Series and parallel capacitors — derivations: Visual walkthrough

1.8.12 · D2 · Physics › Electromagnetism › Series and parallel capacitors — derivations

Step 0 — Capacitor kya hota hai, seedha aur saaf

KYA HAI. Capacitor bas do metal plates hain jo ek doosre ke saamne hain aur unke beech ek gap hai. Pehle sign convention fix karte hain. == $Q$ ek positive magnitude hai — "ek plate pe kitne coulombs baithe hain", hamesha $Q \geq 0$ ==. Capacitor ko battery se jodo: ek plate electrons kho deti hai aur positively charged ho jaati hai, jise hum $+ Q$ likhte hain; doosri plate exactly wohi electrons le leti hai aur negatively charged ho jaati hai, jise $- Q$ likhte hain. Toh dono plates signed charges $+ Q$ aur $- Q$ carry karti hain (equal size, opposite sign), jabki akela $Q$ hamesha unsigned magnitude matlab hai — aage laga $\pm$ sirf batata hai ki kaun si plate ki baat ho rahi hai. Plates ab alag alag electric potentials pe hain — unke beech ek "voltage" $V$ hai, jaise height ka difference.

YEH TEEN SYMBOLS KYUN. Poore gadget ko teen quantities describe karti hain:

$Q$ = har plate pe parked charge ki magnitude (coulombs, C, $Q \geq 0$ ).
$V$ = gap ke across potential difference (volts, V) — ise pressure ki tarah socho jo charge ko alag dhakelta hai.
$C$ = capacitance — plate kitna charge store karti hai pressure ke har volt pe. Bada $C$ = zyada greedy plate.

Ye teen ek single relation se bande hain jis par hum poore page rely karte hain, Q = CV se:

$charge magnitude Q = greed factor C \times pressure across gap V$

PICTURE. Ek plate signed charge $+ Q$ (amber) hold karti hai, saamne wali plate $- Q$ (cyan) hold karti hai; common magnitude $Q$ hai. Gap ke across white arrow voltage $V$ hai.

Figure — Series and parallel capacitors — derivations

Step 1 — Goal: ek capacitor jo poore network ki nakal kare

KYA HAI. Hum capacitors ki ek ulajhi hui network phenkna chahte hain aur ek single replacement $C_{e q}$ daalna chahte hain jise battery original se alag na bata sake — same total charge pull out ho, same total voltage terminals ke across ho.

KYUN. "Equivalent" ko define karna zaroori hai, warna word ka koi matlab nahi. Pehle nayi symbol $Q_{total}$ ko pin down karo: yeh wo saara charge hai jo battery ne network ko supply kiya, yani individual capacitors ko deliver kiye gaye charges ka sum,

$Q_{total} \equiv Q_{1} + Q_{2} + \dots + Q_{n}$

(kisi bhi ek $Q_{i}$ se alag, jo bas ek capacitor ka share hai). Isi tarah $V_{total}$ battery ka terminals ke across voltage hai. Phir hum terminals A aur B pe behaviour se equivalence define karte hain:

$C_{e q} \equiv \frac{Q _{total}}{V _{total}}$

Agar koi black box dono $Q_{total}$ aur $V_{total}$ match kare, toh woh baahri duniya ke liye wahi network hai.

PICTURE. Left: terminals A aur B ke beech capacitors ka ek messy box. Right: ek mota $C_{e q}$ capacitor. Battery dono se identical charge-supplied aur voltage across read karti hai.

KYA HAI. Parallel mein, har capacitor ki left plate wire A se milti hai aur har right plate wire B se milti hai. Koi capacitor doosre ke "downstream" nahi hai; sab ek hi rail pe kapdon ki tarah side by side latkte hain.

