Series and parallel capacitors — derivations: Worked examples

1.8.12 · D3 · Physics › Electromagnetism › Series and parallel capacitors — derivations

Yeh page ek drill in coverage hai. Parent note ne do rules banaye the; yahan hum har tarah ke questions dhundte hain jo woh produce kar sakte hain — including weird edge cases (ek zero-gap capacitor, ek infinitely large wala, ek open branch) jo exams mein log galat kar dete hain.

Kuch bhi compute karne se pehle, hum saare cases ka ek map banate hain. Phir har worked example ko us map-cell ke saath tag kiya gaya hai jo woh cover karta hai, toh end tak aapne har cell ko kam se kam ek baar hit kar liya hoga.

Scenario matrix

Har row ek class of situation hai jo yeh topic aap par throw kar sakta hai. Last column us example ka naam deta hai jo use cover karta hai.

#	Case class	Kya cheez tricky banati hai	Covered by
A	Pure parallel — total aur split charges nikalo	kaun sa capacitor zyada charge leta hai	Ex 1
B	Pure series — charge nikalo aur voltages split karo	kaun sa capacitor zyada voltage leta hai	Ex 2
C	Mixed (series + parallel) reduction	pehle innermost group reduce karo	Ex 3
D	Combination mein Energy	energy charge ki tarah share nahi hoti	Ex 4
E	Degenerate: short-circuit plate ( $C \to \infty$ )	ek bada cap: series mein wire, parallel mein short	Ex 5
F	Degenerate: open gap ( $C \to 0$ )	ek chota cap: series mein break, parallel mein dead branch	Ex 5
G	Limiting: equal capacitors, $n$ of them	$n$ ke saath series/parallel scaling	Ex 6
H	Real-world word problem	words ko → circuit mein translate karo	Ex 7
I	Exam twist — charge redistribution reconnecting ke baad	charge conservation, naya common $V$	Ex 8
J	Dielectric slab ek $C$ ko change karta hai	network ko recompute karo	Ex 9

Cells D, E, F, I woh hain jo students skip karte hain — hum inhe poora attention dete hain.

Example 1 — Cell A: pure parallel, charge split

Worked example Do capacitors side by side

$C_{1} = 3 μ F$ , $C_{2} = 6 μ F$ parallel mein $V = 10 V$ ke across. $C_{e q}$ , total charge, aur har capacitor ka charge nikalo.

Forecast: Donon par same voltage. Bada ya chota capacitor zyada charge lega? Aage padhne se pehle $Q_{t o t a l}$ ka ek number guess karo.

Step 1 — Combine. $C_{e q} = C_{1} + C_{2} = 3 + 6 = 9 μ F$ . Yeh step kyun? Parallel ka matlab same nodes hai → capacitances directly add hoti hain.

Step 2 — Total charge. $Q_{t o t a l} = C_{e q} V = 9 \times 10 = 90 μ C$ . Yeh step kyun? Equivalent capacitor apni definition ke hisaab se battery se utna hi total charge kheenchta hai.

Step 3 — Split. Dono par same $V = 10 V$ , toh $Q_{1} = C_{1} V = 30 μ C$ , $Q_{2} = C_{2} V = 60 μ C$ . Yeh step kyun? Har branch independently shared voltage ke saath $Q = C V$ obey karta hai.

Verify: $Q_{1} + Q_{2} = 30 + 60 = 90 = Q_{t o t a l}$ ✓. Bada cap ( $C_{2}$ ) ne zyada charge liya: $Q \propto C$ jab $V$ common ho. Units: $μ F \times V = μ C$ ✓.

Neeche figure — aise padho: do branches ek hi pair of rails ke beech latki hain (visually yahi "parallel" ka matlab hai), toh donon ko poora $10 V$ milta hai. Pink arrow follow karo: wider capacitor $C_{2}$ bada charge ( $60 μ C$ ) collect karta hai, $Q \propto C$ at fixed voltage confirm karta hai.

