Kirchhoff's current law (KCL), Kirchhoff's voltage law (KVL): Visual walkthrough

1.8.18 · D2 · Physics › Electromagnetism › Kirchhoff's current law (KCL), Kirchhoff's voltage law (KVL)

Hume sirf do everyday truths chahiye: charge appear ya vanish nahi ho sakta aur ek pahaad par round trip karne ke baad tum same height par wapas aate ho. Baaki sab isi se nikalta hai.

Step 1 — "Current" kya hota hai? Charges ko count karo jo past jaate hain

KYA kiya humne: "bijli bahna" ki vague idea ko ek precise count mein badla — charge per second ek chosen gate ke through, ek chosen positive direction ke saath.

KYU: "current in = current out" kehne se pehle hum current ko count kar sake aur uski sign jaante hain. Arrow hi poora point hai — ek rightward arrow ke through $- 3$ A ka current bas matlab hai ki $3$ A actually leftward bah raha hai.

PICTURE (s01): blue arrow labelled "chosen + direction" humara reference hai; chhote red charges yellow gate se drift karte hain, aur wire ke neeche caption $I = Δ Q /Δ t$ dikhata hai — arrow hi us $I$ ki sign fix karta hai.

Figure — Kirchhoff's current law (KCL), Kirchhoff's voltage law (KVL)

Step 2 — Node sirf ek meeting point hai jo charge store nahi kar sakta

KYA kiya humne: junction ko naam diya aur uski ek khaas property state ki — zero storage.

KYU: yahi ek fact ("charge hoard nahi kar sakta") do steps mein KCL ban jaayega. Sab kuch isi par depend karta hai, isliye hum ise explicit banate hain aur picture mein dikhate hain.

PICTURE (s02): teen wires red node dot par milti hain; do blue arrows ( $I_{1}, I_{2}$ ) andar point karte hain, yellow arrow ( $I_{3}$ ) bahar point karta hai. Dashed circle node ke around ek tiny "accounting box" hai — hum agले step mein uske andar charge ka hisaab karte hain.

Step 3 — Box ke andar charge ka hisaab karna

Dashed box ke andar trapped charge $Q_{node}$ ki matra dekho. Yeh tabhi change ho sakti hai jab zyada andar aaye than bahar jaaye:

$kitni tezi se stored charge badhta hai \frac{d Q _{node}}{d t} = coulombs/sec enter karte hue \sum I_{in} - coulombs/sec nikaalte hue \sum I_{out}$

Ek ek term padhte hain:

$\frac{d Q _{node}}{d t}$ — stored charge ke change hone ki rate. " $d / d t$ " bas "kitni tezi se, per second" hai. Agar positive hai, charge pile ho raha hai; negative, drain ho raha hai.
$\sum I_{in}$ — har current ka sum jiska arrow box ke andar point karta hai.
$\sum I_{out}$ — har current ka sum jiska arrow bahar point karta hai.

KYA kiya humne: box ke liye exact charge balance likha.

KYU: yeh honest accounting ke siwa kuch nahi — andar ki pile tabhi badh sakti hai jab inflow outflow se zyada ho.

PICTURE (s03): box ke saath ek blue " $\sum I_{in}$ " pipe left se enter karti hai aur ek red " $\sum I_{out}$ " pipe right se nikalti hai; beech mein yellow " $Q$ inside" meter hai jo box ke neeche likhi equation $d Q / d t = \sum I_{in} - \sum I_{out}$ track kar rahi hai.

