Kirchhoff's current law (KCL), Kirchhoff's voltage law (KVL): Worked examples

1.8.18 · D3 · Physics › Electromagnetism › Kirchhoff's current law (KCL), Kirchhoff's voltage law (KVL)

Shuru karne se pehle, ek sentence mein har symbol earn karo:

Neeche har worked example ek global rule pe tika hua hai. Ise ek baar state karo, har jagah maano:

Scenario matrix

Har problem jo Kirchhoff throw kar sakta hai un cells mein se kisi ek mein aati hai. Neeche ke examples mein cell(s) label kiye hue hain.

Cell	Kya tricky banata hai	Example
A. Single loop, ek battery	baseline — sirf sign discipline	Ex 1
B. Node, in vs out signs	kaun se currents $+$ hain, kaun se $-$	Ex 2
C. Assumed direction GALAT	answer negative aata hai	Ex 3
D. Do batteries aiding	EMFs add hoti hain	Ex 4a
E. Do batteries opposing	EMFs subtract hoti hain, current reverse ho sakta hai	Ex 4b
F. Degenerate: $R = 0$ (short)	limiting behaviour, $I \to \infty$ ?	Ex 5
G. Degenerate: $R = \infty$ (open)	koi current nahi, saari voltage gap ke across	Ex 6
H. Dead battery: $ε = 0$	battery ek plain wire/resistor ban jaati hai	Ex 6
I. Do loops (KCL + KVL saath mein)	simultaneous equations	Ex 7
J. Real-world word problem	words ko circuit mein translate karo	Ex 8
K. Exam twist: "open gap ke across $V$ dhundho"	voltage jahan koi current nahi flow hota	Ex 9

Cells D–E ek figure share karte hain; F–H ek share karte hain; two-loop cell ko apna milta hai.

Ex 1 — Cell A: baseline single loop

Forecast: aage padhne se pehle current guess karo — yeh $9$ A se zyada hogi ya kam?

Figure s01 — ek rectangular loop: battery ( $ε = 9$ V) left pe hai, resistors $R_{1}$ aur $R_{2}$ top pe hain, aur red arrow assumed clockwise current $I$ dikhata hai.

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

Assume karo $I$ clockwise (figure mein red arrow). Yeh step kyun? KVL ek equation hai $I$ mein; humein ek direction naam dena padega taaki "current ke saath" aur "uske against" ka matlab ho.
Loop mein clockwise chalo battery ke $-$ terminal se, signed changes likhte hue: $+ ε - I R_{1} - I R_{2} = 0.$ Yeh step kyun? Battery mein $-$ se $+$ jaana ek rise hai ( $+ ε$ ); har resistor current ke saath traverse karna ek drop hai ( $- I R$ ).
$I$ ke liye solve karo: $I = \frac{ε}{R _{1} + R _{2}} = \frac{9}{1 + 2} = 3 A .$ Yeh step kyun? Pure algebra; series resistances add hoti hain, isliye yeh $I = ε / R_{eq}$ jaisa hi hai.

Verify: total drop $= I R_{1} + I R_{2} = 3 (1) + 3 (2) = 9 V = ε$ . ✓ Units: $V /Ω = A$ . ✓

Ex 2 — Cell B: node signs

Forecast: kya $I_{3}$ positive hoga (sach mein andar) ya negative (secretly bahar)?

KCL likho in $= +$ , out $= -$ ke saath: $I_{1} + I_{2} + I_{3} - I_{4} = 0.$ Yeh step kyun? Ek node mein charge store nahi hota, isliye jo andar aata hai woh bahar jaana chahiye: $\sum I_{k} = 0$ .
Numbers plug karo: $4 + 1 + I_{3} - 6 = 0 \Rightarrow I_{3} = 1 A .$ Yeh step kyun? Linear equation solve karo. Positive ⇒ $I_{3}$ sach mein andar flow karta hai, jaisa assume kiya gaya.

Verify: in $= 4 + 1 + 1 = 6 A$ , out $= 6 A$ . Balanced. ✓

Ex 3 — Cell C: galat guess (negative answer)

Forecast: is baar $I$ ka sign kya hoga?

Assume karo $I$ counter-clockwise. Ab loop ko counter-clockwise traverse karte hue, battery $+$ se $-$ enter hoti hai (ek drop) aur resistors is nayi current ke saath traverse hote hain. $- ε - I R_{1} - I R_{2} = 0.$ Yeh step kyun? Physics nahi badla; sirf humara labelling flip hua, isliye signs flip ho gaye.
Solve karo: $I = \frac{- ε}{R _{1} + R _{2}} = \frac{- 9}{3} = - 3 A .$ Yeh step kyun? Minus sign yahi batata hai: asli current $3 A$ opposite (clockwise) direction mein hai — exactly Ex 1 jaisa.

Verify: magnitude $∣ I ∣ = 3 A$ Ex 1 se match karta hai; direction expected ke anusaar reversed hai. ✓

Ex 4 — Cells D & E: do batteries, pehle aiding phir opposing

Figure s02 — do side-by-side panels. Left: dono EMF arrows same direction mein point karte hain, isliye woh $10$ V tak add hote hain (red mein dikhaya). Right: EMF arrows opposite directions mein point karte hain, isliye woh subtract karke $2$ V hote hain (red mein dikhaya).

Ex 4a — Cell D: batteries aiding

Forecast: effective push $6 + 4$ honi chahiye ya $6 - 4$ ?

