RL circuit — growth and decay of current: Worked examples

1.8.29 · D3 · Physics › Electromagnetism › RL circuit — growth and decay of current

Shuru karne se pehle, do symbols ka ek reminder jo hum baar baar use karte hain:

$τ = L / R$ — time constant, circuit ki natural "clock tick" (seconds mein).
$I_{m a x} = ε / R$ — steady current, woh value jis ki taraf growth badhti hai aur jahan se decay girta hai.

Scenario matrix

Har RL problem ko is grid ke ek cell mein utarte hue socho. Columns hain kaun si quantity unknown hai; rows hain circuit kis phase mein hai.

Phase → / Unknown ↓	Ek time par current dhundho	Time dhundho	Ek component dhundho ( $L$ , $R$ , $ε$ )	Voltage / rate / energy
Growth (badhta hua)	Ex 1	Ex 2	Ex 3	Ex 6 (voltage), Ex 8 (energy)
Decay (girta hua)	Ex 4	Ex 5	Ex 7	—
Degenerate / limiting	Ex 9 ( $t = 0$ , $t \to \infty$ , $L \to 0$ )	—	—	—
Word problem / exam twist	Ex 10 (two-stage: charge karo phir dump)

Har woh cell jisme ek alag idea aa sakta hai, neeche uska apna example hai. Jaise jaise tum har ek padhte ho, label batata hai kaunsa cell fill ho raha hai taaki tum coverage badhte dekh sako.

Figure — RL circuit — growth and decay of current

Growth-phase examples

Worked example Example 1 — Growth: ek diye gaye time par current

(cell: Growth / find current) $ε = 20 V$ , $R = 5 Ω$ , $L = 10 H$ . $t = 4$ s par $I$ dhundho.

Forecast: Pehle andaza lagao ki $t = 4$ s "early" hai ya "late". Kyunki $τ = L / R = 2$ s hai, $4$ s $= 2 τ$ hai. Do time constants matlab hum 63% mark ke baad hain lekin abhi saturated nahi hue — max ka karib 85% expect karo.

Step 1 — $τ = L / R = 10/5 = 2$ s compute karo. Yeh step kyun? Har exponential ko uski clock chahiye. $τ$ ke bina exponent mein " $- t / τ$ " ka koi matlab nahi.

Step 2 — $I_{m a x} = ε / R = 20/5 = 4$ A compute karo. Yeh step kyun? Growth ek ceiling ka fraction hai. Pehle ceiling chahiye.

Step 3 — Growth mein plug karo: $I = 4 (1 - e^{- 4/2}) = 4 (1 - e^{- 2})$ . Yeh step kyun? Yahi actual growth law hai; $t / τ = 4/2 = 2$ .

Step 4 — Evaluate karo: $e^{- 2} = 0.1353$ , to $I = 4 (1 - 0.1353) = 4 (0.8647) = 3.459$ A.

Verify: $3.459/4 = 86.5%$ of max — bilkul wahi "around 85%" range mein jo humne forecast kiya tha. Units: (V/Ω) = A ✓.

Worked example Example 2 — Growth:

time ke liye solve karo (cell: Growth / find time) Same circuit ( $τ = 2$ s, $I_{m a x} = 4$ A). Current pehli baar $3$ A kab pahunche ga?

Forecast: $3$ A max ka $75%$ hai. Yeh $τ$ (63%) aur $2 τ$ (86.5%) ke beech hai, to expect karo $t$ $2$ s aur $4$ s ke beech hoga — probably $2.8$ s ke karib.

Step 1 — $3 = 4 (1 - e^{- t /2})$ likho, phir divide karo: $1 - e^{- t /2} = 0.75$ . Yeh step kyun? Hum exponential ko isolate karte hain taaki unknown $t$ akela uske andar baitha rahe.

Step 2 — Rearrange karo: $e^{- t /2} = 0.25$ . Yeh step kyun? Ab exponential ek bare number ke barabar hai; ek hi tool hai jo $e^{(\cdot)}$ ko "undo" kar sake — uska inverse, natural log $ln$ . Yahi reason hai ki $ln$ enter hota hai — yeh exactly woh sawaal hai "किस power par $e$ ko raise karna $0.25$ dega?"