Step 4 — Storage ko zero karo → KCL paida hota hai

Ideal node charge store nahi kar sakta (Step 2). Toh andar ki pile kabhi badhti ya ghatti nahi:

$\frac{d Q _{node}}{d t} = 0.$

Ise Step 3 ke balance mein daalo:

$0 = \sum I_{in} - \sum I_{out} ⟹ \sum I_{in} = \sum I_{out}$

Ab woh promised concrete sign rule jo ise ek compact sum mein badal deta hai. Jab tumne har wire par arrow draw kar liya (arrows kisi bhi direction mein point kar sakte hain — maths sach pata kar lega), har current ko is node par ek sign assign karo:

Toh upar wali boxed equation ke dono sides ek clean sum mein merge ho jaate hain:

$k \sum I_{k} = 0$

$\sum_{k}$ — "node ko touch karne wali har wire $k$ par add karo".
$I_{k}$ — wire $k$ mein current, upar ke rule se $\pm$ lekar (positive agar uska arrow enter karta hai, negative agar woh nikalta hai).

Sign ka worked example. Maano teen arrows draw hain: do node ke andar ( $I_{1}, I_{2}$ ) aur ek bahar ( $I_{3}$ ). Rule deta hai $+ I_{1} + I_{2} - I_{3} = 0$ . Agar tumne $I_{3}$ ka arrow bhi node ke andar draw kiya hota, toh rule deta $+ I_{1} + I_{2} + I_{3} = 0$ — aur solve karne par tumhe negative $I_{3}$ milta, jo sahi se batata ki current actually bahar jaati hai.

KYA kiya humne: $d Q / d t = 0$ force kiya, phir ek explicit arrow→sign rule diya taaki kisi bhi set ke drawn arrows $\sum_{k} I_{k} = 0$ mein collapse ho jaayein.

KYU: kyunki "no storage" ek ideal junction ke baare mein physical sach hai — KCL mein koi naya physics nahi hai, yeh hai conservation of charge circuit ke kapde mein; sign rule sirf arrows ka consistent bookkeeping hai.

PICTURE (s04): water-junction jisme sign tags dikhaye hain — $+ 3 A$ aur $+ 2 A$ enter karte hain (blue, " $+$ "), toh exactly $5 A$ nikalna chahiye (green, " $-$ "); caption $\sum I_{in} = \sum I_{out} \Rightarrow 3 + 2 = 5$ seedha arrows se padha jaata hai.

Step 5 — Energy ki taraf jao: potential ek pahaad ki height hai

Ab doosra law. Pehle hume samajhna hoga ki ek point par "voltage" ka matlab kya hai.

Key property: har point ki exactly ek hi height hoti hai. Point $A$ ek saath do altitudes par nahi ho sakta.

KYA kiya humne: voltage ko height ke roop mein reinterpret kiya, har location ke liye ek number.

KYU: KVL round trips ke baare mein hai, aur round trips ko landscape par reason karna sabse aasaan hota hai — baaho, wapas aao, aur tum wahi altitude par land karoge jahan se shuru kiya tha. Woh "har point ke liye single height" fact KVL ka beej hai.

PICTURE (s05): ek hilly altitude profile. Green ramp labelled "battery = lift $(+ ε)$ " height badhata hai; har red ramp ("R1 slide", "R2 slide") ise ghataata hai. Vertical axis literally "height = potential $V$ " hai, aur yellow dot starting point $A$ mark karta hai.

Step 6 — Battery se milo: EMF asliyat mein kya hota hai

Steps sign karne se pehle, hume formally lift ko naam dena hoga.

Ab har step ka sign. Jab hum ek chosen direction mein loop mein chalte hain, har element humari height ko kisi $Δ V$ se badlaata hai:

Battery mein − se + tak: height badhti hai → $Δ V = + ε$ . ( $ε$ = EMF / lift height, abhi define ki gayi.)
Resistor mein current ke saath (Ohm's law $V = I R$ ke saath): height ghatti hai → $Δ V = - I R$ .
Same element ke against chalo: sign flip karo — drop rise ban jaata hai aur vice-versa.

Resistor step $Δ V = - I R$ ke liye term by term:

$I$ — resistor ke through current.
$R$ — uski resistance.
$I R$ — potential drop ka size (yeh hai Ohm's law passive sign convention ke under Step 1 se).
minus — kyunki current ke saath move karte hue hum downhill jaate hain.

