Exercises — RL circuit — growth and decay of current

WHAT: Read off $\tau =L/R$. WHY: $\tau$ is the only number that sets the timescale — everything else scales against it. $\tau =\frac{L}{R}=\frac{6}{3}=2\text{s}.$ Units check: $\frac{\text{H}}{\Omega }=\frac{\text{V⋅s/A}}{\text{V/A}}=\text{s}$ ✓. So $\tau =2\text{s}$.

WHAT: After a long time the inductor acts like a plain wire, so only $R$ limits the current. WHY: As $t\to \infty$, $e^{-t/\tau }\to 0$, so $I\to \epsilon /R$. $I_{\text{max}}=\frac{\epsilon }{R}=\frac{12}{3}=4\text{A}.$

WHAT: At $t=\tau$, decay drops to $e^{-1}\approx 0.368$ of the start. WHY: $I_{\text{decay}}(\tau )=I_0{e}^{-1}$; the "37%" number is the decay signature (growth's is 63%). $I=5\times 0.368=1.84\text{A}.$

Step 1 — $\tau$: $\tau =L/R=10/5=2$ s. Why: fixes the clock; $t=4$ s is $2\tau$. Step 2 — $I_{\text{max}}$: $\epsilon /R=20/5=4$ A. Why: the ceiling the curve approaches. Step 3 — plug in growth formula: we use the growth engine because the battery is connected and current is rising from zero toward the ceiling. $I=4\left(1-e^{-4/2}\right)=4\left(1-e^{-2}\right)=4(1-0.1353)=4(0.8647)=3.46\text{A}.$ Why plug in $t/\tau =2$: the formula only cares about time measured in units of $\tau$, so we feed it the ratio, not raw seconds. Sanity: at $2\tau$ growth should be about $86.5\%$ of $4$ A ✓.

Step 1 — $\tau$: $\tau =4/8=0.5$ s. Why: $t=1$ s is $2\tau$; the clock is what turns raw seconds into the exponent. Step 2 — use the decay formula: why decay and not growth? The battery is removed and the loop is shorted, so there is no ceiling to climb toward — current simply bleeds away from its starting value $I_0$. That is exactly the blue curve of figure s01. $I=3e^{-1/0.5}=3e^{-2}=3(0.1353)=0.406\text{A}.$

WHAT: Use $V_L=\epsilon e^{-t/\tau }$ (starts at the full EMF, decays to zero as the inductor gives up). WHY this formula: during growth $V_L=LdI/dt$ (our definition of $V_L$); differentiating the growth curve $I=\frac{\epsilon }{R}(1-e^{-t/\tau })$ gives $\frac{dI}{dt}=\frac{\epsilon }{R\tau }{e}^{-t/\tau }$, and multiplying by $L=R\tau$ leaves exactly $\epsilon e^{-t/\tau }$. Here $dI/dt>0$, so $V_L>0$ — the inductor is resisting a rise, as the sign convention promised. $V_L=20e^{-2/2}=20e^{-1}=20(0.368)=7.36\text{V}.$

WHAT: Set $I/I_{\text{max}}=0.8$ and solve for $t$. WHY the logarithm: the unknown $t$ is trapped in the exponent, and $\ln$ is the only tool that pulls an exponent down to ground level (it undoes $e^{(\cdot )}$). $0.8=1-e^{-t/\tau }\Rightarrow e^{-t/\tau }=\mathrm{0.2.}$ Take $\ln$: $-t/\tau =\ln 0.2=-1.609$. Why: $\ln$ of both sides converts "$e$ to some power" into just that power, freeing $t$. $t=0.5\times 1.609=0.805\text{s}.$

Step 1 — form the ratio: $\frac{I}{I_0}=\frac{1}{4}=e^{-t/\tau }$, so $-t/\tau =\ln (0.25)=-1.386$. Why the ratio: dividing kills the unknown starting current and leaves a clean exponential we can log. Step 2 — solve for $\tau$: $\tau =\frac{t}{1.386}=\frac{0.6}{1.386}=0.4328$ s. Why: $t$ is known ($0.6$ s), so dividing it by the log isolates the timescale. Step 3 — recover $R$: $\tau =L/R\Rightarrow R=L/\tau =2/0.4328=4.62\Omega .$ Why: the definition $\tau =L/R$ rearranges to give $R$ once $\tau$ and $L$ are both known.

WHAT: At $t=0$ the current is $0$, so KVL $\epsilon -IR-LdI/dt=0$ becomes $\epsilon =LdI/dt$. WHY: with $I=0$ the resistor drops nothing, so the entire EMF goes into changing the current — this is the steepest the curve ever is. ${\frac{dI}{dt}|}_0=\frac{\epsilon }{L}=\frac{10}{5}=2\text{A/s}.$ Geometric meaning (figure s03): if the current kept this slope it would hit $I_{\text{max}}$ in exactly one $\tau$ — that is the classic "tangent-at-origin" reading of $\tau$.