Figure — Series and parallel capacitors — derivations

Example 2 — Cell B: pure series, voltage split

Worked example Do capacitors ek chain mein

$C_{1} = 2 μ F$ , $C_{2} = 4 μ F$ series mein $V = 12 V$ ke across. $C_{e q}$ , common charge, aur har voltage nikalo.

Forecast: Dono par same charge. Bada voltage kaun leta hai — chota ya bada capacitor? Abhi ek answer decide karo.

Step 1 — Combine. $\frac{1}{C _{e q}} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4} \Rightarrow C_{e q} = \frac{4}{3} \approx 1.33 μ F$ . Yeh step kyun? Series → reciprocals add hote hain (voltages add hote hain, charge common hai).

Step 2 — Common charge. $Q = C_{e q} V = \frac{4}{3} \times 12 = 16 μ C$ — dono plates par. Yeh step kyun? Trapped middle island identical $Q$ sabko force karta hai; equivalent capacitor woh $Q$ deta hai.

Step 3 — Voltages split karo. $V_{1} = \frac{Q}{C _{1}} = \frac{16}{2} = 8 V$ , $V_{2} = \frac{16}{4} = 4 V$ . Yeh step kyun? Common $Q$ ke saath har capacitor par $V = Q / C$ apply karo.

Verify: $V_{1} + V_{2} = 8 + 4 = 12 = V$ ✓ (loop ke around KVL). Chote cap $C_{1}$ ne bada voltage liya: $V \propto 1/ C$ .

Neeche figure — aise padho: yellow arrow do caps ke beech "isolated island" point karta hai — woh trapped conductor hi reason hai ki same $Q = 16 μ C$ dono se guzarta hai. Phir blue callout dekho: narrower cap $C_{1}$ bada voltage label ( $8 V$ ) dikhata hai, $V \propto 1/ C$ ka geometric proof hai.

Example 3 — Cell C: mixed reduction

Worked example Series-with-a-parallel-block

$C_{1} = 2 μ F$ series mein parallel pair $C_{2} = 2 μ F$ , $C_{3} = 4 μ F$ ke saath. Battery $= 18 V$ . Har charge aur voltage nikalo.

Forecast: Pehle kya reduce karte ho — series part ya parallel block? Aur block ke andar $C_{2}$ ya $C_{3}$ zyada charge rakhega?

Step 1 — Inner parallel. $C_{23} = C_{2} + C_{3} = 2 + 4 = 6 μ F$ . Yeh step kyun? Hamesha pehle innermost group collapse karo taaki baaki network simple series/parallel ban jaye.

Step 2 — Outer series. $\frac{1}{C _{e q}} = \frac{1}{C _{1}} + \frac{1}{C _{23}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6} \Rightarrow C_{e q} = 1.5 μ F$ . Yeh step kyun? $C_{1}$ aur block $C_{23}$ ab ek single series chain mein hain.

Step 3 — Series charge. $Q = C_{e q} V = 1.5 \times 18 = 27 μ C$ , $C_{1}$ aur poore $C_{23}$ block par. Yeh step kyun? Series → $C_{1}$ aur block se same charge.

Step 4 — Series voltages split karo. $V_{1} = \frac{27}{2} = 13.5 V$ , $V_{23} = \frac{27}{6} = 4.5 V$ . Yeh step kyun? Har series member par $V = Q / C$ . Check: $13.5 + 4.5 = 18$ ✓.

Step 5 — Block ke andar. $C_{2}$ aur $C_{3}$ par same $V_{23} = 4.5 V$ : $Q_{2} = 2 \times 4.5 = 9 μ C$ , $Q_{3} = 4 \times 4.5 = 18 μ C$ . Yeh step kyun? Parallel → shared voltage; har ek $Q = C V$ obey karta hai.

Verify: $Q_{2} + Q_{3} = 9 + 18 = 27 = Q$ (block ka total charge series charge ke barabar hai) ✓.