KYA kiya humne: EMF ko properly define kiya, phir har tarah ke element ko ek signed height-change assign ki.

KYU: bina firm sign rule ke sum meaningless hai — circuits mein aadhe "voltage errors" bas ek dropped minus sign hote hain.

PICTURE (s06): ek chhota loop flat draw kiya; green node tags " $+ ε$ (EMF, up)" aur do red tags " $- I R_{1}$ (down)", " $- I R_{2}$ (down)" har element ki signed $Δ V$ mark karte hain, ek blue "walk dir" arrow dikhata hai jis direction mein woh signs padhe gaye.

Step 7 — Loop orientation fix karo, round trip lo → KVL paida hota hai

Sum karne se pehle, hume ek loop orientation fix karni hogi — aur uspar commit karna hoga.

Ab point $A$ par height $V_{A}$ se shuru karo. Chosen orientation mein poora loop chalo, har signed step add karo, aur $A$ par wapas aao:

$V_{A} + Δ V_{1} + Δ V_{2} + \dots + Δ V_{n} = V_{A}$

Baya $V_{A}$ — tumhari starting height.
$Δ V_{1}, \dots, Δ V_{n}$ — har element ke signed rises aur drops jise tum pass karte ho, chosen orientation mein padhe gaye.
Daya $V_{A}$ — tumhari ending height, jo same point hai, isliye same height (Step 5!).

Dono sides se $V_{A}$ minus karo — woh cancel ho jaata hai kyunki dono sides par identical hai:

$k \sum Δ V_{k} = 0$

$\sum_{k} V_{k} = 0, V_{k} \equiv Δ V_{k} (branch change, node potential nahi) .$

Agar tum apni orientation ke against traverse karo toh? Tum nahi kar sakte — ek lap ke liye tum ek hi direction par commit karte ho. Lekin agar tum same loop ko opposite orientation ke saath re-solve karo, har $Δ V_{k}$ ek saath sign flip karta hai (har rise drop ban jaata hai aur vice-versa), toh poori equation bas $- 1$ se multiply hoti hai: $\sum_{k} (- Δ V_{k}) = 0$ same equation hai. Physics — aur jo current tum compute karte ho — identical hai; sirf paperwork sign convention flip hua. Illustrated example (s07) dono orientations ko same current par land karte dikhata hai.

KYA kiya humne: ek loop orientation fix ki, har signed height-change ka sum liya, "same point ⇒ same height" use kiya, phir carefully $Δ V_{k}$ ko $V_{k}$ mein rename kiya jagte hue ki yeh branch changes hain, node potentials nahi.

KYU: round trip hi trick hai — messy $V_{A}$ precisely isi liye gayab hota hai kyunki tum ghar wapas aaye, sirf ups aur downs bachte hain, jo cancel hone chahiye.

PICTURE (s07): loop ek altitude profile ki tarah — $+ ε$ se pump up (green), $- I R_{1}$ aur $- I R_{2}$ se slide down (red), starting altitude par exactly wapas $A$ par land karte hue (blue dots); inset reversed orientation ko same current dete dikhata hai.

Step 8 — Jab KVL ko fix ki zaroorat hoti hai: changing magnetic fields

Poora round-trip argument secretly assume karta tha ki potential single-valued hai — tab sach jab electric field conservative ho, yaani $\oint E \cdot d l = 0$ .

Common mistake KVL

universally valid nahi hai Faraday's law kehta hai ki loop ke through time-varying magnetic flux $Φ_{B}$ ek non-conservative electric field induce karta hai: $\oint E \cdot d l = - \frac{d Φ _{B}}{d t} \neq = 0 (jab flux change ho) .$ Jab tumhare loop ke through flux change hota hai (ek inductor, paas mein AC field, ek transformer), round trip same "height" par wapas nahi aata — $\sum_{k} V_{k} \neq = 0$ . Do honest fixes:

Flux ke source ko ek element ki tarah model karo. Inductor ki voltage $V_{L} = L d I / d t$ ko loop mein apne $Δ V$ ki tarah daalo; tab $\sum_{k} V_{k} = 0$ phir kaam karta hai kyunki tumne flux term ko element side par move kar diya.
Ya raw field form $\oint E \cdot d l = - d Φ_{B} / d t$ rakho aur $\sum V_{k} = 0$ ki jagah usse use karo.

