Exercises — Understand RC charging - discharging time constants

WHAT: we just multiply, because $\tau =RC$ by definition. WHY the powers of ten: "k" means $\times 10^3$ and "$\mu$" (micro) means $\times 10^{-6}$. We MUST convert to base SI or the answer is off by factors of a million. $\tau =(4.7\times 10^3)(220\times 10^{-6})=4.7\times 220\times 10^{-3}=1034\times 10^{-3}=1.034\text{ s}.$ Answer: $\tau \approx 1.03\text{ s}$.

WHERE these come from: plug $t=\tau$ so $t/\tau =1$. Charging: $V_C=V_s(1-e^{-1})=V_s(1-0.368)=0.632V_s$ → 63.2%. Discharging: $V_C=V_0{e}^{-1}=0.368V_0$ → 36.8%. Mnemonic: "63 up, 37 down."

WHY 5: at $5\tau$ the charge fraction is $1-e^{-5}=0.993$ → $>99\%$, close enough that engineers call it done. $t_{\text{full}}\approx 5\tau =5\times 2=10\text{ s}.$

STEP 1 — work in units of $\tau$: $t/\tau =1.5/0.5=3$. (Working in "how many time constants" keeps arithmetic clean.) STEP 2 — evaluate: $V_C=5(1-e^{-3})=5(1-0.0498)=5(0.9502)=4.75\text{ V}.$ Sanity check: $t=3\tau$ should give the $95.0\%$ table row — and $4.75/5=0.950$. ✔ Answer: $V_C\approx 4.75\text{ V}$.

WHICH formula: discharging falls, so use $V_C=V_0{e}^{-t/\tau }$ (not the $1-e^{-t/\tau }$ form). $t/\tau =0.8/0.5=1.6$. $V_C=9e^{-1.6}=9\times 0.2019=1.82\text{ V}.$ Answer: $V_C\approx 1.82\text{ V}$.

WHY at $t=0$ the current is largest: the capacitor starts empty ($V_C=0$), so the full source voltage sits across $R$. From $I(t)=\frac{V_s}{R}{e}^{-t/\tau }$: $I(0)=\frac{12}{3000}=4\times 10^{-3}\text{ A}=4\text{ mA}.$ After one $\tau$: $I(\tau )=4\text{ mA}\times e^{-1}=4\times 0.368=1.47\text{ mA}.$ Answer: $4\text{ mA}$ initially, $\approx 1.47\text{ mA}$ after one $\tau$.

STEP 1 — set up: $7=10(1-e^{-t/\tau })$, so $e^{-t/\tau }=1-0.7=0.3$. STEP 2 — WHY take $\ln$: the unknown $t$ is trapped in the exponent; $\ln$ is the exact tool that "brings it down," because $\ln (e^x)=x$. $-\frac{t}{\tau }=\ln (0.3)=-1.204\Rightarrow t=1.204\tau =1.204\times 2=2.41\text{ s}.$ Answer: $t\approx 2.41\text{ s}$.

$2=16e^{-t/\tau }\Rightarrow e^{-t/\tau }=2/16=\mathrm{0.125.}$ $-\frac{t}{\tau }=\ln (0.125)=-2.079\Rightarrow t=2.079\times 3=6.24\text{ s}.$ Interpretation: $0.125=\frac{1}{8}$, and halving happens every $\ln 2\approx 0.693\tau$; three halvings ($16\to 8\to 4\to 2$) take $3\times 0.693=2.079\tau$. Same answer, different lens. ✔ Answer: $t\approx 6.24\text{ s}$.

WHAT to look at: both curves aim at the same dashed ceiling ($5\text{ V}$), but the orange curve rises more gently and takes longer to reach the $63.2\%$ line (the gray marker at $3.16\text{ V}$). WHY that means bigger $\tau$: the value of $t$ where a curve crosses $63.2\%$ is $\tau$ (that's the definition). Orange crosses later ⇒ larger $\tau$. Answer: the orange curve has the larger time constant. (Here blue $\tau =1\text{ s}$, orange $\tau =2.5\text{ s}$.)

