Exercises — Series and parallel capacitors — derivations

WHAT: parallel → capacitances add. WHY: dono capacitors same do nodes ke across baithe hain (same voltage), toh charges bas saath mein pile up ho jaate hain. $C_{eq}=C_1+C_2=4+12=16\mu F$ Sanity check: sabse bade member ($12$) se bada hai — parallel ke liye correct hai.

WHAT: series → reciprocals add. Do-capacitor shortcut use karo $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$. WHY: same charge dono se guzarta hai, voltages add hote hain, toh $1/C$ values add hote hain. $C_{eq}=\frac{4\times 12}{4+12}=\frac{48}{16}=3\mu F$ Sanity check: sabse chote member ($4$) se bhi chota hai — series ke liye correct hai.

(a) Series shortcut $C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$: $C_{eq}=\frac{2\times 6}{2+6}=\frac{12}{8}=1.5\mu F$ Yeh formula kyun? Kyunki yeh series mein hain, same charge $Q$ dono se guzarta hai jabki unke voltages add hote hain ($V=V_1+V_2$). $V_1=Q/C_1$ aur $V_2=Q/C_2$ us sum mein daalne se $\frac{1}{C_{eq}}=\frac{1}{C_1}+\frac{1}{C_2}$ milta hai, jo do capacitors ke liye product-over-sum mein rearrange hota hai. Shortcut same-charge idea ka hi disguise hai. (b) Series mein har jagah same charge hota hai; use equivalent capacitor se nikalo: $Q=C_{eq}V=1.5\times 16=24\mu C$ Equivalent kyun? $C_{eq}$ by definition same voltage par same total charge draw karta hai, toh $Q=C_{eq}V$ hume woh single charge deta hai jo phir dono capacitors par baithta hai. (Units: $\mu F\times V=\mu C$.) (c) Har ek par $V=Q/C$ use karo (Capacitance and the relation Q = CV se): $V_1=\frac{24}{2}=12V,V_2=\frac{24}{6}=4V$ Kirchhoff's voltage law se check: $12+4=16V$ ✓. Chota capacitor ($C_1$) bada voltage leta hai, kyunki $V\propto 1/C$.

(a) Parallel → add karo: $C_{eq}=3+5=8\mu F$. Add kyun? Dono capacitors same do nodes ke across wired hain, toh unhe same voltage $V$ milta hai; unke charges $Q_1=C_{1}V$ aur $Q_2=C_{2}V$ add hote hain, $Q_{total}=(C_1+C_2)V$ dete hain — yaani capacitances add hoti hain. (b) $Q_{total}=C_{eq}V=8\times 20=160\mu C$. $C_{eq}$ kyun? By definition equivalent capacitor pore network ka charge applied voltage par store karta hai. (Units: $\mu F\times V=\mu C$.) (c) Har ek par same voltage; har branch par $Q=CV$ apply karo: $Q_1=3\times 20=60\mu C,Q_2=5\times 20=100\mu C$ Check (Charge conservation): $60+100=160\mu C$ ✓. Bada capacitor zyada charge leta hai ($Q\propto C$). (d) Energy stored in a capacitor se equivalent use karke energy. $C$ ko farads mein convert karo taaki answer joules mein aaye: $U=\frac{1}{2}{C}_{eq}{V}^2=\frac{1}{2}(8\times 10^{-6})(20{)}^2=1.6\times 10^{-3}J=1.6mJ$

Step 1 — innermost group reduce karo (parallel $C_2\parallel C_3$): $C_{23}=C_2+C_3=2+4=6\mu F$ Pehle kyun, aur add kyun? $C_2$ aur $C_3$ same nodes A aur B ke beech hang karte hain (figure dekho), toh woh voltage share karte hain aur unki capacitances add hoti hain (parallel rule). Is group ko pehle collapse karna puri cheez ko ek clean series pair mein badal deta hai. Step 2 — ab $C_1$ series mein $C_{23}$ ke saath: $C_{eq}=\frac{C_1{C}_{23}}{C_1+C_{23}}=\frac{4\times 6}{4+6}=\frac{24}{10}=2.4\mu F$ Product-over-sum kyun? $C_1$ aur block $C_{23}$ ab same charge carry karte hain aur voltages add karte hain — do-capacitor series shortcut apply hota hai. Step 3 — battery se total charge: $Q=C_{eq}V=2.4\times 24=57.6\mu C$ Kyun? Equivalent capacitor $24V$ par network ka total charge draw karta hai; aur kyunki $C_1$ aur block series mein hain, yahi same $Q$ $C_1$ par aur $C_{23}$ block par baithta hai. (Units: $\mu F\times V=\mu C$.) Step 4 — $V=Q/C$ se voltages: $V_1=\frac{57.6}{4}=14.4V,V_{23}=\frac{57.6}{6}=9.6V$ Check (Kirchhoff's voltage law): $14.4+9.6=24V$ ✓. Step 5 — parallel block ke andar charge split karo (dono $V_{23}=9.6V$ feel karte hain): $Q_2=2\times 9.6=19.2\mu C,Q_3=4\times 9.6=38.4\mu C$ Dono ke liye same $V_{23}$ kyun? Woh parallel hain — same do nodes A, B — toh har ek common $V_{23}$ ke saath $Q=CV$ obey karta hai. Check: $Q_2+Q_3=57.6\mu C=Q$ ✓ (block ka charge series charge ke barabar hai).

