RC circuits — charging, discharging, time constant τ = RC: Worked examples

1.8.19 · D3 · Physics › Electromagnetism › RC circuits — charging, discharging, time constant τ = RC

Shuru karne se pehle, ek symbol reminder taaki kuch bhi bina samjhe use na ho:

$E$ (script E) = battery ki emf, volts mein uski steady push.
$R$ = ohms mein resistance ( $Ω$ ); yeh current ko $V_{R} = I R$ ke zariye limit karta hai (dekho Ohm's Law and Resistance).
$C$ = farads (F) mein capacitance; cap charge $Q$ ko $V_{C} = Q / C$ ke saath hold karta hai (dekho Capacitors and Capacitance).
$τ = R C$ = time constant, seconds mein measure hota hai.
$e \approx 2.718$ = exponential base. $e^{- 1} \approx 0.368$ , aur $e^{- t / τ}$ yeh bataata hai ki "time $t$ ke baad kya fraction bacha hai?"

The scenario matrix

Har RC problem jo tumse pucha ja sakta hai, in cells mein se kisi ek mein hota hai. Right column us example ka naam deta hai jo use cover karta hai.

#	Case class	Kya alag hai isme	Covered by
A	Charging — diye gaye $t$ par $V$ ya $Q$ nikalo	$t$ ko $1 - e^{- t / τ}$ mein plug karo	Ex 1
B	Charging — target $V$ ke liye $t$ nikalo	$ln$ se invert karo	Ex 2
C	Discharging — diye gaye $t$ par $V$ nikalo	$e^{- t / τ}$ mein plug karo	Ex 3
D	Discharging — target $V$ ke liye $t$ nikalo	$ln$ se invert karo	Ex 4
E	$t = 0$ par current vs baad mein	current dono modes mein same-shape decay hai	Ex 5
F	Limiting / degenerate inputs	$t \to 0$ , $t \to \infty$ , $R \to 0$ , $C \to 0$	Ex 6
G	Half-life style ("aadha hone ka time")	special target $V_{0} /2$ → $τ ln 2$	Ex 7
H	Real-world word problem	ek story (camera flash) ko $R, C$ mein translate karo	Ex 8
I	Exam twist: do resistors / hidden $C$ nikalo	kisi chupi hui quantity ke liye rearrange karo	Ex 9

Do master laws jo hum baar baar use karte rahenge:

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

Upar wala figure tumhara map hai: rising teal curve charging hai, falling orange curve discharging hai, aur dono $t = τ$ line ko 63.2% aur 36.8% par cross karte hain — yeh do numbers hain jinke around yeh poora page ghoomta hai.

The worked examples

Cell A — charging, kisi time par voltage nikalo

Worked example Example 1 — 3 seconds baad kitna bhara hai?

$E = 9$ V, $R = 200 k Ω$ , $C = 10 μ$ F, capacitor khali shuru hota hai. $t = 3$ s par $V_{C}$ nikalo.

Forecast: $τ$ kuch seconds ka hoga, toh 3 s par hum ek time constant se aage hain — guess karo around 70–80% of 9 V. Apna number likh lo.

$τ = R C = (2 \times 1 0^{5}) (1 \times 1 0^{- 5}) = 2$ s. Yeh step kyun? Har cheez $τ$ ke multiples mein measure hoti hai; iske bina hum nahi keh sakte ki 3 s "zyada" hai ya "thoda."
$t / τ = 3/2 = 1.5$ . Yeh step kyun? Formula sirf ratio $t / τ$ dekhta hai, kabhi $t$ akela nahi.
$V_{C} = 9 (1 - e^{- 1.5}) = 9 (1 - 0.2231) = 9 (0.7769) = 6.99$ V. Yeh step kyun? Case A charging law ka direct plug-in hai.

Verify: $1.5 τ$ ko $1 τ$ value ( $0.632 \times 9 = 5.69$ V) aur $2 τ$ value ( $0.865 \times 9 = 7.78$ V) ke beech hona chahiye. Hamara $6.99$ V theek unke beech hai. ✓ Units: volts × (dimensionless) = volts. ✓

Cell B — charging, target voltage ke liye time nikalo

Cell C — discharging, kisi time par voltage nikalo

Cell D — discharging, target voltage ke liye time nikalo

Cell E — $t = 0$ par current vs baad mein (dono modes)

Worked example Example 5 — current ki kahani

Charging circuit: $E = 6$ V, $R = 2 k Ω$ , $C = 500 μ$ F. $t = 0$ , $t = τ$ , aur $t = 2 τ$ par current nikalo.

Forecast: Current bilkul shuru mein sabse zyada hoti hai (khali cap = ek saadi wire) aur same $τ$ ke saath decay karti hai jaise baaki sab. Guess hai ki yeh kuch milliamps se shuru hoti hai aur roughly har $τ$ mein ek-tihaai ho jaati hai.

