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1.2.7Circuit Analysis Fundamentals

Understand RC charging - discharging time constants

1,560 words7 min readdifficulty · medium
🇮🇳 Yeh note Hinglish mein padho →

1. What is happening physically

2. Derivation from first principles (charging)

Series loop: a source VsV_s, resistor RR, capacitor CC. Kirchhoff's Voltage Law says voltages around the loop sum to the source:

Vs=VR+VC=IR+VCV_s = V_R + V_C = I R + V_C

Substitute I=CdVCdtI = C\dfrac{dV_C}{dt}:

Vs=RCdVCdt+VCV_s = RC\frac{dV_C}{dt} + V_C

Solve it. Rearrange:

dVCdt=Vs−VCRC\frac{dV_C}{dt} = \frac{V_s - V_C}{RC}

Separate variables (why? so each side has only one variable to integrate):

dVCVs−VC=dtRC\frac{dV_C}{V_s - V_C} = \frac{dt}{RC}

Integrate both sides, with VC(0)=0V_C(0)=0:

−ln⁡(Vs−VC)=tRC+k-\ln(V_s - V_C) = \frac{t}{RC} + k

At t=0t=0: −ln⁡(Vs)=k-\ln(V_s) = k. Substitute back and solve for VCV_C:

ln⁡ ⁣Vs−VCVs=−tRC\ln\!\frac{V_s - V_C}{V_s} = -\frac{t}{RC}

Discharging

Now no source; a charged capacitor (VC(0)=V0V_C(0)=V_0) drives current through RR. KVL gives 0=IR+VC0 = IR + V_C, i.e. RCdVCdt+VC=0RC\dfrac{dV_C}{dt}+V_C=0. Same integration with Vs→0V_s\to 0:

3. Meaning of τ\tau

Time Charged fraction Discharged fraction
1τ1\tau 63.2% 36.8%
2τ2\tau 86.5% 13.5%
3τ3\tau 95.0% 5.0%
4τ4\tau 98.2% 1.8%
5τ5\tau 99.3% 0.7%
Figure — Understand RC charging - discharging time constants

4. Worked examples

5. Common mistakes

6. Flashcards

What is the RC time constant formula?
τ=RC\tau = RC (ohms × farads = seconds)
What fraction is charged after 1τ1\tau?
63.2%
What fraction remains after discharging 1τ1\tau?
36.8%
After how many τ\tau is a capacitor considered fully charged?
About 5τ5\tau (>99%)
Charging voltage equation?
VC(t)=Vs(1−e−t/τ)V_C(t)=V_s(1-e^{-t/\tau})
Discharging voltage equation?
VC(t)=V0e−t/τV_C(t)=V_0 e^{-t/\tau}
Why can't capacitor voltage change instantly?
It would require infinite current (I=C dV/dtI=C\,dV/dt), which R forbids
What ODE governs an RC charging circuit?
RC dVCdt+VC=VsRC\,\frac{dV_C}{dt}+V_C=V_s
Does increasing R speed up or slow down charging?
Slows it down (τ\tau increases)
What is the charging current at t=0t=0?
I=Vs/RI=V_s/R (max, then decays as e−t/τe^{-t/\tau})
Recall Feynman: explain to a 12-year-old

Imagine filling a bucket (capacitor) with a thin straw (resistor). The fuller the bucket gets, the less the water pushes in, so it fills fast at first and then slower and slower. It never quite overflows instantly — it takes about "5 sips" (5τ5\tau) to look full. The time constant is how long one "sip" takes: use a thinner straw (bigger R) or a bigger bucket (bigger C) and each sip takes longer.

Connections

  • 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

Concept Map

defines

current

combined with KVL

substituted into

solve, VC 0 = 0

solve, no source

dVC/dt

at t = tau

rule of thumb

voltage cannot jump

R and C in series

Time constant tau = RC

Capacitor Q = C VC

I = C dVC/dt

Resistor VR = I R

RC dVC/dt + VC = Vs

VC = Vs 1 - e^-t/tau

VC = V0 e^-t/tau

I = Vs/R e^-t/tau

63.2% charged / 36.8% left

5 tau = essentially done

VC ramps smoothly

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Socho ek resistor aur capacitor series me lage hain. Capacitor ek chhoti "battery-jaisi" storage hai jo charge jama karta hai, aur resistor decide karta hai ki charge kitni tezi se andar-bahar jaayega. Inka combination ek natural speed set karta hai jise hum time constant kehte hain: τ=RC\tau = RC. Yaad rakho, capacitor ka voltage kabhi ekdum jump nahi karta — kyunki instant change ke liye infinite current chahiye, jo resistor allow nahi karta.

Charging ke time voltage ek smooth curve me upar chadhta hai: VC=Vs(1−e−t/τ)V_C = V_s(1-e^{-t/\tau}). Ek τ\tau ke baad sirf 63.2% hi charge hota hai, poora full hone me lagbhag 5τ\tau lagte hain (99% se zyada). Discharging me ulta hota hai — voltage e−t/τe^{-t/\tau} ke hisaab se girta hai, ek τ\tau me 36.8% bacha reh jaata hai.

Ek common galti: log samajhte hain "bada R matlab fast charging" — galat! Bada RR current ko rok deta hai, isliye τ\tau badhta hai aur charging slow ho jaati hai. Doosri galti units ki hai: farad me hi count karo, μF\mu F me 10−610^{-6} multiply karna mat bhoolo, warna τ\tau galat aayega.

Yeh concept har jagah kaam aata hai — timers, button debounce, filters, aur digital signals ka rise-time. Formula ratne se pehle KVL se derive karke dekho, tabhi Feynman-style samajh pakki hogi. Rule of thumb yaad rakho: "63 upar, 37 neeche, 5 me settle."

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