STEP 1 — find the required $\tau$: from $3.3=5(1-e^{-t/\tau })$ at $t=2$: $e^{-2/\tau }=1-\frac{3.3}{5}=1-0.66=0.34\Rightarrow -\frac{2}{\tau }=\ln (0.34)=-\mathrm{1.0788.}$ $\tau =\frac{2}{1.0788}=1.854\text{ s}.$ STEP 2 — solve $\tau =RC$ for $R$: (this is why we needed $\tau$ first — it's the bridge between "timing spec" and "component value"). $R=\frac{\tau }{C}=\frac{1.854}{100\times 10^{-6}}=1.854\times 10^4=18.54\text{ k}\Omega .$ Answer: $R\approx 18.5\text{ k}\Omega$ (nearest standard value $18\text{ k}\Omega$ gives $\tau =1.8\text{ s}$, close enough).

WHY add the resistors: the same current flows through both (series), so their voltage drops add — they behave as one resistor $R=R_1+R_2$. $R=10+22=32\text{ k}\Omega ,\tau =(32\times 10^3)(47\times 10^{-6})=1.504\text{ s}.$ Time to $90\%$: $0.90=1-e^{-t/\tau }\Rightarrow e^{-t/\tau }=0.10\Rightarrow t=\tau \ln (10)=1.504\times 2.3026=3.46\text{ s}.$ Answer: $\tau \approx 1.50\text{ s}$, and $90\%$ reached at $t\approx 3.46\text{ s}$ ($\approx 2.3\tau$, matching the standard "$90\%$ takes $2.3\tau$" fact).

Part (a) — end of charging: $V_C(1.5)=10(1-e^{-1.5})=10(1-0.2231)=7.769\text{ V}.$ Part (b) — WHY $V_0$ changes: the discharge doesn't start from $10\text{ V}$; it starts from wherever charging left off, so $V_0=7.769\text{ V}$. Now reset the clock to $t=0$ for the discharge: $V_C=7.769e^{-0.5/1}=7.769\times e^{-0.5}=7.769\times 0.6065=4.71\text{ V}.$ Answer: (a) $\approx 7.77\text{ V}$, (b) $\approx 4.71\text{ V}$.

Set them equal: $5(1-e^{-t/\tau })=5e^{-t/\tau }.$ Divide by $5$: $1-e^{-t/\tau }=e^{-t/\tau }\Rightarrow 1=2e^{-t/\tau }\Rightarrow e^{-t/\tau }=\frac{1}{2}.$ $-\frac{t}{\tau }=\ln \frac{1}{2}=-0.693\Rightarrow t=0.693\tau =0.693\text{ s}.$ What it looks like: both curves pass through $V_C=2.5\text{ V}$ (exactly half of $5$) at that instant — the crossing point in the figure. This time, $0.693\tau =\tau \ln 2$, is the half-way time, a number worth memorising. Answer: they cross at $t\approx 0.693\text{ s}$, at $2.5\text{ V}$.

WHY not half the time: energy $\propto V^2$, so half-energy does not mean half-voltage. Link the two: $\frac{1}{2}C{V}_C^2=\frac{1}{2}\left(\frac{1}{2}C{V}_s^2\right)\Rightarrow V_C^2=\frac{1}{2}{V}_s^2\Rightarrow V_C=\frac{V_s}{\sqrt{2}}=0.7071V_s.$ Now find the time for $V_C=0.7071V_s$ during charging: $0.7071=1-e^{-t/\tau }\Rightarrow e^{-t/\tau }=0.2929\Rightarrow t=-\tau \ln (0.2929)=1.228\tau .$ Answer: half the energy is stored at $t\approx 1.23\tau$, later than the half-voltage time (which would be $\approx 0.693\tau$). See Capacitors and stored energy for the $\frac{1}{2}C{V}^2$ law.

The four moves you practised, in order ::: (1) compute $\tau =RC$, (2) evaluate $V_C$ with $t/\tau$, (3) invert with $\ln$ to solve for $t$, (4) chain stages by carrying the ending voltage as the next $V_0$.

Half-voltage time vs. half-energy time ::: half-voltage at $0.693\tau$; half-energy later, at $1.23\tau$, because energy $\propto V^2$.