1.2.7
Ek resistor R C voltage ko instantly change nahi kar sakta . Capacitor charge store karta hai, resistor limit karta hai ki charge kitni tezi se flow ho sakta hai. Dono milke circuit ki ek natural "clock speed" set karte hain jise time constant τ = R C
YE kyun matter karta hai: har real wire mein resistance hoti hai aur har real component mein capacitance hoti hai, isliye RC behaviour har jagah dikhta hai — button debouncers, timers (555), filters, oscilloscope probes, aur yahi reason hai ki digital signals ki finite rise times hoti hain.
Intuition Voltage kyun jump nahi kar saktiV C d t d V C = ∞ smoothly ramp karni padti hai, kabhi step nahi leti.
Series loop: ek source V s R C Kirchhoff's Voltage Law kehta hai ki loop ke around voltages source ke barabar sum hote hain:
V s = V R + V C = I R + V C
I = C d t d V C
V s = R C d t d V C + V C
Solve karo. Rearrange karo:
d t d V C = R C V s − V C
Variables separate karo (kyun? taaki har side mein sirf ek variable ho integrate karne ke liye):
V s − V C d V C = R C d t
Dono sides integrate karo, V C ( 0 ) = 0
− ln ( V s − V C ) = R C t + k
t = 0 − ln ( V s ) = k V C
ln V s V s − V C = − R C t
Ab koi source nahi; ek charged capacitor (V C ( 0 ) = V 0 R 0 = I R + V C R C d t d V C + V C = 0 V s → 0
τ t = τ V C = V s ( 1 − e − 1 ) = V s ( 1 − 0.368 ) = 0.632 V s ek time constant ke baad capacitor apne target ka 63.2% . Discharging mein ek τ 36.8%
Time
Charged fraction
Discharged fraction
1 τ 63.2%
36.8%
2 τ 86.5%
13.5%
3 τ 95.0%
5.0%
4 τ 98.2%
1.8%
5 τ 99.3%
0.7%
Worked example Example 1 — Time constant nikaloR = 10 k Ω C = 100 μ F τ
τ = R C = ( 10 × 1 0 3 ) ( 100 × 1 0 − 6 ) = 1 s Kyun: ohms × farads = seconds (dimensionally Ω ⋅ F = A V ⋅ V C = A C = s 5 τ = 5 s
Worked example Example 2 — Diye gaye time par voltageSame circuit, V s = 5 V t = 2 s V C
t / τ = 2/1 = 2 Kyun: hum hamesha τ V C = 5 ( 1 − e − 2 ) = 5 ( 1 − 0.135 ) = 4.32 V
Worked example Example 3 — Time ke liye solve karo (ek delay design karo)V C 3 V V s = 5 τ = 1
3 = 5 ( 1 − e − t ) ⇒ e − t = 1 − 0.6 = 0.4 Kyun ln lete hain: t − t = ln 0.4 = − 0.916 ⇒ t = 0.916 s
Worked example Example 4 — DischargingCap 12 V R = 2 k Ω C = 470 μ F τ = 2000 × 470 × 1 0 − 6 = 0.94 s 1 s V C = 12 e − 1/0.94 = 12 e − 1.064 = 12 ( 0.345 ) = 4.14 V
τ Kyun sahi lagta hai: τ the special time hai, isliye finish line lagta hai.
Fix: 1 τ 63.2% ho jaate ho. Full (99%+) ke liye ≈ 5 τ
R Kyun sahi lagta hai: intuitively "bada" matlab "stronger" lagta hai.
Fix: bada R limit karta hai, isliye τ = R C badhta hai → charging slower hoti hai. Faster charge karne ke liye, R C
Common mistake Units mix karna.Kyun sahi lagta hai: formula τ = R C Fix: base SI use karo — ohms aur farads (µF nahi!). 100 μ F = 100 × 1 0 − 6 F 1 0 − 6 τ
Common mistake Discharging ke liye( 1 − e − t / τ ) Fix: charging upar jaati hai → ( 1 − e − t / τ ) neeche aati hai → e − t / τ
RC time constant formula kya hai? τ = R C
1 τ 63.2%
1 τ 36.8%
Kitne τ About 5 τ
Charging voltage equation? V C ( t ) = V s ( 1 − e − t / τ )
Discharging voltage equation? V C ( t ) = V 0 e − t / τ
Capacitor voltage instantly kyun nahi change ho sakti? Iske liye infinite current chahiye hogi (I = C d V / d t
RC charging circuit ko kaun sa ODE govern karta hai? R C d t d V C + V C = V s
R badhane se charging speed up hoti hai ya slow down? Slow down hoti hai (τ
t = 0 I = V s / R e − t / τ
Recall Feynman: 12-saal ke bachche ko explain karoSocho ek bucket (capacitor) ko patli straw (resistor) se bharna. Jitna zyada bucket bharta hai, utna kam paani andar push hota hai, isliye pehle tezi se bharta hai aur phir dheere dheere. Yeh kabhi ekdum nahi bhar jaata — isko "full" dikhne mein lagbhag "5 ghoonts" (5 τ
Mnemonic Numbers yaad rakho"63 upar, 37 neeche, 5 mein settle."
Ek τ 63% charge hota hai, 37% tak discharge hota hai, aur 5 time constants ke baad kaam khatam.
Capacitors and stored energy Kirchhoff's Voltage Law First-order differential equations RC low-pass and high-pass filters 555 timer circuits Digital signal rise time and integrity
63.2% charged / 36.8% left
