RL circuit — growth and decay of current — Hinglish notes

1.8.29 · Physics › Electromagnetism

Physically KYA ho raha hai

Current instantly kyun nahi jump karta? Ek inductor ek back-EMF $- L \frac{d I}{d t}$ produce karta hai. Agar current infinitely fast change karne ki koshish kare, to $d I / d t \to \infty$ ho jaayega, jo infinite voltage demand karta hai — yeh possible nahi. Isliye current smoothly change hoti hai.

Growth equation KAISE derive karein (scratch se)

Loop ke around Kirchhoff's voltage law apply karo. Loop mein jaate hue: battery potential ko $ε$ se raise karti hai; resistor $I R$ drop karta hai; inductor $L d I / d t$ se oppose karta hai.

$ε - I R - L \frac{d I}{d t} = 0$

Rate isolate karne ke liye rearrange karo:

$\frac{d I}{d t} = \frac{ε - I R}{L}$

Variables separate karo. $I_{0} = ε / R$ likhte hain (final steady current):

$\frac{d I}{ε - I R} = \frac{d t}{L}$

Yeh step kyun? Hum chahte hain saara $I$ ek side pe aur $t$ doosri side pe, taaki hum dono ko independently integrate kar sakein.

$I = 0$ at $t = 0$ se $I$ at $t$ tak integrate karo:

$\int_{0}^{I} \frac{d I}{ε - I R} = \int_{0}^{t} \frac{d t}{L}$

Left integral: $u = ε - I R$ maano, $d u = - R d I$ , to $\int \frac{d I}{ε - I R} = - \frac{1}{R} ln (ε - I R)$ .

$- \frac{1}{R} [ln (ε - I R) - ln ε] = \frac{t}{L}$

$ln (\frac{ε - I R}{ε}) = - \frac{R t}{L}$

Exponentiate karo:

$\frac{ε - I R}{ε} = e^{- R t / L} \Rightarrow I R = ε (1 - e^{- R t / L})$

Figure — RL circuit — growth and decay of current

Decay equation KAISE derive karein

Ab battery remove karo aur loop short karo ( $ε = 0$ ):

$0 - I R - L \frac{d I}{d t} = 0 \Rightarrow L \frac{d I}{d t} = - I R$

$\frac{d I}{I} = - \frac{R}{L} d t$

Yeh step kyun? Phir se $I$ aur $t$ separate karo integrate karne ke liye. Initial current $I_{0}$ se $I$ tak integrate karo:

$ln \frac{I}{I _{0}} = - \frac{R t}{L}$

Time constant — 80/20 core

Worked examples

Common mistakes

Active recall

Recall Khud test karo (answers chhupao)

RL circuit ke liye $τ$ kya hai, units check ke saath? ::: $τ = L / R$ ; H/Ω = s.
Growth ke dauran $t = 0$ aur $t \to \infty$ pe inductor ka behavior? ::: $t = 0$ : open circuit ( $I = 0$ ); $t \to \infty$ : wire ( $I = ε / R$ ).
$t = τ$ pe max current ka fraction (growth)? ::: $1 - e^{- 1} \approx 63.2%$ .

Growth-of-current formula in RL circuit

I (t) = \frac{ε}{R} (1 - e^{- t / τ})

Decay-of-current formula in RL circuit

I (t) = I_{0} e^{- t / τ}

RL time constant aur uska matlab

τ = L / R

; 63.2% tak pahunchne ka time (growth) ya 36.8% tak girne ka (decay)

KVL equation for RL growth loop

ε - I R - L \frac{d I}{d t} = 0

Growth ke dauran inductor voltage

V_{L} = ε e^{- t / τ}

(max

ε

t = 0

, 0 tak decay karta hai)

Inductor mein current instantly kyun nahi jump kar sakta

Isse

d I / d t = \infty

chahiye hoga, matlab infinite back-EMF

L d I / d t

— yeh possible nahi

Steady state mein inductor mein stored energy

U = \frac{1}{2} L I_{0}^{2}

Kitne

τ

ke baad transient ~khatam hota hai?

5 τ

(

e^{- 5} \approx 0.7%

)

Recall Ek 12-saal ke bachche ko samjhao (Feynman)

Socho ek bhaari shopping cart dhakka de rahe ho. Jab tum dhakka dena shuru karte ho (switch flip karo), woh full speed pe instantly nahi jaata — uska weight usse dheere dheere speed up karata hai. Inductor electric current ke liye wahi bhaari pan hai: yeh electrons ke flow ki changing speed ko resist karta hai. Jab tum dhakka dena band karte ho (battery remove karo), cart roll karta rehta hai aur dheere dheere rukta hai — wahi current decay ho rahi hai. Number $τ = L / R$ batata hai cart kitna "bhaari aur sticky" hai: bada matlab shuru hone mein slow aur rukne mein bhi slow.

Connections

Inductance and self-induction — $L d I / d t$ term ka source.
Faraday's law of electromagnetic induction — back-EMF ki origin.
Lenz's law — kyun induced EMF change oppose karta hai.
RC circuit — charging and discharging — dual structure, $τ = R C$ , capacitor inductor ka opposite.
Energy stored in a magnetic field — $U = \frac{1}{2} L I^{2}$ .
Kirchhoff's voltage law — woh loop equation jisse humne shuru kiya.