Exercises — RL circuit — growth and decay of current

KYA: $\tau =L/R$ read karo. KYUN: $\tau$ woh akela number hai jo timescale set karta hai — baaki sab iske against scale hota hai. $\tau =\frac{L}{R}=\frac{6}{3}=2\text{s}.$ Units check: $\frac{\text{H}}{\Omega }=\frac{\text{V⋅s/A}}{\text{V/A}}=\text{s}$ ✓. Isliye $\tau =2\text{s}$.

KYA: Kaafi time baad inductor ek plain wire ki tarah act karta hai, isliye sirf $R$ current limit karta hai. KYUN: Jab $t\to \infty$, $e^{-t/\tau }\to 0$, isliye $I\to \epsilon /R$. $I_{\text{max}}=\frac{\epsilon }{R}=\frac{12}{3}=4\text{A}.$

KYA: $t=\tau$ par, decay start ka $e^{-1}\approx 0.368$ tak gir jaati hai. KYUN: $I_{\text{decay}}(\tau )=I_0{e}^{-1}$; "37%" number decay ka signature hai (growth ka 63% hai). $I=5\times 0.368=1.84\text{A}.$

Step 1 — $\tau$: $\tau =L/R=10/5=2$ s. Kyun: clock fix karta hai; $t=4$ s matlab $2\tau$ hai. Step 2 — $I_{\text{max}}$: $\epsilon /R=20/5=4$ A. Kyun: woh ceiling jis tak curve approach karta hai. Step 3 — growth formula mein plug karo: hum growth engine use karte hain kyunki battery connected hai aur current zero se ceiling ki taraf badh rahi hai. $I=4\left(1-e^{-4/2}\right)=4\left(1-e^{-2}\right)=4(1-0.1353)=4(0.8647)=3.46\text{A}.$ $t/\tau =2$ kyun plug kiya: formula sirf $\tau$ ki units mein measured time ki parwah karta hai, isliye hum ratio feed karte hain, raw seconds nahi. Sanity: $2\tau$ par growth $4$ A ka lagbhag $86.5\%$ honi chahiye ✓.

Step 1 — $\tau$: $\tau =4/8=0.5$ s. Kyun: $t=1$ s matlab $2\tau$ hai; clock hi raw seconds ko exponent mein convert karta hai. Step 2 — decay formula use karo: kyun decay aur growth nahi? Battery remove ho gayi hai aur loop short hai, isliye koi ceiling nahi hai chadne ke liye — current bas apni starting value $I_0$ se bleed away hoti hai. Yahi figure s01 ki blue curve hai. $I=3e^{-1/0.5}=3e^{-2}=3(0.1353)=0.406\text{A}.$

KYA: $V_L=\epsilon e^{-t/\tau }$ use karo (full EMF se start hota hai, zero tak decay karta hai jab inductor haar jaata hai). YE FORMULA KYUN: growth ke dauran $V_L=LdI/dt$ (humari $V_L$ ki definition); growth curve $I=\frac{\epsilon }{R}(1-e^{-t/\tau })$ ko differentiate karne par $\frac{dI}{dt}=\frac{\epsilon }{R\tau }{e}^{-t/\tau }$ milta hai, aur $L=R\tau$ se multiply karne par exactly $\epsilon e^{-t/\tau }$ bachta hai. Yahaan $dI/dt>0$ hai, isliye $V_L>0$ — inductor rise ko resist kar raha hai, jaisa sign convention ne promise kiya tha. $V_L=20e^{-2/2}=20e^{-1}=20(0.368)=7.36\text{V}.$

KYA: $I/I_{\text{max}}=0.8$ set karo aur $t$ ke liye solve karo. LOGARITHM KYUN: unknown $t$ exponent mein fasa hua hai, aur $\ln$ woh akela tool hai jo exponent ko ground level par laata hai (yeh $e^{(\cdot )}$ ko undo karta hai). $0.8=1-e^{-t/\tau }\Rightarrow e^{-t/\tau }=\mathrm{0.2.}$ $\ln$ lo: $-t/\tau =\ln 0.2=-1.609$. Kyun: dono sides ka $\ln$ "$e$ to some power" ko sirf woh power mein convert karta hai, $t$ ko azaad karta hai. $t=0.5\times 1.609=0.805\text{s}.$

Step 1 — ratio banao: $\frac{I}{I_0}=\frac{1}{4}=e^{-t/\tau }$, isliye $-t/\tau =\ln (0.25)=-1.386$. Ratio kyun: divide karne se unknown starting current khatam ho jaata hai aur ek clean exponential milti hai jise hum log kar sakte hain. Step 2 — $\tau$ solve karo: $\tau =\frac{t}{1.386}=\frac{0.6}{1.386}=0.4328$ s. Kyun: $t$ known hai ($0.6$ s), isliye ise log se divide karne par timescale isolate ho jaata hai. Step 3 — $R$ recover karo: $\tau =L/R\Rightarrow R=L/\tau =2/0.4328=4.62\Omega .$ Kyun: definition $\tau =L/R$ rearrange hoti hai $R$ dene ke liye jab $\tau$ aur $L$ dono known hon.

KYA: $t=0$ par current $0$ hai, isliye KVL $\epsilon -IR-LdI/dt=0$ ban jaata hai $\epsilon =LdI/dt$. KYUN: $I=0$ ke saath resistor kuch nahi drop karta, isliye poori EMF current change karne mein jaati hai — yeh curve ki sabse steep jagah hoti hai. ${\frac{dI}{dt}|}_0=\frac{\epsilon }{L}=\frac{10}{5}=2\text{A/s}.$ Geometric meaning (figure s03): agar current is slope ko maintain karti toh exactly ek $\tau$ mein $I_{\text{max}}$ hit karti — yeh $\tau$ ki classic "tangent-at-origin" reading hai.

