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

KYA chahiye: $\tau =RC$ — woh akela number jo kehta hai yeh circuit kitni fast respond karta hai. KYUN sirf multiply karo: bada $R$ current ko throttle karta hai, bada $C$ matlab zyada charge move karna hai; dono cheezein slow karti hain, isliye multiply hote hain. Pehle base units mein convert karo (yahan sabse zyada galtiyan hoti hain): $R=220,000\Omega =2.2\times 10^5\Omega ,C=4.7\times 10^{-6}\text{F}$ $\tau =RC=(2.2\times 10^5)(4.7\times 10^{-6})=1.034\text{s}$ Answer: $\tau \approx 1.03\text{s}$.

KYA: $t=\tau$ ko charging law mein daalo. $V_C(\tau )=\mathcal{E}\left(1-e^{-\tau /\tau }\right)=\mathcal{E}\left(1-e^{-1}\right)$ KYUN $\tau$ cancel hote hain: $t/\tau$ literally count karta hai "kitne time constants guzre hain," toh $t=\tau$ par woh count $1$ hai. $1-e^{-1}=1-0.3679=0.6321$ Answer: $63.2\%$.

Step 1 — timescale. $\tau =RC=(1.5\times 10^4)(2.2\times 10^{-4})=3.3\text{s}$. Step 2 — time ko time-constants mein express karo. $\frac{t}{\tau }=\frac{5}{3.3}=1.515$. Kyun: exponential sirf yahi ratio "dekhta" hai. Step 3 — charging law. $V_C=9\left(1-e^{-1.515}\right)=9(1-0.2198)=9(0.7802)=7.02\text{V}$ Answer: $V_C\approx 7.02\text{V}$.

Step 1. $\tau =(3.3\times 10^4)(1.0\times 10^{-4})=3.3\text{s}$. Step 2 — decay law mein daalo. $6=24e^{-t/\tau }⟹e^{-t/\tau }=\frac{6}{24}=0.25$ Step 3 — $\ln$ se exponential undo karo. Kyun $\ln$: unknown $t$ exponent mein fansa hua hai; $\ln$ woh akela tool hai jo exponent ko ground level par kheench laata hai, kyunki $\ln (e^x)=x$. $-\frac{t}{\tau }=\ln (0.25)=-1.386⟹t=1.386\tau =1.386\times 3.3=4.58\text{s}$ Answer: $t\approx 4.58\text{s}$.

Strategy: humein charging curve par ek point diya gaya hai aur hidden $\tau$, phir $C$ maanga gaya hai. Step 1 — charging law mein se exponential solve karo. $7.6=12\left(1-e^{-t/\tau }\right)⟹1-e^{-t/\tau }=\frac{7.6}{12}=0.6333$ $e^{-t/\tau }=0.3667$ Step 2 — $\tau$ free karne ke liye $\ln$ lo. $-\frac{t}{\tau }=\ln (0.3667)=-1.0034⟹\tau =\frac{t}{1.0034}=\frac{4}{1.0034}=3.986\text{s}$ Step 3 — $C$ extract karo. Kyun divide: $\tau =RC\Rightarrow C=\tau /R$. $C=\frac{\tau }{R}=\frac{3.986}{5\times 10^4}=7.97\times 10^{-5}\text{F}\approx 79.7\mu \text{F}$ Answer: $C\approx 79.7\mu \text{F}$ (essentially $\tau =4\text{s}$, kyunki $7.6\text{V}$ ek time-constant ki worth of charging hai).

Step 1 — timescale. $\tau =(2\times 10^3)(5\times 10^{-4})=1.0\text{s}$, toh $t/\tau =1.5$. (a) Current apni starting value $I_0=\mathcal{E}/R$ se decay karta hai. $I_0=\frac{6}{2000}=3.0\times 10^{-3}\text{A}=3\text{mA}$ $I(1.5)=3e^{-1.5}=3\times 0.2231=0.6694\text{mA}$ (b) Charge. Final charge $Q_{\max }=\mathcal{E}C=6\times 5\times 10^{-4}=3.0\times 10^{-3}\text{C}=3\text{mC}$. $Q(1.5)=Q_{\max }\left(1-e^{-1.5}\right)=3(1-0.2231)=2.33\text{mC}$ Answers: $I\approx 0.669\text{mA}$, $Q\approx 2.33\text{mC}$.

Neeche figure dekho: $t=1.5\tau$ par blue voltage curve high hai (almost pahunch gayi) jabki red current curve already low hai — yeh dono mirror partners hain.