SAME VOLTAGE KYUN. Wire ek conductor hai jo ek single potential pe hota hai (ek equipotential — plain wire ke along koi pressure difference nahi). Toh node A ek height pe hai, node B doosri height pe, aur har capacitor exactly same do heights ko bridge karta hai:

$V_{1} = V_{2} = V_{3} = V (the battery’s own voltage)$

Har $V_{i}$ ko $V$ ke barabar force kiya jaata hai kyunki har capacitor literally same do nodes ko connect karta hai.

PICTURE. Teen capacitors, saari left plates cyan A-rail pe, saari right plates amber B-rail pe. Identical white $V$ arrow teeno ke across copy-paste hai.

Step 3 — Parallel: charges node pe add ho jaate hain

KYA HAI. Har capacitor pe same $V$ ke saath $Q = C V$ apply karo, phir collect karo ki battery ne kya supply kiya hoga.

KYUN. Charge conservation se node A pe, battery se nikla charge teen left plates pe split ho jaata hai. Kuch bhi lost nahi hota, toh $Q_{total}$ (Step 1 mein $Q_{1} + Q_{2} + Q_{3}$ define kiya) sum hai:

$Q_{total} = C_{1} V Q_{1} + C_{2} V Q_{2} + C_{3} V Q_{3} = (C_{1} + C_{2} + C_{3}) V$

Last bracket padho: kyunki $V$ common hai, woh factor out ho jaata hai, aur jo bacha woh plain sum of capacitances hai jo use multiply kar raha hai. $Q_{total} = C_{e q} V$ se compare karo aur $V$ cancel ho jaata hai:

$C_{e q} = C_{1} + C_{2} + \dots + C_{n}$

Har term $C_{i}$ ek alag greedy plate hai; unhe side by side rakhne se charge ko zyada plate area milta hai baithe rehne ke liye, toh total greed add ho jaati hai.

PICTURE. Battery ka charge stream node A pe teen coloured streams mein fork hota hai jo teen plates pe land karte hain; B pe wapas merge dikhata hai $Q_{total} = Q_{1} + Q_{2} + Q_{3}$ .

Step 4 — Series wiring: metal ka trapped island

KYA HAI. Series mein capacitors ek chain banate hain: $C_{1}$ ki right plate sirf $C_{2}$ ki left plate se judi hoti hai, aur aage bhi aise hi. Woh connecting piece kisi aur cheez ko touch nahi karta — yeh conductor ka ek isolated island hai.

SAME CHARGE KYUN. Island electrically neutral shuru hua. Jab battery $C_{1}$ ki far-left plate pe $+ Q$ force karti hai, woh $+ Q$ $C_{1}$ ki right plate pe $- Q$ kheench leta hai. Lekin island ( $C_{1}$ ki right plate plus $C_{2}$ ki left plate) net-neutral rehna chahiye — yeh kisi charge source se connected nahi hai. Toh agar island ke ek end pe $- Q$ aata hai, toh doosre end pe $+ Q$ force ho jaata hai. Woh $+ Q$ $C_{2}$ ka charge ban jaata hai. Isliye wahi $Q$ har capacitor mein se guzarta hai:

$Q_{1} = Q_{2} = Q_{3} = Q$

PICTURE. Chain jisme middle conductor ko amber mein circle kiya gaya hai, "isolated island, net charge = 0" label ke saath, induced $- Q$ / $+ Q$ ka balance dikhata hai.

Step 5 — Series: voltages poore loop mein add ho jaate hain

KYA HAI. Har capacitor ko uska apna voltage do $Q = C V$ rearrange karke, phir circuit ke around drops add karte hue chalo.