Step 3 — $ln$ lo: $- t /2 = ln 0.25 = - 1.386$ . Yeh step kyun? $ln$ ne $t$ ko exponent se neeche khींch liya taaki hum linearly solve kar sakein.

Step 4 — Solve karo: $t = 2 \times 1.386 = 2.77$ s.

Verify: Wapas plug karo: $4 (1 - e^{- 2.77/2}) = 4 (1 - e^{- 1.386}) = 4 (1 - 0.25) = 3$ A ✓. Aur $2.77$ s, $2$ aur $4$ s ke beech baitha hai jaise forecast kiya tha.

Worked example Example 3 — Growth: inductor

$L$ dhundho (cell: Growth / find component) $R = 8 Ω$ wala ek circuit $0.6$ s mein apni maximum current ka $50%$ reach kar leta hai. $L$ dhundho.

Forecast: $50%$ tak pahunchne mein karib $0.69 τ$ lagta hai (kyunki $e^{- 0.69} \approx 0.5$ ). Agar $0.69 τ = 0.6$ s to $τ \approx 0.87$ s, isliye $L = R τ \approx 8 \times 0.87 \approx 7$ H.

Step 1 — $0.5 = 1 - e^{- t / τ} \Rightarrow e^{- t / τ} = 0.5$ . Yeh step kyun? "Max ka percentage" wala statement exponential form mein convert karo.

Step 2 — $- t / τ = ln 0.5 = - 0.693 \Rightarrow τ = t /0.693 = 0.6/0.693 = 0.866$ s. Yeh step kyun? $ln$ ne $τ$ ko free kiya; ek exponential ki "half-life" hamesha $0.693 τ$ hoti hai.

Step 3 — $L = R τ = 8 \times 0.866 = 6.93$ H. Yeh step kyun? $τ = L / R \Rightarrow L = R τ$ — sirf definition par algebra.

Verify: $τ = 6.93/8 = 0.866$ s; $t = 0.6$ s par, $1 - e^{- 0.6/0.866} = 1 - e^{- 0.693} = 1 - 0.5 = 0.5$ ✓. $\approx 7$ H forecast se match karta hai.

Decay-phase examples

Component & energy examples

Worked example Example 6 — Growth: inductor par voltage

(cell: Growth / voltage) $ε = 24$ V, $τ = 2$ s. $t = 0$ , $t = 2$ s, aur $t = 6$ s par inductor $V_{L}$ par voltage dhundho.

Forecast: Growth ke dauran inductor ki voltage ek decaying exponential $V_{L} = ε e^{- t / τ}$ hai (current ke opposite behaviour!). $t = 0$ par yeh puri $24$ V hai; ek $τ$ par yeh $37%$ hai ( $\approx 9$ V); $3 τ$ par yeh $\approx 5%$ ( $\approx 1.2$ V) hai.

Step 1 — $t = 0$ par: $V_{L} = 24 e^{0} = 24$ V. Yeh step kyun? Switching ke us instant par, current jump nahi kar sakti, isliye saari EMF coil par appear hoti hai — yahi reason hai ki inductor $t = 0$ par open circuit ki tarah behave karta hai.

Step 2 — $t = 2$ s par: $V_{L} = 24 e^{- 1} = 24 \times 0.3679 = 8.83$ V. Yeh step kyun? Jaise current badhti hai, resistor zyada share leta hai aur coil ka share girta hai.

Step 3 — $t = 6$ s par: $V_{L} = 24 e^{- 3} = 24 \times 0.0498 = 1.20$ V.

Verify: Kisi bhi time par $V_{L} + I R = ε$ . $t = 2$ s par: $I = \frac{ε}{R} (1 - e^{- 1})$ , to $I R = ε (1 - e^{- 1}) = 24 \times 0.632 = 15.17$ V; phir $V_{L} + I R = 8.83 + 15.17 = 24$ V $= ε$ ✓ (Kirchhoff's voltage law se KVL check).

Worked example Example 7 — Decay: ek measured drop se

$R$ dhundho (cell: Decay / find component) Ek decaying inductor current $0.2$ s mein $4$ A se $1.5$ A tak girti hai. Inductor $L = 0.5$ H hai. $R$ dhundho.

Forecast: Ratio $1.5/4 = 0.375 \approx e^{- 1}$ hai, to $0.2$ s lagbhag ek $τ$ hai. Isliye $τ \approx 0.2$ s aur $R = L / τ \approx 0.5/0.2 = 2.5 Ω$ .