Neeche figure — aise padho: right side ka boxed region collapsed block $C_{23} = 6 μ F$ hai; ise $C_{1}$ ke saath series mein ek single capacitor samjho. Notice karo $27 μ C$ label $C_{1}$ se hokar box mein unchanged jaata hai (series = same charge), phir box ke andar $C_{2}$ aur $C_{3}$ ke beech split hota hai jo dono same $4.5 V$ carry karte hain.

Example 4 — Cell D: energy across a combination

Worked example Energy charge ki tarah split NAHI hoti

Same as Ex 2: $C_{1} = 2 μ F$ , $C_{2} = 4 μ F$ series mein $12 V$ ke across. Har mein aur total mein stored energy nikalo.

Forecast: Kaun sa capacitor zyada energy store karta hai — bade voltage wala ya badi capacitance wala?

Step 1 — Energy formula yaad karo. Energy $U = \frac{1}{2} C V^{2} = \frac{Q ^{2}}{2 C} = \frac{1}{2} Q V$ . Yeh step kyun? Hume stored-energy tool chahiye; in teeno forms mein se koi bhi kaam karta hai. Kyunki series mein $Q$ common hai, $U = \frac{Q ^{2}}{2 C}$ sabse clean hai.

Step 2 — Per capacitor ( $Q = 16 μ C$ Ex 2 se use karke): $U_{1} = \frac{Q ^{2}}{2 C _{1}} = \frac{( 16 ) ^{2}}{2 \cdot 2} = 64 μ J, U_{2} = \frac{( 16 ) ^{2}}{2 \cdot 4} = 32 μ J .$ Yeh step kyun? Same $Q$ , toh $U \propto 1/ C$ : chota cap zyada energy store karta hai.

Step 3 — Total. $U_{t o t} = 64 + 32 = 96 μ J$ . Yeh step kyun? Series parts ki energies add hoti hain.

Verify: Equivalent capacitor se compute karo: $U = \frac{1}{2} C_{e q} V^{2} = \frac{1}{2} \cdot \frac{4}{3} \cdot 1 2^{2} = 96 μ J$ ✓. (Units: $μ F \cdot V^{2} = μ J$ .) Chota cap $C_{1}$ zyada energy store karta hai — uske bade voltage ko mirror karta hai.

Example 5 — Cells E & F: degenerate capacitors (series AUR parallel)

Numbers se pehle, notation of extremes par ek quick word, kyunki hum ise aage heavily use karne wale hain.

Definition Yahan

$\to$ , $\infty$ aur $0^{+}$ ka matlab kya hai

Arrow $\to$ ka matlab hai "approaches" — jaise $C \to \infty$ ka matlab hai "let $C$ grow larger and larger, without limit". Hum kabhi actually infinity nahi pahunchte; hum dekhte hain ki formula kahan jaata hai.
$\infty$ ("infinity") ek number nahi hai balki "arbitrarily large" ka shorthand hai. Ek capacitor jiske plates almost touching hain (near-zero gap) ka enormous $C$ hota hai, toh hum ise $C \to \infty$ model karte hain.
$0^{+}$ ka matlab hai "zero from above" — ek positive number jo zero ki taraf shrink ho raha hai (ek almost-infinite gap wala capacitor). Hum $0$ ki jagah $0^{+}$ likhte hain yeh signal karne ke liye ki value positive rehti hai, toh isse divide karne par ek large positive result milta hai, sign ambiguity nahi.
Hum jo rule use karte hain: $\frac{1}{\infty} \to 0$ (ek bada denominator fraction ko vanish karta hai) aur $\frac{1}{0 ^{+}} \to \infty$ (ek tiny positive denominator fraction ko blow up karta hai).

Ek capacitor jiske plates touching hain (zero gap) ka $C \to \infty$ hota hai; ek jiska infinite gap / broken plate hai uska $C \to 0^{+}$ hota hai. Hume dikhana hai ki dono wiring styles mein kya hota hai — series ka behavior parallel se opposite hai.

