RC circuits — charging, discharging, time constant τ = RC

1.8.19 · Physics › Electromagnetism

Intuition Badi picture (WHY yeh exist karta hai)

Ek capacitor charge store karta hai, lekin woh instantly nahi bhar sakta — current ko ek resistor ke through flow karna padta hai charge deliver karne ke liye, aur resistor limit karta hai ki charge kitni tezi se pahunche.

KYA hota hai: jab tum ek battery ko $R$ aur $C$ ke saath series mein connect karte ho, capacitor dheere dheere charge hota hai; jab battery hata dete ho aur loop short karte ho, toh woh dheere dheere discharge hota hai.
KYU dheere dheere: jaise jaise capacitor ka voltage badhta hai, woh battery ke against push karta hai, toh $R$ ke across driving voltage kam hota hai → current kam hoti hai → charging slow hoti hai. Yeh ek self-limiting race hai.
KAISE hum isko describe karte hain: ek exponential approach jo SIRF EK number se govern hoti hai, woh hai time constant $τ = R C$ .

1. Setting up from first principles

KYU ek differential equation? Kyunki unknown ( $Q$ ) aur uska rate of change ( $I = d Q / d t$ ) saath mein appear hote hain. Hum Kirchhoff's voltage law (KVL) use karte hain: kisi bhi loop ke around, voltages ka sum zero hota hai.

Figure — RC circuits — charging, discharging, time constant τ = RC

2. Charging — law ko scratch se derive karna

Ek battery (emf $E$ ), resistor $R$ , aur uncharged capacitor $C$ series mein. Switch $t = 0$ par close hota hai.

Loop ke around KVL (battery rise minus drops = 0): $E - I R - \frac{Q}{C} = 0$

Yeh step kyun? Battery push karti hai ( $+ E$ ); resistor $I R$ drop karta hai; capacitor $Q / C$ roke rakhta hai. Sab kuch balance hona chahiye.

$I = \frac{d Q}{d t}$ substitute karo: $E = R \frac{d Q}{d t} + \frac{Q}{C}$

Rearrange karo rate isolate karne ke liye: $\frac{d Q}{d t} = \frac{E}{R} - \frac{Q}{R C} = \frac{1}{R C} (E C - Q)$

Yeh step kyun? Isko $\frac{1}{R C} (Q_{m a x} - Q)$ ki tarah likhne se pata chalta hai ki rate kitna baaki reh gaya hai uske proportional hai. Define karo $Q_{m a x} = E C$ (final charge, jab $V_{C} = E$ aur current ruk jaati hai).

Variables separate karo aur integrate karo: $\int_{0}^{Q} \frac{d Q ^{'}}{Q _{m a x} - Q ^{'}} = \int_{0}^{t} \frac{d t ^{'}}{R C}$ $- ln (Q_{m a x} - Q^{'})_{0}^{Q} = \frac{t}{R C}$ $- ln (\frac{Q _{m a x} - Q}{Q _{m a x}}) = \frac{t}{R C} ⟹ \frac{Q _{m a x} - Q}{Q _{m a x}} = e^{- t / R C}$

3. Discharging — scratch se derive karna

Ab battery hata di gayi hai; charged capacitor ( $Q_{0}$ ) $R$ ke through current drive karta hai. KVL (koi source nahi): $\frac{Q}{C} + I R = 0 ⟹ R \frac{d Q}{d t} = - \frac{Q}{C}$

Minus sign kyun? Capacitor charge kho raha hai, isliye $d Q / d t < 0$ .

$\frac{d Q}{Q} = - \frac{d t}{R C} ⟹ ln \frac{Q}{Q _{0}} = - \frac{t}{R C}$

4. τ ka actual matlab

5. Worked examples

6. Steel-manned mistakes

7. The 80/20 core

Recall Feynman: ek 12-saal ke bachche ko samjhao

Socho ek bucket (capacitor) ko paani se bharna hai, lekin tap (battery) ek patli straw (resistor) ke through push karta hai. Shuru mein bucket khaali hai toh paani tezi se aata hai. Jaise jaise bucket bharta hai, andar ka paani tap ke against push karta hai, toh aahista aahista trickle hone lagta hai — kabhi bhi instantly overflow nahi hota. " $τ$ " yeh hai ki tumhari straw kitni moti hai multiply karo bucket kitna bada hai: moti straw ya chhota bucket = jaldi bharo; patli straw ya bada bucket = dheere bharo. Usi straw se bucket khaali karna mirror image hai: pehle tezi se nikalta hai, phir dribble karta hai.

8. Active-recall flashcards

RC circuit ka time constant kya hai?

τ = R C

, final charge ka 63.2% reach karne ka time (charging) ya 36.8% tak girne ka (discharging).

Capacitor par charge ke liye charging equation?

Q (t) = Q_{m a x} (1 - e^{- t / τ})

jisme

Q_{m a x} = E C

Discharging equation?

Q (t) = Q_{0} e^{- t / τ}

t = τ

par charging mein kaun sa fraction reach hota hai?

1 - e^{- 1} \approx 63.2%

t = τ

par discharging mein kaun sa fraction bachta hai?

e^{- 1} \approx 36.8%

Charging start hone par initial current?

I_{0} = E / R

— capacitor

t = 0

par plain wire ki tarah act karta hai.

Charging current decay kyun karta hai?

Net driving voltage

E - V_{C}

kam hota hai jaise

V_{C}

badhta hai, isliye

I = (E - V_{C}) / R

gir jaata hai.

Kitne τ ke baad capacitor ~fully charged hota hai?

Lagbhag

5 τ

(

e^{- 5} \approx 0.7%

bacha hota hai).

Dikhao ki RC ki units seconds hain.

Ω \cdot F = (V / A) (C / V) = C / (C / s) = s

Charging RC loop ke liye KVL?

E - I R - Q / C = 0

jisme

I = d Q / d t

9. Connections

Capacitors and Capacitance — $V_{C} = Q / C$ ka source.
Ohm's Law and Resistance — $V_{R} = I R$ deta hai.
Kirchhoff's Voltage Law — loop equation jisse humne shuru kiya.
Exponential Decay and Differential Equations — same math jaise radioactive decay, Newton's Law of Cooling.
Energy Stored in a Capacitor — charging ke dauran battery ki aadhi energy $R$ mein dissipate hoti hai.
LR Circuits — inductive analogue jisme $τ = L / R$ .