Step 1 — $\frac{1.5}{4} = e^{- t / τ} = e^{- 0.2/ τ}$ , to $0.375 = e^{- 0.2/ τ}$ . Yeh step kyun? Final current ko initial se divide karne par $I_{0}$ cancel ho jaata hai, sirf unknown $τ$ mein ek pure exponential bachta hai.

Step 2 — $ln 0.375 = - 0.9808 = - 0.2/ τ \Rightarrow τ = 0.2/0.9808 = 0.2039$ s. Yeh step kyun? $ln$ ne $τ$ ko exponent se free kiya.

Step 3 — $R = L / τ = 0.5/0.2039 = 2.452 Ω$ . Yeh step kyun? $τ = L / R$ ko invert karo $R = L / τ$ paane ke liye.

Verify: $τ = 0.5/2.452 = 0.2039$ s; $4 e^{- 0.2/0.2039} = 4 e^{- 0.9808} = 4 \times 0.375 = 1.5$ A ✓. $2.5 Ω$ forecast ke karib.

Worked example Example 8 — Steady state par stored energy

(cell: Growth / energy) $ε = 12$ V, $R = 3 Ω$ , $L = 2$ H. Jab current poori tarah grow ho jaaye tab stored magnetic energy dhundho, aur $t = τ$ par energy dhundho.

Forecast: Full current $I_{m a x} = 4$ A, to full energy $\frac{1}{2} L I^{2} = \frac{1}{2} (2) (16) = 16$ J. $t = τ$ par current max ka $63.2%$ hai, aur energy $\propto I^{2}$ , to energy $\approx 0.63 2^{2} = 40%$ of $16 \approx 6.4$ J.

Step 1 — $I_{m a x} = ε / R = 12/3 = 4$ A. Yeh step kyun? Energy current par depend karti hai, isliye pehle final current dhundho.

Step 2 — $U_{\infty} = \frac{1}{2} L I_{m a x}^{2} = \frac{1}{2} (2) (4^{2}) = 16$ J. Yeh step kyun? Yeh Energy stored in a magnetic field formula hai — energy us field mein chupi hai jo coil build karti hai.

Step 3 — $t = τ$ par: $I = 4 (1 - e^{- 1}) = 2.528$ A, to $U = \frac{1}{2} (2) (2.52 8^{2}) = 6.39$ J. Yeh step kyun? Same energy formula, current-at- $τ$ plug in kiya.

Verify: $6.39/16 = 39.9% \approx 0.63 2^{2} = 0.399$ ✓. Units: H·A² = (V·s/A)·A² = V·A·s = W·s = J ✓.

Degenerate & limiting cases

Har achhe student ko pata hona chahiye ki edges par kya hota hai: bilkul pehla instant, door ka future, aur pathological components. Yeh ek example un sab ko sweep karta hai taaki tum kabhi kisi "trick" limit par freeze na ho.

Worked example Example 9 — Teen limits

(cell: degenerate / limiting) Ek growth circuit $ε = 10$ V, $R = 2 Ω$ , $L = 4$ H ke liye, current evaluate karo (a) $t = 0$ par, (b) $t \to \infty$ par, aur (c) degenerate limit $L \to 0$ discuss karo.

Forecast: (a) current jump nahi kar sakti, to $I = 0$ . (b) coil ek plain wire ban jaati hai, to $I = ε / R = 5$ A. (c) koi inductance nahi to current ko slow karne wala kuch nahi — yeh turant $5$ A par jump kar jaani chahiye.

Step 1 — $t = 0$ par: $I = 5 (1 - e^{0}) = 5 (1 - 1) = 0$ A. Yeh step kyun? $e^{0} = 1$ hai, bracket ko zero kar deta hai. Yeh "inductor = $t = 0$ par open circuit" rule confirm karta hai.

Step 2 — $t \to \infty$ par: $e^{- t / τ} \to 0$ , to $I \to 5 (1 - 0) = 5$ A. Yeh step kyun? Shrink factor vanish ho jaata hai; jab $d I / d t = 0$ hota hai to coil koi opposition offer nahi karti, isliye woh bare wire ki tarah act karti hai aur current sirf $R$ se set hoti hai.