Worked example Do extreme members, series mein test kiye gaye

Ek chain: $C_{1} = 2 μ F$ series mein $C_{2}$ ke saath. (E) $C_{2} \to \infty$ hone do (ek shorted capacitor). (F) $C_{2} \to 0^{+}$ hone do (ek open gap). Har case mein $C_{e q}$ nikalo.

Forecast: Ek bada capacitor charge ko barely resist karta hai — kya yeh wire ki tarah act karta hai ya break ki tarah? Aur ek zero capacitor?

Step 1 — Case E, $C_{2} \to \infty$ . $\frac{1}{C _{e q}} = \frac{1}{2} + \frac{1}{\infty} \to \frac{1}{2} + 0 = \frac{1}{2}$ , toh $C_{e q} \to 2 μ F = C_{1}$ . Yeh step kyun? $1/\infty \to 0$ : ek giant series capacitor koi "difficulty" contribute nahi karta, yani yeh plain wire ki tarah behave karta hai. Network sirf $C_{1}$ hai.

Step 2 — Case F, $C_{2} \to 0^{+}$ . $\frac{1}{C _{e q}} = \frac{1}{2} + \frac{1}{0 ^{+}} \to \infty$ , toh $C_{e q} \to 0$ . Yeh step kyun? Ek vanishing series capacitor ek open gap hai — koi charge chain se guzar nahi sakta, toh poora series combo kuch store nahi karta.

Verify (sanity): Series mein, $C_{e q}$ hamesha smallest member se neeche hoti hai. Case E: smallest finite member $2$ hai, aur near-infinite wala add karne par hum essentially $2$ par hi rehte hain ✓. Case F: smallest member $\approx 0$ hai, aur sach mein $C_{e q} \to 0$ ✓.

Worked example SAME extremes, ab parallel mein

Ab $C_{1} = 2 μ F$ ko parallel mein $C_{2}$ ke saath rakkho. (E) $C_{2} \to \infty$ . (F) $C_{2} \to 0^{+}$ . Har case mein $C_{e q}$ nikalo.

Forecast: Parallel capacitances ko directly add karta hai. Predict karo kaun sa extreme dominate karta hai aur kaun sa vanish hota hai — roles series case ke versus swap hone chahiye.

Step 1 — Case E, $C_{2} \to \infty$ (parallel). $C_{e q} = C_{1} + C_{2} = 2 + \infty \to \infty$ . Yeh step kyun? Parallel add karta hai. Do nodes par rakha ek infinite capacitor kisi bhi charge ko no voltage drop ke saath absorb kar sakta hai: kyunki $V = Q / C$ aur $C \to \infty$ , kisi bhi finite $Q$ ke liye uska apna $V \to 0$ . Toh yeh un do nodes ke beech zero-voltage-drop link (ek short) ki tarah behave karta hai, aur pair ki capacitance blow up ho jaati hai. Precise statement note karo: yeh us branch ke across zero voltage enforce karta hai, "jo chahe voltage le sakta hai" nahi. Series case se compare karo, jahan same $C_{2} \to \infty$ ek harmless wire tha.

Step 2 — Case F, $C_{2} \to 0^{+}$ (parallel). $C_{e q} = 2 + 0 = 2 μ F = C_{1}$ . Yeh step kyun? Ek vanishing parallel branch koi charge store nahi karta; yeh ek dead / open branch hai jise aap erase kar sakte ho. Network sirf $C_{1}$ hai. Series case se compare karo, jahan same $C_{2} \to 0^{+}$ ne poori chain ko kill kar diya tha.

Verify (sanity): Parallel mein, $C_{e q}$ hamesha largest member se upar hoti hai. Case E: ek near-infinite wala add karne par $C_{e q} \to \infty$ drive hota hai ✓ (sab se upar). Case F: ek $\approx 0$ member add karne par largest, $2 μ F$ bachta hai ✓.