Rule of thumb: plain KVL ( $\sum V_{k} = 0$ ) DC ke liye aur lumped circuits ke liye exact hai jahan saara magnetic flux labelled inductor elements ke andar confined ho. Yeh un bare loops ke liye fail karta hai jisme changing external flux thread karta ho.

KYA kiya humne: woh ek situation naam di jahan Step 7 ka premise toot jaata hai.

KYU: honesty — jo reader baad mein inductors ya transformers se milega use yeh jaanna chahiye ki " $\sum V = 0$ " conservative-field ka special case hai, universe ka law nahi.

PICTURE (s08): blue circuit loop ko badhta hua red magnetic flux $Φ_{B}$ pierce karta hai (dots = field into page); yellow arrow induced non-conservative field mark karta hai, aur caption $\oint E \cdot d l = - d Φ_{B} / d t \neq = 0$ exactly wajah hai ki trip height mein close nahi hoti.

Step 9 — Degenerate & edge cases (kabhi surprise mat ho)

Yahan chaar cases hain jo "weird lagte hain" — har ek unhi do lines ka paalan karta hai, aur figure (s09) chaaon ko labelled panels ki tarah dikhata hai.

Worked example Chaar edge cases, spelled out

(a) Two-wire node. Sirf ek current andar aur ek bahar ⇒ $I_{in} = I_{out}$ . Isi liye current series resistors mein same hoti hai — ek resistor storage wala node nahi hai, isliye kuch split off nahi hota. (Panel a: ek blue arrow andar, ek green arrow red dot se bahar.)

(b) Bina source wala loop. EMF hata do ( $ε = 0$ ) lekin real resistance $R > 0$ rakho. KVL deta hai $\sum (- I R) = 0$ , yaani $- I (R_{1} + R_{2}) = 0$ , jo force karta hai $I = 0$ — koi pump nahi, koi persistent current nahi. (Aur zero resistance par nonzero EMF unphysical hai — infinite current — isliye ek real loop mein hamesha kuch $R > 0$ hota hai.) (Panel b: ek closed loop labelled "I = 0".)

(c) Negative answer. Agar KCL/KVL equations solve karne par $I = - 2 A$ milta hai, tumhara assumed arrow ulta drawn tha. Magnitude $2 A$ correct hai; bas arrow flip karo aur sign positive ho jaayega. Kuch galat nahi hua — algebra ne tumhe sahi direction bataya. (Panel c: ek arrow ek taraf assumed, annotation "got −2 A ⇒ reverse".)

(d) Ideal wire / zero-ohm branch ( $R = 0$ ). Tab uska drop hai $Δ V = - I R = 0$ : dono ends same height par hain, chahe kitni bhi current bahe. Saari current khushi se is zero-resistance path (ek "short") mein pour hoti hai. (Panel d: ek thick green wire, "R = 0, same height".)

PICTURE (s09): chaar mini-panels (a)–(d), upar ke har case ke liye ek, har ek apna verdict lekar taaki tum panel → paragraph directly match kar sako.

Ek-picture summary

Dono laws ek single diagram par: red node par signed arrows balance karte hain ( $\sum I_{k} = 0$ ); blue loop ke around yellow height profile ghar wapas aata hai ( $\sum V_{k} = 0$ , conservative case). Node arrows ko Step-4 sign rule se padho aur loop ko Step 7 ki fixed orientation se.

