Exercises — Understand RL transient behavior

"milli" means $\times 10^{-3}$, so $20\text{ mH}=0.020\text{ H}$. $\tau =\frac{L}{R}=\frac{0.020}{4}=0.005\text{ s}=5\text{ ms}.$ Why $L/R$ and not $RL$? Units force it: $\frac{\text{henry}}{\text{ohm}}=\frac{\text{V⋅s/A}}{\text{V/A}}=\text{seconds}$. A product $RL$ would give ohm·henry, not seconds.

After a long time the current stops changing, so $di/dt=0$, so $v_L=Ldi/dt=0$ — the inductor becomes a plain wire. Only $R$ limits the current: $I_{\text{final}}=\frac{V}{R}=\frac{12}{4}=3\text{ A}.$

The inductor forbids sudden jumps in current, so $i(0^{+})=i(0^{-})=0\text{ A}$. All the source voltage momentarily sits across the inductor.

$\tau =L/R=0.004/2=2\text{ ms}$, so $t=2\text{ ms}=1\tau$. $I_{\text{final}}=V/R=10/2=5\text{ A}$. $i=5\left(1-e^{-1}\right)=5(1-0.3679)=5(0.6321)=3.16\text{ A}.$ On the figure: read the magenta rise curve at $t=1\tau$ — the navy dot sits at $63.2\%$ of the $5\text{ A}$ final, i.e. $3.16\text{ A}$.

Using $v_L=V{e}^{-t/\tau }$ (derived above from $v_L=Ldi/dt$). At $t=1\tau$: $v_L=10e^{-1}=10(0.3679)=3.68\text{ V}.$ Polarity: current is rising ($di/dt>0$), so by the passive sign convention $v_L>0$ — the inductor is absorbing, eating part of the source. KVL check: $v_R=iR=3.16\times 2=6.32\text{ V}$; $v_R+v_L=6.32+3.68=10\text{ V}=V$. ✓ (This is Kirchhoff's Voltage Law closing the loop.)

$\tau =0.005/10=0.5\text{ ms}$, so $t=1\text{ ms}=2\tau$. Discharge is pure exponential $i=I_0{e}^{-t/\tau }$: $i=8e^{-2}=8(0.1353)=1.08\text{ A}.$ On the figure: read the violet decay curve at $t=2\tau$ — the height is $13.5\%$ of the starting $8\text{ A}$. Polarity note: here $di/dt<0$, so $v_L$ flips sign — the inductor now acts as a source pushing the current through $R$, with $|v_L|=iR=1.08\times 10=10.8\text{ V}$.

This is the figure read backward: given a height ($4/5$ of final) on the magenta curve, find the horizontal position. Start from $i=I_{\text{final}}(1-e^{-t/\tau })$ and solve for $\tau$ (this "undoes" the exponential with a logarithm — the tool built in First-Order Differential Equations). $\frac{i}{I_{\text{final}}}=1-e^{-t/\tau }\Rightarrow e^{-t/\tau }=1-\frac{4}{5}=\mathrm{0.2.}$ Take the natural log of both sides: $-\frac{t}{\tau }=\ln (0.2)=-1.609\Rightarrow \tau =\frac{t}{1.609}=\frac{3\text{ ms}}{1.609}=1.86\text{ ms}.$ Why $\ln$? The exponential $e^{-t/\tau }$ hides $\tau$ inside a power; $\ln$ is the exact question "to what power do I raise $e$?" — it pulls $\tau$ out.

This reads the violet decay curve backward: at what horizontal position does the height drop to $0.10$? $i/I_0=e^{-t/\tau }=0.10$. $-\frac{t}{\tau }=\ln (0.10)=-2.303\Rightarrow \frac{t}{\tau }=2.30.$ So about 2.3 τ. (Cross-check with the parent's table: at $2\tau$ we have $13.5\%$, at $3\tau$ we have $5\%$ — $10\%$ sits between, closer to $2\tau$. ✓)

Same $\tau$ and $\tau =L/R$ means $L=\tau R$, so $L_A=0.002(5)=10\text{ mH}$, $L_B=0.002(20)=40\text{ mH}$. Final currents: $I_A=10/5=2\text{ A}$, $I_B=10/20=0.5\text{ A}$. Energies: $E_A=\frac{1}{2}(0.010)(2{)}^2=0.020\text{ J}=20\text{ mJ},$ $E_B=\frac{1}{2}(0.040)(0.5{)}^2=0.005\text{ J}=5\text{ mJ}.$ Circuit A stores more (4×). Same settling speed, very different stored energy — $R$ controls both current and energy here.