Dono cases mein battery voltage same hai, toh $U=\frac{1}{2}{C}_{eq}{V}^2$ use karo — energy directly $C_{eq}$ ke saath scale karti hai. Parallel: $C_{eq}^{\text{par}}=10+10=20\mu F$. Add kyun? Same do nodes → same voltage → capacitances add. $U_{\text{par}}=\frac{1}{2}(20\times 10^{-6})(100{)}^2=0.10J$ Series: $C_{eq}^{\text{ser}}=\frac{10\times 10}{10+10}=5\mu F$. Product-over-sum kyun? Same charge, adding voltages → reciprocals add. $U_{\text{ser}}=\frac{1}{2}(5\times 10^{-6})(100{)}^2=0.025J$ Ratio: $\frac{U_{\text{par}}}{U_{\text{ser}}}=\frac{20}{5}=4$ Parallel 4× zyada energy store karta hai. Sahi lagta hai: same voltage, chaar times bada equivalent capacitance ($20$ vs $5$). (Dono energies ne $C$ farads mein use kiya toh results joules mein hain.)

Step 1 — reciprocals add: $\frac{1}{C_{eq}}=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}=\frac{6+3+2}{6}=\frac{11}{6}(\mu F{)}^{-1}\Rightarrow C_{eq}=\frac{6}{11}\mu F\approx 0.545\mu F$ Reciprocals kyun? Teeno series mein hain → same charge har ek se guzarta hai, voltages add hote hain; $V_i=Q/C_i$ ko $V=\sum V_i$ mein substitute karne se $\frac{1}{C_{eq}}=\sum \frac{1}{C_i}$ milta hai. Step 2 — common charge: $Q=C_{eq}V=\frac{6}{11}\times 11=6\mu C$ Kyun? Equivalent capacitor woh single charge set karta hai jo series mein har member par identical hota hai. (Units: $\mu F\times V=\mu C$.) Step 3 — $V=Q/C$ se individual voltages: $V_1=\frac{6}{1}=6V,V_2=\frac{6}{2}=3V,V_3=\frac{6}{3}=2V$ Check: $6+3+2=11V$ ✓. Sabse chota capacitor $C_1=1\mu F$ sabse zyada voltage leta hai ($6V$) — bilkul isliye kyunki $V\propto 1/C$.

$C_{eq}=\frac{C_1{C}_2}{C_1+C_2}$ use karo (series shortcut — valid hai kyunki pair ek charge share karta hai adding voltages ke saath). Jab $C_2\to 0$: numerator denominator se tezi se $\to 0$ hota hai, toh $C_{eq}\to 0$. Vanishing capacitor ek badi "difficulty" hai ($1/C_2\to \infty$); series difficulties add hoti hain, toh puri chain charge karna impossible ho jaata hai — $C_{eq}\to 0$. Jab $C_2\to \infty$: top aur bottom ko $C_2$ se divide karo: $C_{eq}=\frac{C_1}{C_1/C_2+1}\stackrel{C_2\to \infty }{\to }\frac{C_1}{0+1}=C_1=5\mu F$ Ek giant $C_2$ zero difficulty contribute karta hai ($1/C_2\to 0$); woh ek plain wire ki tarah behave karta hai, toh sirf $C_1$ chain ko limit karta hai — $C_{eq}\to C_1$. Lesson: series mein, sabse chota capacitor $C_{eq}$ ko dominate karta hai.

(a) Voltage rating series force karta hai. Ek $2\mu F$ sirf $50V$ survive karta hai, toh $150V$ hold karne ke liye voltage split honi chahiye, yaani series. Identical capacitors ke series mein voltage equally split hota hai, toh hume chahiye $n=\frac{150V}{50V}=3\text{capacitors in series per string,}$ aur har ek phir exactly $150/3=50V$ dekhta hai (rating par, safe). String ki capacitance: $\frac{1}{C_{\text{string}}}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=\frac{3}{2}\Rightarrow C_{\text{string}}=\frac{2}{3}\mu F\approx 0.667\mu F.$ 3 kyun aur kam kyun nahi? $n=2$ har capacitor par $75V$ dalega — $50V$ rating se zyada. $n=3$ minimum hai jo har capacitor ko $50V$ ya usse neeche rakhta hai. (b) Parallel capacitance badhata hai, aur yeh unbounded hai. Strings ko parallel mein add karne se koi bhi capacitor jo voltage dekhta hai woh nahi badalta (har string phir bhi $150V$ ko $50/50/50$ mein split karti hai), toh yeh hamesha safe hai, aur parallel capacitances add hoti hain: $\boxed{C_{\text{block}}(m)=m\times \frac{2}{3}\mu F.}$ Yeh $m$ ke saath linearly grow karta hai bina kisi upper bound ke — unlimited capacitors ke saath tum block capacitance jitni chahte ho utni bada sakte ho. "Largest possible" ka koi finite answer nahi hai; design balki capacitors ko capacitance ke liye trade karta hai ($m$ strings of 3 = $3m$ capacitors $\frac{2m}{3}\mu F$ dete hain). (c) Sirf voltage rating meet karne ke liye minimum count: ek single string of $n=3$ capacitors, $C_{\text{string}}=\frac{2}{3}\mu F$ deta hai — total 3 capacitors.

1. $4\mu F$ aur $12\mu F$ ki series? ::: $3\mu F$ (product over sum).
2. $16V$ ke across series mein $C_1=2,C_2=6\mu F$ ke saath, kaun zyada voltage leta hai? ::: $C_1$, chota wala, $12V$ vs $4V$.
3. Fixed battery: parallel ya series zyada energy store karta hai? ::: Parallel — bada $C_{eq}$, aur $U=\frac{1}{2}{C}_{eq}{V}^2$.
4. Series mein $C_2\to \infty$, $C_1=5\mu F$ fixed: $C_{eq}\to$? ::: $C_1=5\mu F$ (huge $C_2$ wire ki tarah act karta hai).