KYUN. $Q = C V$ ko $V = Q / C$ mein rearrange karna yeh jawaab deta hai ki "is capacitor ko charge $Q$ hold karne ke liye kitna pressure chahiye?" Common $Q$ ke saath:

$V_{1} = \frac{Q}{C _{1}}, V_{2} = \frac{Q}{C _{2}}, V_{3} = \frac{Q}{C _{3}}$

Kirchhoff's voltage law se, agar aap ek full loop chalte ho — battery ke $+$ terminal se shuru karo, $C_{1}$ ke across step down karo, phir $C_{2}$ , phir $C_{3}$ , aur wapas wahan pahuncho jahan se chale — ups aur downs cancel hone chahiye. Battery ne aapko $V$ raise kiya; teen capacitors ko aapko total mein exactly $V$ drop karna chahiye. Neeche ki picture us loop ko ek arrow ke roop mein draw karti hai jisme har drop $V_{1}, V_{2}, V_{3}$ exactly usi step pe tag kiya gaya hai jahan woh hota hai, toh:

$V = \frac{Q}{C _{1}} + \frac{Q}{C _{2}} + \frac{Q}{C _{3}} = Q (\frac{1}{C _{1}} + \frac{1}{C _{2}} + \frac{1}{C _{3}})$

Common $Q$ factor out ho jaata hai; bracket ke andar capacitance ke reciprocals hain — yeh har capacitor ki "hardness to fill" hai. Ab $V = Q / C_{e q}$ use karo aur common $Q$ cancel karo:

$\frac{1}{C _{e q}} = \frac{1}{C _{1}} + \frac{1}{C _{2}} + \dots + \frac{1}{C _{n}}$

PICTURE. Loop explicitly drawn (circuit ke around curved amber arrow), aur potential staircase: left pe $V$ se shuru karo, $V_{1}$ drop karo, phir $V_{2}$ , phir $V_{3}$ , grounded reference wire pe $0 V$ pe land karo — total fall battery ki height ke barabar.

Step 6 — Edge & degenerate cases (kabhi surprise mat ho)

KYA HAI. Formulas ko unki limits tak push karo taaki kuch bhi tumhe ambush na kar sake.

KYUN. Jo formula sirf "nice" middle mein trust karte ho woh ek aisa formula hai jo aap samjhe nahi. Extremes check karo.

Series extremes ( $\frac{1}{C _{e q}} = \sum \frac{1}{C _{i}}$ use karo):

Ek capacitor bahut chota ( $C_{1} ≪ C_{2}$ ): $\frac{1}{C _{e q}} \approx \frac{1}{C _{1}}$ , toh $C_{e q} \approx C_{1}$ (aur hamesha $C_{1}$ se strictly neeche). Chota capacitor dominate karta hai — sabse kamzor link chain control karta hai.
Dono equal ( $C_{1} = C_{2} = C$ ): $C_{e q} = C /2$ . Gaps stack karne se effective thickness double ho jaati hai → half capacitance.
Chain mein ek short ( $C_{2} \to \infty$ , almost ideal fat wire ki tarah treat karo): $1/ C_{2} \to 0$ drop out ho jaata hai, $C_{e q} \to C_{1}$ . Ek "wire" ek aisa capacitor ki tarah behave karta hai jo itna greedy hai ki essentially koi voltage nahi leta. ✓
Open circuit / missing capacitor ( $C_{2} \to 0$ ): ab $1/ C_{2} \to \infty$ , toh $\frac{1}{C _{e q}} \to \infty$ aur isliye $C_{e q} \to 0$ . Poori chain kuch nahi store karti. Yahi ambush case hai: ek single broken (zero-capacitance) link pipe mein ek gap hai — charge march nahi kar sakta, toh poora series combination khatam ho jaata hai. Series mein, sabse chota member hamesha $C_{e q}$ ko neeche kheenchta hai, aur zero use poori tarah zero tak le jaata hai.

Parallel extremes ( $C_{e q} = \sum C_{i}$ use karo):

Ek missing member ( $C_{2} \to 0$ ): woh simply kuch add nahi karta, $C_{e q} = C_{1} + C_{3}$ . Parallel mein ek dead branch harmless hai — baaki sab chalte rehte hain.
Ek member bahut bada ( $C_{2} \to \infty$ ): tab $C_{e q} \to \infty$ bhi. Terminals ke across ek extremely greedy plate reference rail ke short ki tarah behave karti hai — woh kisi bhi voltage pe near-unlimited charge accept karti hai, finite doosron ko swamp kar deti hai. Yeh exactly series-open case ka dual hai: parallel mein sabse bada member dominate karta hai aur ek bada value $C_{e q}$ ko blow up karta hai.