Recall Ek table mein charo degenerate corners

Series, $C \to \infty$ ::: plain wire ki tarah act karta hai; $C_{e q}$ isse affect nahi hoti. Series, $C \to 0^{+}$ ::: open break ki tarah act karta hai; poori chain ko kill karta hai ( $C_{e q} \to 0$ ). Parallel, $C \to \infty$ ::: us branch ke across zero voltage enforce karta hai (ek short); $C_{e q} \to \infty$ . Parallel, $C \to 0^{+}$ ::: dead/open branch ki tarah act karta hai; ise delete karo, $C_{e q}$ unchanged.

Example 6 — Cell G: $n$ identical capacitors (limiting behaviour)

Example 7 — Cell H: real-world word problem

Worked example Camera flash bank

Ek camera flash ko $12 μ F$ storage capacitor chahiye jo $300 V$ par charged ho, lekin aapke paas sirf $4 μ F$ units hain jo $300 V$ rated hain. Kitne, aur kaise wire karke, $12 μ F$ reach karte hain? Kitni energy stored hai?

Forecast: Total capacitance badhane ke liye parallel jaate ho ya series? Kitne $4 μ F$ units $12 μ F$ banate hain?

Step 1 — Topology choose karo. Badi capacitance chahiye → parallel (parallel add karta hai). Yeh step kyun? Series capacitance shrink karta; parallel ise grow karta hai. Har unit tab full $300 V$ bhi dekhta hai, uski rating se match karta hai.

Step 2 — Count karo. $n \times 4 = 12 \Rightarrow n = 3$ units parallel mein. Yeh step kyun? $n$ equal caps ka parallel $n C$ hota hai (Ex 6).

Step 3 — Energy. $U = \frac{1}{2} C_{e q} V^{2} = \frac{1}{2} \times (12 \times 1 0^{- 6}) \times 30 0^{2} = 0.54 J$ . Yeh step kyun? Total stored energy equivalent capacitance ko applied voltage par use karta hai.

Verify: Har unit store karta hai $\frac{1}{2} \times (4 \times 1 0^{- 6}) \times 30 0^{2} = 0.18 J$ ; teeno $3 \times 0.18 = 0.54 J$ ✓. Voltage rating respected ( $300 V$ har ek) ✓.

Example 8 — Cell I: charge redistribution (exam twist)

Classic trap: ek capacitor charge karo, battery disconnect karo, phir ise uncharged capacitor se wire karo. Ab koi battery voltage fix nahi kar rahi — instead charge conserve hota hai aur dono ek naya common voltage dhundte hain.

Worked example Do capacitors ke beech charge share karna

$C_{1} = 2 μ F$ $V_{0} = 10 V$ par charged. Battery remove ki. Phir $C_{1}$ ko uncharged $C_{2} = 3 μ F$ ke saath parallel mein connect kiya. Final voltage aur final charges nikalo.

Forecast: Total charge nahi badal sakta. Voltage $0$ aur $10 V$ ke beech koi value par settle hoga — guess karo $5 V$ se upar ya neeche?

Step 1 — Initial charge. $Q_{0} = C_{1} V_{0} = 2 \times 10 = 20 μ C$ ; $C_{2}$ $0$ se start karta hai. Yeh step kyun? Conserved quantity fix karo: total charge $= 20 μ C$ pehle aur baad mein.

Step 2 — Connect karne ke baad, common voltage. Parallel mein dono ek voltage $V_{f}$ share karte hain. Total charge $Q_{1}^{'} = C_{1} V_{f}$ , $Q_{2}^{'} = C_{2} V_{f}$ mein split hota hai, aur $Q_{1}^{'} + Q_{2}^{'} = Q_{0}$ : $(C_{1} + C_{2}) V_{f} = Q_{0} \Rightarrow V_{f} = \frac{20}{2 + 3} = 4 V .$ Yeh step kyun? Koi battery nahi → charge conservation "voltage fixed by battery" ko replace karta hai. Parallel connection ek shared $V_{f}$ force karta hai.

Step 3 — Final charges. $Q_{1}^{'} = 2 \times 4 = 8 μ C$ , $Q_{2}^{'} = 3 \times 4 = 12 μ C$ . Yeh step kyun? Har ek naye common voltage par $Q = C V_{f}$ obey karta hai.