$I_{0} = \frac{E}{R} = \frac{6}{2000} = 3 \times 1 0^{- 3}$ A $= 3$ mA. Yeh step kyun? $t = 0$ par cap zero volts hold karta hai, toh poora $E$ , $R$ par pada hai; Ohm's law peak current deta hai.
$τ = R C = (2000) (5 \times 1 0^{- 4}) = 1$ s (sirf times label karne ke liye use hua).
$I (τ) = I_{0} e^{- 1} = 3 (0.3679) = 1.10$ mA. Yeh step kyun? Current $I_{0} e^{- t / τ}$ follow karti hai — same decay shape chahe charging ho ya discharging.
$I (2 τ) = I_{0} e^{- 2} = 3 (0.1353) = 0.406$ mA.

Verify: ratio $I (τ) / I_{0} = 0.368$ aur $I (2 τ) / I_{0} = 0.135$ — exactly standard $e^{- 1}$ aur $e^{- 2}$ fractions. ✓ Note karo ki current decay karti hai jabki charge rise karta hai: koi contradiction nahi, kyunki rising $V_{C}$ , driving voltage $E - V_{C}$ ko kha jaata hai.

Cell F — degenerate aur limiting inputs

Worked example Example 6 — edge cases (calculator ki zarurat nahi)

Charging law $V_{C} = E (1 - e^{- t / τ})$ aur decay $V = V_{0} e^{- t / τ}$ lo. Har extreme ko reason karo.

Forecast: step padhne se pehle har limit guess karo.

$t \to 0$ (charging): $e^{0} = 1$ , toh $V_{C} = E (1 - 1) = 0$ . Kyun? Switch close hone ke instant mein cap abhi bhi khali hai — yeh jump nahi kar sakta.
$t \to \infty$ (charging): $e^{- \infty} = 0$ , toh $V_{C} = E (1 - 0) = E$ . Kyun? Aakhirkar cap battery ko poori tarah oppose karta hai; current rukti hai; $V_{C}$ source ke equal ho jaata hai.
$R \to 0$ : $τ = R C \to 0$ , toh kisi bhi $t > 0$ ke liye $e^{- t / τ} \to 0$ instantly. Charging immediate ho jaati hai. Kyun? Koi resistor nahi matlab current ko throttle karne wala kuch nahi — lekin physically $I_{0} = E / R \to \infty$ , ek warning hai ki ideal wires current limit nahi kar sakti.
$C \to 0$ : $τ \to 0$ bhi, toh again instant — lekin store karne ke liye essentially koi charge nahi ( $Q_{m a x} = E C \to 0$ ).
$R \to \infty$ ya $C \to \infty$ : $τ \to \infty$ , toh sabhi finite $t$ ke liye $e^{- t / τ} \to 1$ : charging kabhi perceptibly hoti nahi lagti.

Verify (numeric spot-check): $E = 10$ , $τ = 1$ lo. $t = 0.001$ s par, $V_{C} = 10 (1 - e^{- 0.001}) = 10 (0.0009995) = 0.009995$ V — essentially 0, step 1 confirm. $t = 20$ s par, $V_{C} = 10 (1 - e^{- 20}) = 9.99999998$ V — essentially $E$ , step 2 confirm. ✓

Cell G — "aadha hone ka time" special target

Worked example Example 7 — discharge ka half-life

Ek capacitor discharge ho raha hai. Dikhao ki voltage aadhi hone ka time $t_{1/2} = τ ln 2$ hai, aur $τ = 4$ s ke liye compute karo.

Forecast: $ln 2 \approx 0.693$ , toh half-life ek $τ$ se thodi kam honi chahiye — yahan around 2.8 s.

Target set karo: $\frac{V _{0}}{2} = V_{0} e^{- t / τ}$ . Yeh step kyun? "Aadha" matlab ratio $V / V_{0} = 1/2$ ; actual voltages cancel ho jaate hain.
$V_{0}$ cancel karo: $\frac{1}{2} = e^{- t / τ}$ . Yeh step kyun? Yahi reason hai ki half-life starting voltage se independent hai — same trick jaise radioactive decay mein Exponential Decay and Differential Equations aur Newton's Law of Cooling mein.
$- t / τ = ln (1/2) = - ln 2 \Rightarrow t_{1/2} = τ ln 2$ .
$τ = 4$ s ke liye: $t_{1/2} = 4 (0.6931) = 2.77$ s.

Verify: $t = 2.77$ s par, $e^{- 2.77/4} = e^{- 0.6931} = 0.5000$ — exactly aadha. ✓ Kyunki $ln 2 < 1$ , half-life $τ$ se chhoti hai, jo sahi lagta hai: 50% tak pahunchna 36.8% tak girne se pehle hota hai.

Cell H — real-world word problem

Worked example Example 8 — ek camera flash capacitor

Ek camera flash $1000 μ$ F capacitor use karta hai jo $300$ V tak charged hai. Yeh apna charge flash tube ke through dump karta hai, jiska effective resistance $10 Ω$ hai, aur flash "khatam" hoti hai jab voltage peak ka $37%$ ho jaata hai. Flash kitni der chalti hai, aur peak current kya hai?