Yaad rakhne ke guardrails: series hamesha sabse chhote member se neeche land karta hai (zero agar koi member zero ho); parallel hamesha sabse bade se upar land karta hai (unbounded agar koi member bina limit ke bade).

PICTURE. Do number lines: parallel result $max (C_{i})$ ke upar pinned, series result $min (C_{i})$ ke neeche pinned, ek worked $2 μ F$ & $3 μ F$ pair marked ke saath ( $5$ aur $1.2$ ).

Ek-picture summary

PICTURE. Left half = parallel: shared $V$ arrow, charges $Q_{1}, Q_{2}, Q_{3}$ fan out karte hue, result $C_{e q} = \sum C_{i}$ . Right half = series: shared $Q$ through march karta hua, voltages $V_{1}, V_{2}, V_{3}$ staircase ke roop mein stack hote hue, result $\frac{1}{C _{e q}} = \sum \frac{1}{C _{i}}$ . Ek canvas, dono stories.

Related energy accounting energy stored in a capacitor mein hai; ek slab insert karna har $C_{i}$ ko dielectrics ke according change karta hai — lekin upar ke dono rules kabhi nahi badlte.

Parallel mein kaun si quantity shared hoti hai?

Har capacitor ke across voltage

V

Series mein kaun si quantity shared hoti hai?

Har capacitor pe charge

Q

Series mein

C_{e q}

sabse chote member se neeche kyun hota hai?

Reciprocals add karne se sirf

1/ C_{e q}

badhta hai, toh

C_{e q}

har term se neeche hi girega.

Series mein kaun sa capacitor sabse zyada voltage leta hai?

Sabse chota wala, kyunki

V = Q / C

mein

Q

common hai.

Agar series chain mein ek capacitor zero (open) ho toh kya hoga?

1/ C_{e q} \to \infty

, toh

C_{e q} \to 0

— poori chain kuch store nahi karti.

Agar parallel bank mein ek capacitor bahut bada ho toh kya hoga?

C_{e q} \to \infty

— woh terminals ke across ek short ki tarah act karta hai.

Recall Feynman retelling — poori walk plain words mein

Do metal plates ko charge ke liye ek bucket socho; voltage paani ka pressure hai aur $C$ bucket kitna bada hai woh hai. Parallel: saare buckets ko same do wires pe latka do. Kyunki wire flat hai (ek potential), har bucket ko same pressure milta hai. Unhe us pressure pe bharo aur charge ki maatrayein simply pile ho jaati hain — toh bucket-sizes add ho jaate hain: $C_{e q} = \sum C_{i}$ . Side by side zyada buckets = ek bada bucket. Agar ek bucket enormously bada hota, poori rail near-unlimited paani le sakti — ek short. Series: buckets ko ek pipe mein stack karo. Do buckets ke beech metal ka lump sealed off hai aur zero charge se shuru hua, toh jo charge ek side pe baitha hai use doosri side pe mirror hona hi padega — wahi same $Q$ har bucket se force hoke guzarta hai. Har bucket phir apna pressure demand karta hai $V = Q / C$ , aur woh pressures stack up hoke battery ke barabar ho jaate hain. Kyunki ek chota bucket bada pressure demand karta hai, aur pressures add ho jaate hain, poora stack bharna mushkil hota hai — total capacitance sabse chote bucket se neeche gir jaata hai, aur yeh "difficulties" $1/ C$ add hoti hain: $\frac{1}{C _{e q}} = \sum \frac{1}{C _{i}}$ . Aur agar ek link toot jaaye (zero bucket), pipe block ho jaata hai — poori chain kuch store nahi karti. Do lines, do pictures, dono formulas — kuch yaad nahi karna.