Verify: $Q_{1}^{'} + Q_{2}^{'} = 8 + 12 = 20 = Q_{0}$ ✓ (charge conserved). $V_{f} = 4 V < 5 V$ kyunki bada uncharged cap voltage ko zyada neeche kheenchta hai. Energy check (bonus): initial $U_{i} = \frac{1}{2} \cdot 2 \cdot 1 0^{2} = 100 μ J$ ; final $U_{f} = \frac{1}{2} (2 + 3) \cdot 4^{2} = 40 μ J$ . Energy drop gayi — kuch heat/radiation mein loss hoti hai jab charge flow karta hai. Charge conserve hota hai, energy nahi.

Example 9 — Cell J: dielectric ek capacitor ko change karta hai

Constant $κ$ (dielectric constant, ek pure number $\geq 1$ jo batata hai ki ek insulator vacuum se kitna behtar charge store karta hai) ka dielectric insert karna us capacitor ki value multiply karta hai: $C \to κ C$ . Phir network re-solve karo.

Worked example Series pair mein slab

$C_{1} = 2 μ F$ series mein $C_{2} = 2 μ F$ ke saath $12 V$ ke across. Ek dielectric $κ = 3$ $C_{2}$ ko fill karta hai, ise $C_{2}^{'} = 6 μ F$ banata hai. Naya $C_{e q}$ aur naya voltage split nikalo.

Forecast: $C_{2}$ ko fill karna ise bada banata hai. Series mein, bada cap kam voltage leta hai — toh kya $V_{2}$ rise karta hai ya fall?

Step 1 — Nai value. $C_{2}^{'} = κ C_{2} = 3 \times 2 = 6 μ F$ . Yeh step kyun? Constant $κ$ ka dielectric capacitance multiply karta hai.

Step 2 — Recombine. $\frac{1}{C _{e q}} = \frac{1}{2} + \frac{1}{6} = \frac{4}{6} \Rightarrow C_{e q} = 1.5 μ F$ . Yeh step kyun? Ab bhi series hai; sirf naye $C_{2}^{'}$ ke saath reciprocals dobara add karo.

Step 3 — Charge & voltages. $Q = 1.5 \times 12 = 18 μ C$ ; $V_{1} = \frac{18}{2} = 9 V$ , $V_{2} = \frac{18}{6} = 3 V$ . Yeh step kyun? Series common charge, phir har cap par $V = Q / C$ .

Verify: $V_{1} + V_{2} = 9 + 3 = 12$ ✓. Slab se pehle ( $C_{2} = 2 μ F$ ) split $6 V$ each tha; insert karne ke baad, $C_{2}$ (ab bada) $3 V$ par drop karta hai aur $C_{1}$ $9 V$ par rise karta hai — $V \propto 1/ C$ se consistent ✓. Units: $μ C / μ F = V$ ✓.

Active recall

Recall Har cell ke liye one-liners

Parallel: bada cap zyada...? ::: charge leta hai (same $V$ , $Q \propto C$ ). Series: bada cap zyada...? ::: kam voltage leta hai ( $V \propto 1/ C$ ); charge barabar hota hai. $C \to \infty$ wala series capacitor...? ::: plain wire ki tarah behave karta hai ( $1/ C \to 0$ ). $C \to 0^{+}$ wala series capacitor...? ::: open break ki tarah (koi charge nahi guzarta). $C \to \infty$ wala parallel capacitor...? ::: us branch ke across zero voltage enforce karta hai (ek short); $C_{e q} \to \infty$ . $C \to 0^{+}$ wala parallel capacitor...? ::: dead/open branch ki tarah jise aap delete kar sakte ho. $n$ equal caps $C$ series mein $C_{e q} =$ ? ::: $C / n$ . Battery disconnect karke caps reconnect karne ke baad kya conserve hota hai? ::: charge (voltage nahi, energy nahi). Dielectric $κ$ capacitor ki value kaise change karta hai? ::: ise multiply karta hai: $C \to κ C$ .