(a) $I_0=\epsilon /R=24/6=4$ A. Kyun: steady state par $dI/dt=0$, isliye $V_L=0$ aur Ohm's law akele current fix karta hai. (b) Magnetic energy Energy stored in a magnetic field use karta hai. Yeh formula kyun: ek inductor mein energy jo uski current chadne ke waqt daali jaati hai woh field mein $\frac{1}{2}L{I}^2$ ke roop mein store hoti hai, final current par evaluate ki jaati hai. $U=\frac{1}{2}L{I}_0^2=\frac{1}{2}(3)(4{)}^2=\frac{1}{2}(3)(16)=24\text{J}.$ (c) Steady state par $dI/dt=0$, isliye inductor ek wire hai aur saari power Ohmic hai. $I^{2}R$ kyun: inductor aur zyaada energy store nahi kar raha, isliye battery ka har watt $R$ mein heat ke roop mein dissipate hota hai. $P=I_0^{2}R=(4{)}^2(6)=96\text{W}.$ Note: stored energy (24 J, fixed) aur dissipated power (96 W, ongoing) alag cheezein hain — ek bank balance hai, doosra kharch karne ki rate.

(a) Decay ke dauran Half-life: $\frac{1}{2}=e^{-t/\tau }$ set karo. Kyun: "halves" matlab decay ratio $1/2$ hai; $t$ ko azaad karne ke liye log karo. $t=\tau \ln 2=0.4(0.6931)=0.277\text{s}.$ (b) Energy $I^2$ ke saath scale karti hai, isliye halved current $I=1$ A par $\frac{1}{2}L{I}^2$ evaluate karo. Square kyun: field energy current par squared depend karti hai, linearly nahi. $U=\frac{1}{2}L{I}^2=\frac{1}{2}(0.5)(1{)}^2=0.25\text{J}.$ Sanity: current half karne par energy quarter ho jaati hai, kyunki initial energy $\frac{1}{2}(0.5)(2{)}^2=1$ J thi aur $0.25=1/4$ hai ✓.

(a) $0.95=1-e^{-t/\tau }\Rightarrow e^{-t/\tau }=0.05\Rightarrow t/\tau =\ln 20=2.996$. Log kyun: required $\tau$ exponent mein chhupi hai, isliye hum log karte hain use bahar laane ke liye. Phir $\tau =t/\ln 20$ (known time $t$ upar). $\tau =\frac{0.3}{2.996}=0.1001\text{s}.$ (b) $L=R\tau$. Kyun: $\tau =L/R$ invert karo; spec se $R$ fixed hai, isliye $L$ force ho jaata hai. $L=20\times 0.1001=2.00\text{H}.$ (c) $t=0.15$ s par exponent $t/\tau =0.15/0.1001=1.498$ hai. Ratio kyun feed karein: formula $\tau$ ki units mein time padhta hai. $\frac{I}{I_{\text{max}}}=1-e^{-1.498}=1-0.2236=0.776\Rightarrow 77.6\%.$ Insight: aakhri $5\%$ tak pahunchna (from $77.6\%$ to $95\%$) utna hi time leta hai jitna pehle $77.6\%$ tak pahunchna — exponential ceiling ke paas brutally slow hoti hai.

(a) $I_{\text{max}}=\epsilon /R=100/50=2$ A. Kyun: woh ceiling jis ki taraf current chadti hai; kyunki trip threshold $1.5$ A, $2$ A se neeche hai, current eventually wahan pahunch jaati hai — relay trip hogi. (b) $\tau =L/R=0.2/50=0.004$ s. Growth formula mein $I=1.5$ A set karo aur $t$ ke liye solve karo. Log kyun: trip time exponent mein hai, isliye use azaad karne ke liye log karte hain. $1.5=2\left(1-e^{-t/\tau }\right)\Rightarrow e^{-t/\tau }=0.25\Rightarrow t/\tau =\ln 4=\mathrm{1.386.}$ $t=0.004\times 1.386=5.55\times 10^{-3}\text{s}=5.55\text{ms}.$ (c) $V_L=\epsilon e^{-t/\tau }=100\times 0.25=25$ V. Kyun: humara growth $V_L$ formula, trip par evaluate kiya; positive hai kyunki current abhi bhi badh rahi hai ($dI/dt>0$), matlab inductor abhi bhi rise ko resist kar raha hai. KVL check (Kirchhoff's voltage law): resistor drop $=IR=1.5\times 50=75$ V. Phir $75+25=100=\epsilon$ ✓ — resistor drop aur inductor voltage exactly full source EMF tak add ho jaate hain, jaisa loop equation demand karta hai.

• Growth: $t=\tau$ par $I_{\text{max}}$ ka kya fraction, aur $t=2\tau$ par? ::: $63.2\%$ aur $86.5\%$.
• Max ka $95\%$ reach karne ke liye kitne $\tau$? ::: $\ln 20\approx 3.0\tau$.
• $\tau$ ke terms mein decay half-life? ::: $t_{1/2}=\tau \ln 2\approx 0.693\tau$.
• Growth ke dauran $t=\tau$ par inductor voltage? ::: $V_L=\epsilon e^{-1}\approx 0.368\epsilon$.
• Agar current half ho jaaye, energy kya fraction ban jaati hai? ::: one quarter ($\times 1/4$).
• $V_L$ ka sign tumhe kya batata hai? ::: positive = inductor rise ko resist kar raha hai; negative = inductor fall ko prop up kar raha hai.