Recall Feynman: poora walkthrough plain words mein

Do ideas, kuch nahi iske alaawa. Charge paani ki tarah hota hai aur kabhi gayab nahi hota. Toh kisi bhi junction par, andar aata paani bahar jaane wale paani ke barabar hona chahiye — woh pile nahi ho sakta kyunki junction ke paas koi bucket nahi hai. Har pipe par ek arrow draw karo: in-flows ko $+$ count karo, out-flows ko $-$ , aur total exactly zero hoga. Woh hai KCL, aur agar koi number negative nikle toh bas woh arrow ulta draw kiya tha. Voltage ek pahaad par height ki tarah hai. Battery ek lift hai jo tumhe upar le jaati hai; har resistor ek slide hai jo tumhe neeche le jaata hai. Loop mein chalne ke liye ek direction chuno aur uspar stick karo. Poora lap chalo aur wahan wapas aao jahan se shuru kiya — tum zaroor same height par hoge, toh saare ups aur downs zero ho jaate hain. Woh hai KVL — jab tak koi changing magnetic field tumhare loop mein thread na kar raha ho. Agar kar raha hai, toh pahaad tumhare pairon ke neeche secretly shift karta hai (Faraday), aur tumhe inductor ki voltage ko apni slide ki tarah add karna hoga taaki trip phir close ho. Baaki sab — minus signs, negative currents, shorts — sirf carefully count karna hai ki paani kis taraf bah raha hai aur tum pahaad par kis taraf chal rahe ho.

Recall

Hum node par

d Q / d t = 0

kyun set kar sakte hain?

Ek ideal junction ke paas charge store karne ki koi capacity nahi hoti (stray capacitance negligible), isliye andar ka charge kabhi change nahi hota.

Jab arrows drawn hain,

\sum_{k} I_{k} = 0

mein har

I_{k}

ko sign kaise karo?

Arrow node ke andar point karta hai →

+ I_{k}

; arrow bahar →

- I_{k}

\pm

purely drawn arrow se aata hai.

KVL ki round-trip derivation mein

V_{A}

kyun cancel hota hai?

Tum same point par wapas aate ho, jiska single height hai, toh start aur end

V_{A}

identical hain aur subtract ho jaate hain.

KVL mein

V_{k}

kya represent karta hai — aur woh kya NAHI hai?

Loop traverse karte hue element

k

par signed potential change

Δ V_{k}

(ek branch voltage change) — yeh kisi node par potential nahi hai.

Tum mid-loop kabhi "apni orientation ke against traverse" kyun nahi kar sakte, aur agar poori tarah reverse karo toh kya hoga?

Tum ek lap ke liye ek direction par commit karte ho; poori orientation reverse karne par har

Δ V_{k}

sign ek saath flip hota hai, equation ko

- 1

se multiply karta hai — same physics, same current.

Plain KVL (

\sum V_{k} = 0

) kab fail karta hai?

Jab ek changing magnetic flux loop mein thread karta hai (

\oint E \cdot d l = - d Φ_{B} / d t \neq = 0

); ise fix karo inductor ki voltage ko ek element ki tarah model karke.

Zero-ohm branch ke across potential difference kitna hota hai?

Zero, kyunki

Δ V = - I R = 0

— dono ends same potential par hain.

Agar ek network solve karne par negative current milta hai, uska matlab kya hai?

Assumed arrow ulta drawn tha; magnitude correct hai, bas arrow reverse karo.

Connections

Ohm's Law — resistor drop $Δ V = - I R$ deta hai jo Step 6 mein use hota hai.
Conservation of Charge — "no storage" fact jo KCL ban jaata hai.
Conservation of Energy — round-trip-returns-home fact jo KVL ban jaata hai (aur Faraday's law, uski limit).
Electric Potential — "har point ke liye ek height" property Step 5 mein.
Series and Parallel Resistors — Step 9(a) same series current explain karta hai.
Wheatstone Bridge · Mesh and Nodal Analysis — jahan dono laws systematically combine hote hain.