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

WHAT we need: $\tau =RC$ — the single number that says how fast this circuit responds. WHY just multiply: bigger $R$ throttles the current, bigger $C$ means more charge to move; both slow things down, so they multiply. Convert to base units first (this is where most slips happen): $R=220,000\Omega =2.2\times 10^5\Omega ,C=4.7\times 10^{-6}\text{F}$ $\tau =RC=(2.2\times 10^5)(4.7\times 10^{-6})=1.034\text{s}$ Answer: $\tau \approx 1.03\text{s}$.

WHAT: put $t=\tau$ into the charging law. $V_C(\tau )=\mathcal{E}\left(1-e^{-\tau /\tau }\right)=\mathcal{E}\left(1-e^{-1}\right)$ WHY the $\tau$'s cancel: $t/\tau$ literally counts "how many time constants have passed," so at $t=\tau$ that count is $1$. $1-e^{-1}=1-0.3679=0.6321$ Answer: $63.2\%$.

Step 1 — timescale. $\tau =RC=(1.5\times 10^4)(2.2\times 10^{-4})=3.3\text{s}$. Step 2 — express time in time-constants. $\frac{t}{\tau }=\frac{5}{3.3}=1.515$. Why: the exponential only ever "sees" this ratio. 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 — insert into the decay law. $6=24e^{-t/\tau }⟹e^{-t/\tau }=\frac{6}{24}=0.25$ Step 3 — undo the exponential with $\ln$. Why $\ln$: the unknown $t$ is stuck in the exponent; $\ln$ is the only tool that pulls an exponent down to ground level, because $\ln (e^x)=x$. $-\frac{t}{\tau }=\ln (0.25)=-1.386⟹t=1.386\tau =1.386\times 3.3=4.58\text{s}$ Answer: $t\approx 4.58\text{s}$.

Strategy: we are given a point on the charging curve and asked for the hidden $\tau$, then $C$. Step 1 — solve the charging law for the exponential. $7.6=12\left(1-e^{-t/\tau }\right)⟹1-e^{-t/\tau }=\frac{7.6}{12}=0.6333$ $e^{-t/\tau }=0.3667$ Step 2 — take $\ln$ to free $\tau$. $-\frac{t}{\tau }=\ln (0.3667)=-1.0034⟹\tau =\frac{t}{1.0034}=\frac{4}{1.0034}=3.986\text{s}$ Step 3 — extract $C$. Why 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 $\tau =4\text{s}$, since $7.6\text{V}$ is one time-constant's worth of charging).

Step 1 — timescale. $\tau =(2\times 10^3)(5\times 10^{-4})=1.0\text{s}$, so $t/\tau =1.5$. (a) Current decays from its start value $I_0=\mathcal{E}/R$. $I_0=\frac{6}{2000}=3.0\times 10^{-3}\text{A}=3\text{mA}$ $I(1.5)=3e^{-1.5}=3\times 0.2231=0.6694\text{mA}$ (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: $I\approx 0.669\text{mA}$, $Q\approx 2.33\text{mC}$.

Look at the figure below: at $t=1.5\tau$ the blue voltage curve is high (nearly there) while the red current curve is already low — they are mirror partners.

Same $\tau$ both phases: $\tau =(1\times 10^5)(1\times 10^{-5})=1.0\text{s}$. Phase 1 — charging for $2\text{s}$ (this sets the starting voltage for phase 2): $V_1=20\left(1-e^{-2/1}\right)=20(1-0.1353)=20(0.8647)=17.29\text{V}$ WHY carry this over: the discharge does not start from $\mathcal{E}$ — it starts from wherever charging left it. So $V_0$ for phase 2 is $17.29\text{V}$, not $20\text{V}$. Phase 2 — discharging for $3\text{s}$: $V_2=V_1{e}^{-3/1}=17.29\times e^{-3}=17.29\times 0.04979=0.861\text{V}$ Answer: $V_C\approx 0.861\text{V}$.

(a) Stored energy (from Energy Stored in a Capacitor): $U_C=\frac{1}{2}C{V}^2$. $U_C=\frac{1}{2}(2.2\times 10^{-3})(5{)}^2=\frac{1}{2}(2.2\times 10^{-3})(25)=2.75\times 10^{-2}\text{J}=27.5\text{mJ}$ (b) Heat in the resistor. Key insight: the battery delivers total energy $Q_{\max }\mathcal{E}=(C\mathcal{E})\mathcal{E}=C{\mathcal{E}}^2$. Exactly half ends up stored; the other half is burned in $R$ — no matter what $R$ is. $U_{\text{battery}}=C{\mathcal{E}}^2=(2.2\times 10^{-3})(25)=5.5\times 10^{-2}\text{J}=55\text{mJ}$ $U_R=U_{\text{battery}}-U_C=55-27.5=27.5\text{mJ}$ Answers: stored $=27.5\text{mJ}$; dissipated $=27.5\text{mJ}$ (a perfect 50/50 split).

Set the condition with the charging law and $V_C=\frac{2}{3}\mathcal{E}$: $\frac{2}{3}\mathcal{E}=\mathcal{E}\left(1-e^{-t/\tau }\right)⟹e^{-t/\tau }=\frac{1}{3}$ Notice $\mathcal{E}$ cancels — the timing doesn't depend on supply voltage, only on the fraction. Solve for $\tau$: $-\frac{t}{\tau }=\ln \frac{1}{3}=-1.0986⟹\tau =\frac{t}{1.0986}=\frac{10}{1.0986}=9.102\text{s}$ Get $R$: $R=\frac{\tau }{C}=\frac{9.102}{1\times 10^{-4}}=9.10\times 10^4\Omega =91.0\text{k}\Omega$ Answer: $R\approx 91.0\text{k}\Omega$.

Set up the condition $V_C=\frac{1}{2}{V}_0$ in the decay law: $\frac{1}{2}{V}_0=V_0{e}^{-t/\tau }$ $V_0$ cancels — this is the whole point: the fraction $\frac{1}{2}$ is reached after the same elapsed time no matter how charged you started. (Exactly the logic of a radioactive half-life; see 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: $t_{1/2}=2.5\times 0.6931=1.733\text{s}$. Answer: $t_{1/2}=\tau \ln 2\approx 1.73\text{s}$.

The figure shows why every equal step down (drop to a half, then a quarter, then an eighth) takes the same horizontal distance $\tau \ln 2$ — the signature of exponential decay.

(a) $R\to 0$: $\tau =RC\to 0$, so the exponential collapses instantly — the capacitor jumps to $\mathcal{E}$ at once. Physically, with no resistor there is nothing to limit current, so charge floods in immediately (an idealisation; real wires have tiny $R$). (b) $C\to \infty$: $\tau \to \infty$, so charging takes forever — an infinitely large "bucket" never fills. Final voltage would still aim at $\mathcal{E}$, but you'd wait an eternity to get close. (c) $t\to \infty$: $e^{-t/\tau }\to 0$, so $V_C\to \mathcal{E}$ and $I\to 0$. The capacitor becomes a fully-charged open switch; the loop current stops. This is the steady state — the resting endpoint every charging curve approaches but only reaches at infinity (practically at $\approx 5\tau$).