1.8.19Electromagnetism

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

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1. Setting up from first principles

WHY a differential equation? Because the unknown (QQ) and its rate of change (I=dQ/dtI = dQ/dt) appear together. We use Kirchhoff's voltage law (KVL): around any loop, the voltages sum to zero.

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

2. Charging — deriving the law from scratch

A battery (emf E\mathcal{E}), resistor RR, and uncharged capacitor CC in series. Switch closes at t=0t=0.

KVL around the loop (battery rise minus drops = 0): EIRQC=0\mathcal{E} - IR - \frac{Q}{C} = 0

Why this step? The battery pushes (+E+\mathcal{E}); the resistor drops IRIR; the capacitor holds back Q/CQ/C. Everything must balance.

Substitute I=dQdtI = \dfrac{dQ}{dt}: E=RdQdt+QC\mathcal{E} = R\frac{dQ}{dt} + \frac{Q}{C}

Rearrange to isolate the rate: dQdt=ERQRC=1RC(ECQ)\frac{dQ}{dt} = \frac{\mathcal{E}}{R} - \frac{Q}{RC} = \frac{1}{RC}\big(\mathcal{E}C - Q\big)

Why this step? Writing it as 1RC(QmaxQ)\frac{1}{RC}(Q_{\max} - Q) shows the rate is proportional to how far we still have to go. Define Qmax=ECQ_{\max} = \mathcal{E}C (final charge, when VC=EV_C = \mathcal{E} and current stops).

Separate variables and integrate: 0QdQQmaxQ=0tdtRC\int_0^Q \frac{dQ'}{Q_{\max} - Q'} = \int_0^t \frac{dt'}{RC} ln(QmaxQ)0Q=tRC-\ln(Q_{\max}-Q')\Big|_0^Q = \frac{t}{RC} ln ⁣(QmaxQQmax)=tRC    QmaxQQmax=et/RC-\ln\!\left(\frac{Q_{\max}-Q}{Q_{\max}}\right) = \frac{t}{RC} \implies \frac{Q_{\max}-Q}{Q_{\max}} = e^{-t/RC}


3. Discharging — deriving from scratch

Now the battery is removed; the charged capacitor (Q0Q_0) drives current through RR. KVL (no source): QC+IR=0    RdQdt=QC\frac{Q}{C} + IR = 0 \implies R\frac{dQ}{dt} = -\frac{Q}{C}

Why the minus sign? The capacitor is losing charge, so dQ/dt<0dQ/dt < 0.

dQQ=dtRC    lnQQ0=tRC\frac{dQ}{Q} = -\frac{dt}{RC} \implies \ln\frac{Q}{Q_0} = -\frac{t}{RC}


4. What τ actually means


5. Worked examples


6. Steel-manned mistakes


7. The 80/20 core

Recall Feynman: explain to a 12-year-old

Imagine filling a bucket (capacitor) with water, but the tap (battery) pushes through a thin straw (resistor). At first the bucket is empty so water rushes in fast. As the bucket fills, the water already inside pushes back against the tap, so it trickles in slower and slower — never quite overflowing instantly. The "τ\tau" is how thick your straw is times how big your bucket is: thick straw or small bucket = fast fill; thin straw or huge bucket = slow fill. Emptying the bucket through the same straw is the mirror image: gushes out fast at first, then dribbles.


8. Active-recall flashcards

What is the time constant of an RC circuit?
τ=RC\tau = RC, the time to reach 63.2% of final charge (charging) or fall to 36.8% (discharging).
Charging equation for charge on a capacitor?
Q(t)=Qmax(1et/τ)Q(t)=Q_{\max}(1-e^{-t/\tau}) with Qmax=ECQ_{\max}=\mathcal{E}C.
Discharging equation?
Q(t)=Q0et/τQ(t)=Q_0 e^{-t/\tau}.
What fraction is reached at t=τt=\tau when charging?
1e163.2%1-e^{-1}\approx 63.2\%.
What fraction remains at t=τt=\tau when discharging?
e136.8%e^{-1}\approx 36.8\%.
Initial current when charging starts?
I0=E/RI_0=\mathcal{E}/R — capacitor acts like a plain wire at t=0t=0.
Why does charging current decay?
Net driving voltage EVC\mathcal{E}-V_C shrinks as VCV_C rises, so I=(EVC)/RI=(\mathcal{E}-V_C)/R falls.
After how many τ is a capacitor ~fully charged?
About 5τ5\tau (e50.7%e^{-5}\approx 0.7\% left).
Show units of RC are seconds.
ΩF=(V/A)(C/V)=C/(C/s)=s\Omega\cdot F = (V/A)(C/V)=C/(C/s)=s.
KVL for charging RC loop?
EIRQ/C=0\mathcal{E} - IR - Q/C = 0 with I=dQ/dtI=dQ/dt.

9. Connections

Concept Map

loop sum zero

voltage drop

holds back charge

links Q and rate

separate and integrate

no source

sets rate

sets rate

final charge

start current

decays from

Kirchhoff voltage law

Resistor VR = IR

Capacitor VC = Q over C

Current I = dQ over dt

First-order differential equation

Time constant tau = RC

Charging Q = Qmax times 1 minus exp

Discharging Q = Q0 times exp

Qmax = emf times C

Initial current I0 = emf over R

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, RC circuit ka core idea simple hai: capacitor ko ek bucket samjho aur resistor ko ek patli straw. Battery paani push karti hai, par straw flow ko limit karti hai. Shuru mein capacitor khaali hai, isliye current sabse zyada (I0 = E/R). Jaise-jaise capacitor charge hota hai, uska apna voltage VCV_C badhta hai aur battery ke against "push back" karta hai. Net driving voltage EVC\mathcal{E}-V_C kam ho jaata hai, current girti hai — isliye charging linear nahi, exponential hoti hai.

Saara khel ek number par chalta hai: time constant τ=RC\tau = RC, jiska unit seconds hota hai. Ek τ\tau ke baad charging mein capacitor 63.2% tak pahunchta hai, aur discharging mein 36.8% tak gir jaata hai. Yaad rakho: charging UP to 63, discharging DOWN to 37 — ye dono complementary hain (63+37=100). "Fully charged" maano 5τ5\tau ke baad, kyunki tab sirf 0.7% bacha rehta hai.

Formulae ratt mat maaro — KVL se nikaalo. Charging: EIRQ/C=0\mathcal{E}-IR-Q/C=0, aur I=dQ/dtI=dQ/dt daalo, integrate karo, mil jaata hai Q=Qmax(1et/τ)Q=Q_{max}(1-e^{-t/\tau}). Discharging mein battery nahi, sirf Q/C+IR=0Q/C + IR = 0, jisse Q=Q0et/τQ=Q_0 e^{-t/\tau}. Bada RR ya bada CC matlab dheere — kyunki ya to current chhoti hai ya charge zyada move karna hai. Exam mein bas τ\tau nikaalo, t/τt/\tau count karo, aur exponential lagaa do — 80/20 yahi hai.

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