^{t} dt.
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\]
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The left-hand side gives $\ln(I/I_0)$ and the right-hand side gives $-(R/L)\,t$. Thus,
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\[
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\ln\!\left(\frac{I(t)}{I_0}\right) = -\frac{R}{L}\,t,
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\]
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and exponentiating,
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\[
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I(t) = I_0\,e^{-Rt/L} = I_0\,e^{-t/\tau}.
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\]
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In the limit $t \to \infty$, $I \to 0$. \Qed}
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\cor{Back EMF across the inductor (growth phase)}{Differentiating the growth current gives the self-induced EMF across the inductor:
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\[
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\mathcal{E}_L = -L\,\frac{dI}{dt}
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= -L\,\frac{\mathcal{E}}{R}\,\frac{R}{L}\,e^{-t/\tau}
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= -\mathcal{E}\,e^{-t/\tau}.
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\]
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At $t = 0$, $\mathcal{E}_L = -\mathcal{E}$ (the full battery EMF opposes the change). As $t \to \infty$, $\mathcal{E}_L \to 0$ (the inductor becomes a short circuit). The magnitude of the back EMF decays exponentially with the same time constant $\tau$.}
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\cor{Voltage across the resistor (growth phase)}{The voltage across the resistor is
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\[
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V_R = IR = \mathcal{E}\,\bigl(1 - e^{-t/\tau}\bigr).
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\]
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At $t = 0$, $V_R = 0$. As $t \to \infty$, $V_R \to \mathcal{E}$, so the full battery EMF appears across the resistor in steady state.}
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\mprop{Energy stored in an inductor}{When a current $I$ flows through an inductor of inductance $L$, energy is stored in the magnetic field:
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\[
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U_B = \frac{1}{2}\,L\,I^{2}.
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\]
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The rate at which energy is stored is
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\[
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\frac{dU_B}{dt} = L\,I\,\frac{dI}{dt}.
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\]
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The SI unit of energy is the joule (J). During current growth in an LR circuit, the battery supplies energy at rate $\mathcal{E}I$, part of which is dissipated as Joule heating $I^2R$ in the resistor and part is stored in the inductor's magnetic field.}
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\nt{At $t = \tau = L/R$, the current reaches $I(\tau) = (\mathcal{E}/R)(1 - e^{-1}) \approx 0.632\,(\mathcal{E}/R)$. The inductor EMF has dropped to $e^{-1} \approx 36.8\%$ of its initial value. After $5\tau$, the current is within $1\%$ of its steady-state value and the circuit is effectively in steady state.}
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\ex{Illustrative example}{A series LR circuit with $\mathcal{E} = 12\,\mathrm{V}$, $R = 6.0\,\Omega$, and $L = 3.0\,\mathrm{H}$ has time constant $\tau = L/R = 0.50\,\mathrm{s}$. The maximum current is $I_{\text{max}} = \mathcal{E}/R = 2.0\,\mathrm{A}$.
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\begin{enumerate}[label=(\arabic*)]
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\item At $t = 0$, $I = 0$ and $\mathcal{E}_L = -12\,\mathrm{V}$.
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\item At $t = \tau = 0.50\,\mathrm{s}$, $I = 2.0(1 - e^{-1}) = 1.26\,\mathrm{A}$ and $\mathcal{E}_L = -12\,e^{-1} = -4.41\,\mathrm{V}$.
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\item At $t = 2\tau = 1.0\,\mathrm{s}$, $I = 2.0(1 - e^{-2}) = 1.73\,\mathrm{A}$ and $\mathcal{E}_L = -12\,e^{-2} = -1.62\,\mathrm{V}$.
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\item At $t \to \infty$, $I = 2.0\,\mathrm{A}$ and $\mathcal{E}_L = 0\,\mathrm{V}$.
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\end{enumerate}}
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\qs{Worked example}{A series LR circuit consists of a battery with emf $\mathcal{E} = 24\,\mathrm{V}$, a resistor with resistance $R = 4.0\,\Omega$, and an inductor with inductance $L = 1.0\,\mathrm{H}$, all connected in series with a switch. The switch is closed at time $t = 0$.
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\begin{enumerate}[label=(\alph*)]
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\item Find the time constant $\tau$ of the circuit.
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\item Find the current at $t = 0.25\,\mathrm{s}$.
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\item Find the maximum (steady-state) current.
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\item Find the energy stored in the inductor at $t = 0.25\,\mathrm{s}$.
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\end{enumerate}}
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\sol
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\noindent\textbf{Part (a):} The time constant of a series LR circuit is
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\[
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\tau = \frac{L}{R}.
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\]
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Substituting the given values:
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\[
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\tau = \frac{1.0\,\mathrm{H}}{4.0\,\Omega} = 0.25\,\mathrm{s}.
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\]
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\noindent\textbf{Part (b):} During the growth phase, the current is
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\[
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I(t) = \frac{\mathcal{E}}{R}\,\bigl(1 - e^{-t/\tau}\bigr).
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\]
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At $t = 0.25\,\mathrm{s}$ and with $\tau = 0.25\,\mathrm{s}$:
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\[
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I(0.25) = \frac{24\,\mathrm{V}}{4.0\,\Omega}\,\Bigl(1 - e^{-0.25/0.25}\Bigr)
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= 6.0\,\mathrm{A}\,\bigl(1 - e^{-1}\bigr).
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\]
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Using $e^{-1} \approx 0.3679$:
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\[
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I(0.25) = 6.0\,\mathrm{A} \times (1 - 0.3679)
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= 6.0\,\mathrm{A} \times 0.6321
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= 3.79\,\mathrm{A}.
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\]
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\noindent\textbf{Part (c):} The maximum (steady-state) current is obtained as $t \to \infty$, when the exponential term vanishes:
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\[
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I_{\text{max}} = \frac{\mathcal{E}}{R} = \frac{24\,\mathrm{V}}{4.0\,\Omega} = 6.0\,\mathrm{A}.
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\]
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\noindent\textbf{Part (d):} The energy stored in the inductor's magnetic field is
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\[
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U_B = \frac{1}{2}\,L\,I^{2}.
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\]
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Using the current from part (b):
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\[
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U_B = \frac{1}{2}\,(1.0\,\mathrm{H})\,(3.79\,\mathrm{A})^{2}
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= 0.5 \times 14.36\,\mathrm{J}
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= 7.2\,\mathrm{J}.
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\]
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\noindent Therefore,
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\[
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\tau = 0.25\,\mathrm{s},
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\qquad
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I(0.25\,\mathrm{s}) = 3.79\,\mathrm{A},
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\qquad
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I_{\text{max}} = 6.0\,\mathrm{A},
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\qquad
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U_B(0.25\,\mathrm{s}) = 7.2\,\mathrm{J}.
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\]
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