Dono phases mein same $\tau$: $\tau =(1\times 10^5)(1\times 10^{-5})=1.0\text{s}$. Phase 1 — $2\text{s}$ tak charging (yeh phase 2 ke liye starting voltage set karta hai): $V_1=20\left(1-e^{-2/1}\right)=20(1-0.1353)=20(0.8647)=17.29\text{V}$ KYUN yeh carry over karo: discharge $\mathcal{E}$ se start nahi hoti — yeh wahan se start hoti hai jahan charging ne chhooda. Toh phase 2 ke liye $V_0$, $17.29\text{V}$ hai, $20\text{V}$ nahi. Phase 2 — $3\text{s}$ tak discharging: $V_2=V_1{e}^{-3/1}=17.29\times e^{-3}=17.29\times 0.04979=0.861\text{V}$ Answer: $V_C\approx 0.861\text{V}$.

(a) Stored energy (Energy Stored in a Capacitor se): $U_C=\frac{1}{2}C{V}^2$. $U_C=\frac{1}{2}(2.2\times 10^{-3})(5{)}^2=\frac{1}{2}(2.2\times 10^{-3})(25)=2.75\times 10^{-2}\text{J}=27.5\text{mJ}$ (b) Resistor mein heat. Key insight: battery total energy deliver karti hai $Q_{\max }\mathcal{E}=(C\mathcal{E})\mathcal{E}=C{\mathcal{E}}^2$. Exactly aadhi store hoti hai; baaki aadhi $R$ mein jalti hai — chahe $R$ kuch bhi ho. $U_{\text{battery}}=C{\mathcal{E}}^2=(2.2\times 10^{-3})(25)=5.5\times 10^{-2}\text{J}=55\text{mJ}$ $U_R=U_{\text{battery}}-U_C=55-27.5=27.5\text{mJ}$ Answers: stored $=27.5\text{mJ}$; dissipated $=27.5\text{mJ}$ (perfect 50/50 split).

Condition set karo charging law se aur $V_C=\frac{2}{3}\mathcal{E}$: $\frac{2}{3}\mathcal{E}=\mathcal{E}\left(1-e^{-t/\tau }\right)⟹e^{-t/\tau }=\frac{1}{3}$ Dhyaan do ki $\mathcal{E}$ cancel ho jaata hai — timing supply voltage par depend nahi karta, sirf fraction par. $\tau$ solve karo: $-\frac{t}{\tau }=\ln \frac{1}{3}=-1.0986⟹\tau =\frac{t}{1.0986}=\frac{10}{1.0986}=9.102\text{s}$ $R$ nikalo: $R=\frac{\tau }{C}=\frac{9.102}{1\times 10^{-4}}=9.10\times 10^4\Omega =91.0\text{k}\Omega$ Answer: $R\approx 91.0\text{k}\Omega$.

Condition set up karo $V_C=\frac{1}{2}{V}_0$ decay law mein: $\frac{1}{2}{V}_0=V_0{e}^{-t/\tau }$ $V_0$ cancel ho jaata hai — yahi poora point hai: $\frac{1}{2}$ fraction same elapsed time ke baad reach hota hai chahe tum kitne bhi charged se start karo. (Exactly radioactive half-life ki logic; dekho Exponential Decay and Differential Equations.) $\frac{1}{2}=e^{-t/\tau }⟹-\frac{t}{\tau }=\ln \frac{1}{2}=-\ln 2⟹\boxed{t_{1/2}=\tau \ln 2}$ Evaluate karo: $t_{1/2}=2.5\times 0.6931=1.733\text{s}$. Answer: $t_{1/2}=\tau \ln 2\approx 1.73\text{s}$.

Figure dikhata hai kyun har equal step neeche (half tak drop, phir quarter, phir eighth) same horizontal distance $\tau \ln 2$ leta hai — exponential decay ki pehchaan.

(a) $R\to 0$: $\tau =RC\to 0$, toh exponential instantly collapse ho jaata hai — capacitor turant $\mathcal{E}$ tak jump kar jaata hai. Physically, koi resistor nahi hone par current limit karne wali koi cheez nahi, toh charge immediately flood karta hai (ek idealization; real wires mein thoda $R$ hota hai). (b) $C\to \infty$: $\tau \to \infty$, toh charging forever lagti hai — infinitely bada "bucket" kabhi nahi bharta. Final voltage phir bhi $\mathcal{E}$ ki taraf aim karega, lekin tum eternity wait karoge pass hone ke liye. (c) $t\to \infty$: $e^{-t/\tau }\to 0$, toh $V_C\to \mathcal{E}$ aur $I\to 0$. Capacitor ek fully-charged open switch ban jaata hai; loop current ruk jaata hai. Yeh steady state hai — woh resting endpoint jispar har charging curve approach karti hai lekin sirf infinity par pahunchti hai (practically $\approx 5\tau$ par).