(a) $\tau =0.016/8=2\text{ ms}$; $I_{\text{final}}=24/8=3\text{ A}$. (b) $t=4\text{ ms}=2\tau$. $i=3(1-e^{-2})=3(1-0.1353)=3(0.8647)=2.59\text{ A}.$ $v_L=V{e}^{-t/\tau }=24e^{-2}=24(0.1353)=3.25\text{ V}.$ (c) Current is still rising, so $v_L>0$ (absorbing). $v_R=iR=2.59\times 8=20.75\text{ V}$; $v_R+v_L=20.75+3.25=24\text{ V}=V$. ✓

$\tau =0.050/25=2\text{ ms}$; $I_{\text{final}}=10/25=0.4\text{ A}$. Need $i=0.3\text{ A}$: $0.3=0.4(1-e^{-t/\tau })\Rightarrow e^{-t/\tau }=1-0.75=\mathrm{0.25.}$ $-\frac{t}{\tau }=\ln (0.25)=-1.386\Rightarrow t=1.386\tau =1.386(2\text{ ms})=2.77\text{ ms}.$ Story check: the coil can just barely reach $0.4\text{ A}$, and it needs $0.3\text{ A}$ — that's $75\%$ of final, which takes more than one τ. Answer $>1\tau$: consistent. ✓

Now it's a decay: $\tau =2\text{ ms}$ still, $i=I_0{e}^{-t/\tau }$. $0.1=0.4e^{-t/\tau }\Rightarrow e^{-t/\tau }=0.25\Rightarrow t=1.386\tau =2.77\text{ ms}.$ Same $75\%$ drop → same $1.386\tau$. The symmetry between the $0.3\text{ A}$ rise target and the $0.1\text{ A}$ decay target is not a coincidence — both are the same $25\%$ survival fraction. Polarity: during this decay the coil's voltage flips and it becomes the source keeping current alive.

Fix R from the final current: $R=V/I_{\text{final}}=10/2=5\mathbf{\Omega }$. Reaching $95\%$ takes $3\tau$ (from the parent's table: $95.0\%$ at $3\tau$). We need $3\tau \le 3\text{ ms}$, so $\tau \le 1\text{ ms}$. $\tau =L/R\le 0.001\Rightarrow L\le 0.001\times 5=5\text{ mH}$. Choose $R=5\Omega$, $L=5\text{ mH}$ (gives exactly $\tau =1\text{ ms}$, hits $95\%$ at exactly $3\text{ ms}$). A smaller $L$ would settle faster — also acceptable. Why fix R first? $R$ alone sets the steady value (Steady-State Analysis: inductor is a short at DC); then $L$ is free to tune the speed.

$\tau =L/R\to \infty$, so on any finite time scale the exponential barely moves. But the shape is more revealing than the plateau: near $t=0$ the current is $i\approx \frac{V}{R}\cdot \frac{t}{\tau }=\frac{V}{R}\cdot \frac{Rt}{L}=\frac{V}{L}t$. Notice $R$ cancels — with no resistance the current is a straight ramp $i(t)=\frac{V}{L}t$, climbing forever at constant slope $V/L$ and never settling. Physically: a pure inductor across a battery has nothing to dissipate energy, so current builds without limit (a real coil's tiny resistance eventually caps it). This is the degenerate floor of $\tau =L/R$: remove $R$ and there is no time constant at all — the "settling" never happens.

$\tau =L/R\to 0$: the transient becomes infinitely fast. The final current $I_{\text{final}}=V/R\to 0$ — a huge resistance chokes the current down to nothing, essentially an open circuit. So the inductor's "inertia" becomes irrelevant: with almost no current ever flowing, there is nothing for $L$ to oppose. This is the opposite extreme from L5.2 — here $R$ dominates completely and $L$'s role vanishes. Together L5.2 and L5.3 bracket every real RL circuit: $R$ small → slow ramp dominated by $L$; $R$ large → instant, tiny response dominated by $R$.

Differentiate the rise formula. With $i=\frac{V}{R}(1-e^{-t/\tau })$: $\frac{di}{dt}=\frac{V}{R}\cdot \frac{1}{\tau }{e}^{-t/\tau }=\frac{V}{R}\cdot \frac{R}{L}{e}^{-t/\tau }=\frac{V}{L}{e}^{-t/\tau }.$ At $t=0$: $\frac{di}{dt}{|}_0=\frac{V}{L}$ — no $R$ appears. (Sanity: at $t=0$ the inductor takes the full $V$, and $v_L=Ldi/dt=V$ gives $di/dt=V/L$ directly.) Tangent-line argument: starting at $i=0$ with slope $V/L$, a straight line reaches height $I_{\text{final}}=V/R$ after time $t=\frac{I_{\text{final}}}{\text{slope}}=\frac{V/R}{V/L}=\frac{L}{R}=\tau .$ So the initial tangent crosses the final value exactly at one τ — a clean geometric meaning for the time constant, shown by the dashed line in the figure below.