Step 3 — $L \to 0$ : tab $τ = L / R \to 0$ , to $e^{- t / τ} \to 0$ kisi bhi $t > 0$ ke liye. Current turant $5$ A ho jaati hai — exponential ek step mein collapse ho jaata hai. Yeh step kyun? Zero inductance matlab zero "electrical inertia" — $τ \to 0$ ke saath smooth rise infinitely fast hoti hai, plain resistor circuit recover ho jaata hai (Inductance and self-induction hi tha jo delay create kar raha tha).

Verify: $I (0) = 0$ ✓, $I (\infty) = ε / R = 5$ ✓, aur $lim_{τ \to 0^{+}} 5 (1 - e^{- 1/ τ}) = 5$ ✓ (ek tiny $τ$ par neeche check kiya).

Word problem / exam twist

Worked example Example 10 — Two-stage: charge karo, phir dump karo

(cell: word problem / exam twist) Ek electromagnet coil ( $L = 0.8$ H, $R = 4 Ω$ ) ko ek $16$ V supply se connect kiya gaya hai. Ise $0.2$ s ke liye connected chhoda jaata hai, phir supply ko instantly ek short se replace kar diya jaata hai taaki current decay ho. Switching ke moment par current dhundho, aur short par switch karne ke $0.1$ s baad current dhundho.

Forecast: $τ = 0.2$ s, to charging ke $0.2$ s = ek $τ$ → $I_{m a x} = 4$ A ka karib $63%$ $\approx 2.5$ A. Phir decay ke $0.1$ s half $τ$ hai, factor $e^{- 0.5} \approx 0.61$ , to final $\approx 1.5$ A.

Step 1 — $τ = L / R = 0.8/4 = 0.2$ s; $I_{m a x} = ε / R = 16/4 = 4$ A. Yeh step kyun? Dono phases same $τ$ share karte hain (same $L$ , same $R$ ), isliye hum ise ek baar compute karte hain.

Step 2 — $0.2$ s ke liye Growth: $I_{1} = 4 (1 - e^{- 0.2/0.2}) = 4 (1 - e^{- 1}) = 4 \times 0.6321 = 2.528$ A. Yeh step kyun? Yeh ending current decay phase ki starting current $I_{0}$ ban jaati hai — dono stages is value par stitch hote hain.

Step 3 — $I_{0} = 2.528$ A se shuru hokar $0.1$ s ke liye Decay: $I_{2} = 2.528 e^{- 0.1/0.2} = 2.528 e^{- 0.5} = 2.528 \times 0.6065 = 1.533$ A. Yeh step kyun? Ek baar short ho jaane par, Lenz's law current ko flow karta rakhta hai lekin ab woh jahan bhi thi wahan se decay hoti hai.

Verify: Stage-1 current $2.528$ A $= 63.2%$ of $4$ A ✓. Stage-2 factor $e^{- 0.5} = 0.6065$ ; $2.528 \times 0.6065 = 1.533$ A ✓, $\approx 1.5$ A forecast se match karta hai.

Active recall

Recall In mein se har ek kaun sa cell hai?

"Diye gaye $ε, R, L$ se, $t$ par $I$ dhundho." ::: Growth / find current (Ex 1).
"Current $I_{0}$ se ek value tak girti hai; time dhundho." ::: Decay / find time (Ex 5).
"Time $t$ mein max ka X% reach karta hai; $L$ dhundho." ::: Growth / find component (Ex 3).
" $t$ par coil par voltage." ::: Growth / voltage — $V_{L} = ε e^{- t / τ}$ use karo (Ex 6).
"Thodi der charge karo, phir short karo aur decay karo." ::: Two-stage exam twist — growth ke end ko decay ka $I_{0}$ banao (Ex 10).

Connections

Parent topic (Hinglish) — woh derivations jin par yeh examples practice karte hain.
Kirchhoff's voltage law — Example 6 mein use kiya gaya KVL check.
Inductance and self-induction — kyun $L \to 0$ delay ko khatam kar deta hai (Example 9).
Lenz's law — kyun shorting par current flow karta rehta hai (Example 10).
Energy stored in a magnetic field — Example 8 mein use kiya gaya $\frac{1}{2} L I^{2}$ .
RC circuit — charging and discharging — same log-based time solving, dual $τ = R C$ .