Forecast: $37%$ exactly $e^{- 1}$ point hai, toh flash roughly ek $τ$ tak chalti hai. Aur $τ = R C$ chhote $R$ lekin bade $C$ ke saath — ek second ka chhota fraction guess karo. Peak current: $300/10$ bada hoga, tens of amps.

Peak current: $I_{0} = \frac{V _{0}}{R} = \frac{300}{10} = 30$ A. Yeh step kyun? Fire hone ke instant mein, poore $300$ V tube par pad jaate hain (Ohm's law), ek bada bright burst deta hai.
$τ = R C = (10) (1 \times 1 0^{- 3}) = 0.010$ s $= 10$ ms.
$37%$ voltage matlab $V = 0.37 V_{0} = e^{- t / τ} V_{0}$ , yani $t = τ$ (kyunki $e^{- 1} = 0.368$ ). Yeh step kyun? Problem ne humein special $e^{- 1}$ fraction diya, toh koi logarithm nahi chahiye — ise pehchanna kaam bachata hai.
Toh flash $t \approx τ = 10$ ms tak chalti hai.

Verify: $τ = 10$ ms ek believable flash duration hai (real flashes 1–10 ms ki hoti hain). ✓ Energy sanity via Energy Stored in a Capacitor: $\frac{1}{2} C V_{0}^{2} = \frac{1}{2} (1 0^{- 3}) (300)^{2} = 45$ J release hoti hai — sach mein ek bright pop. ✓

Cell I — exam twist: hidden capacitance nikalo

Worked example Example 9 — measurement se

$C$ reverse-engineer karo Ek engineer $10$ V source se $R = 50 k Ω$ ke through ek unknown capacitor charge karta hai aur $t = 4$ s par $V_{C} = 8$ V measure karta hai. $C$ nikalo.

Forecast: 10 V mein se 8 V, ek $τ$ se aage hai (jo 6.32 V tak pahunchta hai), toh $τ$ 4 s se kam hai → $C$ roughly $80 μ$ F se kam.

Measurement ke saath charging law: $8 = 10 (1 - e^{- 4/ τ})$ . Yeh step kyun? Unknown $τ$ ke andar chhupi hai; ek data point use pin down karta hai.
Exponential isolate karo: $e^{- 4/ τ} = 1 - 0.8 = 0.2$ .
Log lo: $- \frac{4}{τ} = ln 0.2 = - 1.609 \Rightarrow τ = \frac{4}{1.609} = 2.486$ s. $ln$ kyun? Same exponential-ka-inverse tool jaise Cells B aur D mein — har "time ya $τ$ ke liye solve karo" problem ka workhorse.
Ab $τ = R C$ unpack karo: $C = \frac{τ}{R} = \frac{2.486}{5 \times 1 0 ^{4}} = 4.97 \times 1 0^{- 5}$ F $\approx 49.7 μ$ F. Yeh step kyun? $R$ pata hai, toh divide karne se $C$ milta hai.

Verify: $C = 49.7 μ$ F ke saath, $τ = R C = (5 \times 1 0^{4}) (4.97 \times 1 0^{- 5}) = 2.486$ s, aur $V_{C} (4) = 10 (1 - e^{- 4/2.486}) = 10 (1 - 0.2) = 8.00$ V — measurement se match karta hai. ✓

Recall checkpoint

Recall In mein se har ek kaun sa cell hai?

Ek cap discharge ho raha hai — tumhe $V$ bataya gaya hai aur $t$ maanga gaya hai. ::: Cell D — $ln$ se invert karo. Tumhe ek time par $V_{C}$ bataya gaya hai aur $C$ maanga gaya hai. ::: Cell I — $τ$ solve karo phir $R$ se divide karo. "Aadha hone mein kitna time lagega?" ::: Cell G — jawab hai $τ ln 2$ , $V_{0}$ se independent. Switch close hone ke instant mein current. ::: Cell E — $I_{0} = E / R$ , the peak. $R \to \infty$ hone par kya hota hai? ::: Cell F — $τ \to \infty$ , charging kabhi perceptibly complete nahi hoti.

Connections

Parent topic — woh derivations jinhe yeh examples apply karte hain.
Ohm's Law and Resistance — har "initial current" step $I_{0} = V / R$ use karta hai.
Capacitors and Capacitance — $V_{C} = Q / C$ , aur kyun cap shuru mein wire ki tarah aur end mein open switch ki tarah hota hai.
Kirchhoff's Voltage Law — dono laws ke peeche ka loop equation.
Exponential Decay and Differential Equations · Newton's Law of Cooling — Ex 7 jaise same half-life algebra.
Energy Stored in a Capacitor — Ex 8 mein 45 J sanity check.
LR Circuits — $τ = R C$ ko $τ = L / R$ se swap karo aur har example transfer ho jaata hai.

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