(a) $I_0=\epsilon /R=24/6=4$ A. Why: at steady state $dI/dt=0$, so $V_L=0$ and Ohm's law alone fixes the current. (b) Magnetic energy uses Energy stored in a magnetic field. Why this formula: energy poured into an inductor while its current climbs is stored in the field as $\frac{1}{2}L{I}^2$, evaluated at the final current. $U=\frac{1}{2}L{I}_0^2=\frac{1}{2}(3)(4{)}^2=\frac{1}{2}(3)(16)=24\text{J}.$ (c) At steady state $dI/dt=0$, so the inductor is a wire and all power is Ohmic. Why $I^{2}R$: with the inductor storing no further energy, every watt from the battery is dissipated as heat in $R$. $P=I_0^{2}R=(4{)}^2(6)=96\text{W}.$ Note: the stored energy (24 J, fixed) and the dissipated power (96 W, ongoing) are different beasts — one is a bank balance, the other a rate of spending.

(a) Half-life during decay: set $\frac{1}{2}=e^{-t/\tau }$. Why: "halves" means the decay ratio is $1/2$; log it to free $t$. $t=\tau \ln 2=0.4(0.6931)=0.277\text{s}.$ (b) Energy scales with $I^2$, so evaluate $\frac{1}{2}L{I}^2$ at the halved current $I=1$ A. Why square: the field energy depends on current squared, not linearly. $U=\frac{1}{2}L{I}^2=\frac{1}{2}(0.5)(1{)}^2=0.25\text{J}.$ Sanity: halving current quarters the energy, since the initial energy was $\frac{1}{2}(0.5)(2{)}^2=1$ J and $0.25=1/4$ of it ✓.

(a) $0.95=1-e^{-t/\tau }\Rightarrow e^{-t/\tau }=0.05\Rightarrow t/\tau =\ln 20=2.996$. Why the log: the required $\tau$ is hidden in the exponent, so we log to bring it out. Then $\tau =t/\ln 20$ (the known time $t$ on top). $\tau =\frac{0.3}{2.996}=0.1001\text{s}.$ (b) $L=R\tau$. Why: invert $\tau =L/R$; with $R$ fixed by the spec, $L$ is forced. $L=20\times 0.1001=2.00\text{H}.$ (c) At $t=0.15$ s the exponent is $t/\tau =0.15/0.1001=1.498$. Why feed the ratio: the formula reads time in units of $\tau$. $\frac{I}{I_{\text{max}}}=1-e^{-1.498}=1-0.2236=0.776\Rightarrow 77.6\%.$ Insight: reaching the last $5\%$ (from $77.6\%$ to $95\%$) takes as long as reaching the first $77.6\%$ — the exponential is brutally slow near the ceiling.

(a) $I_{\text{max}}=\epsilon /R=100/50=2$ A. Why: the ceiling the current climbs toward; since the trip threshold $1.5$ A is below $2$ A, the current does eventually reach it — the relay will trip. (b) $\tau =L/R=0.2/50=0.004$ s. Set $I=1.5$ A in the growth formula and solve for $t$. Why the log: the trip time sits in the exponent, so we log to free it. $1.5=2\left(1-e^{-t/\tau }\right)\Rightarrow e^{-t/\tau }=0.25\Rightarrow t/\tau =\ln 4=\mathrm{1.386.}$ $t=0.004\times 1.386=5.55\times 10^{-3}\text{s}=5.55\text{ms}.$ (c) $V_L=\epsilon e^{-t/\tau }=100\times 0.25=25$ V. Why: our growth $V_L$ formula, evaluated at the trip; positive because current is still rising ($dI/dt>0$), i.e. the inductor is still resisting the rise. KVL check (Kirchhoff's voltage law): resistor drop $=IR=1.5\times 50=75$ V. Then $75+25=100=\epsilon$ ✓ — the resistor drop plus the inductor voltage add up to the full source EMF exactly, as the loop equation demands.

• Growth: what fraction of $I_{\text{max}}$ at $t=\tau$, and at $t=2\tau$? ::: $63.2\%$ and $86.5\%$.
• To reach $95\%$ of max, how many $\tau$? ::: $\ln 20\approx 3.0\tau$.
• Decay half-life in terms of $\tau$? ::: $t_{1/2}=\tau \ln 2\approx 0.693\tau$.
• Inductor voltage during growth at $t=\tau$? ::: $V_L=\epsilon e^{-1}\approx 0.368\epsilon$.
• If current halves, energy becomes what fraction? ::: one quarter ($\times 1/4$).
• What does the sign of $V_L$ tell you? ::: positive = inductor resisting a rise; negative = inductor propping up a fall.