Assume karo $I$ clockwise; clockwise chalo. Dono batteries $- \to +$ climb hoti hain (dono rise): $+ ε_{1} + ε_{2} - I R = 0.$ Yeh step kyun? Aiding EMFs dono lift karte hain, isliye woh add hote hain.
Solve karo: $I = \frac{ε _{1} + ε _{2}}{R} = \frac{6 + 4}{5} = 2 A .$

Verify: $R$ ke across drop = $I R = 2 (5) = 10 V = ε_{1} + ε_{2}$ . ✓

Ex 4b — Cell E: batteries opposing

Forecast: kaun si battery jeeti hai, aur kya current reverse hoti hai?

Clockwise chalo. $ε_{1}$ ek rise hai ( $- \to +$ ), lekin $ε_{2}$ ab $+ \to -$ climb hoti hai (ek drop): $+ ε_{1} - ε_{2} - I R = 0.$ Yeh step kyun? Opposing EMFs subtract hoti hain — ek push karta hai, doosra pushback karta hai.
Solve karo: $I = \frac{ε _{1} - ε _{2}}{R} = \frac{6 - 4}{5} = 0.4 A .$ Yeh step kyun? Positive ⇒ zyada strong battery ( $ε_{1}$ ) direction set karti hai; weaker battery charge ho rahi hai.

Verify: agar batteries equal hoti ( $ε_{2} = 6$ ), toh $I = 0$ — perfect deadlock. Sanity check pass. ✓

Ex 5 — Cell F: short circuit ( $R \to 0$ )

Forecast: kya current infinity tak blow up hoti hai?

Figure s03 — same rectangular loop, lekin external short wire mota aur red draw hua hai ( $R = 0$ ) jabki battery ki internal resistance $r = 0.5 Ω$ source ke andar hai. Point yeh hai: $r$ hi akela current ko limit kar raha hai.

Loop ke around KVL (sirf $r$ resist karta hai): $ε - I r - I R = 0, R = 0.$ Yeh step kyun? "Perfect" wire loop mein bhi battery ki apni $r$ hoti hai. Truly zero total resistance jaisi koi cheez nahi hoti.
$R = 0$ ke saath solve karo: $I = \frac{ε}{r + R} = \frac{12}{0.5 + 0} = 24 A .$ Yeh step kyun? Limiting value: jaise external $R \to 0$ , current $ε / r$ tak badhti hai, jo uska maximum hai — bada lekin finite, kyunki $r$ hume bachata hai.
Terminal voltage (external short ke across voltage): $V_{term} = ε - I r = 12 - 24 (0.5) = 0 V .$ Yeh step kyun? Saari EMF battery ke andar hi khayi jaati hai, isliye bahar $0 V$ dikhta hai — classic dead-short signature.

Verify: power in $r$ = $I^{2} r = 2 4^{2} \times 0.5 = 288 W$ ; power from EMF = $ε I = 12 \times 24 = 288 W$ . Energy balance hai. ✓ (Isliye battery short karne pe garam ho jaati hai.)

Ex 6 — Cells G & H: open circuit aur dead battery

Forecast: padhne se pehle dono currents guess karo.

Open circuit (G): gap ke saath koi closed path nahi hai, isliye $I = lim_{R \to \infty} \frac{ε}{R} = 0 A .$ Yeh step kyun? Charge gap cross nahi kar sakta; gap pe KCL $I = 0$ force karta hai.
Dead battery (H): $ε = 0$ set karo KVL mein ek loop ke liye jisme resistor $R = 3 Ω$ hai: $0 - I R = 0 \Rightarrow I = 0 A .$ Yeh step kyun? Jab charge ko lift karne ke liye koi EMF nahi, kuch bhi use drive nahi karta — battery sirf ek wire (ya resistor) hai. Koi source nahi, koi current nahi.

Verify: dono $I = 0$ dete hain, lekin alag reasons ke liye: (G) koi path nahi, (H) koi push nahi. Cross-check: (H) mein, power delivered $= ε I = 0$ . ✓

Ex 7 — Cell I: do loops, KCL + KVL saath mein

Forecast: kaun si branch zyada current carry karegi — $4 Ω$ wali ya $6 Ω$ wali?

Figure s04 — left pe battery ek node ko feed karti hai jo do parallel branches mein split hoti hai. Red branch $R_{1} = 4 Ω$ hai, jo zyada current $I_{1} = 3$ A carry karti hai; black branch $R_{2} = 6 Ω$ hai jisme $I_{2} = 2$ A hai. Main current $I = 5$ A top rail pe label hai.

KVL, left loop (battery + $R_{1}$ ): $ε - I_{1} R_{1} = 0 \Rightarrow I_{1} = 12/4 = 3 A$ . Yeh step kyun? $R_{1}$ akela poori EMF span karta hai (parallel branches same voltage share karte hain).
KVL, right loop (battery + $R_{2}$ ): $ε - I_{2} R_{2} = 0 \Rightarrow I_{2} = 12/6 = 2 A$ . Yeh step kyun? Same reasoning; $R_{2}$ bhi poore $12 V$ dekhta hai.
KCL at the node: $I = I_{1} + I_{2} = 3 + 2 = 5 A$ . Yeh step kyun? Main current split hoti hai; junction pe kuch nahi jaata.

Verify: parallel equivalent $R_{eq} = \frac{4 \cdot 6}{4 + 6} = 2.4 Ω$ , isliye $I = 12/2.4 = 5 A$ . ✓ Kam resistance ( $4 Ω$ ) zyada current carry karta hai — jaise predict kiya tha.

Ex 8 — Cell J: real-world word problem

Forecast: total draw — $10 A$ se kam ya zyada?

Translate karo: $12 V$ ke across parallel ⇒ har ek $12 V$ dekhta hai (Ex 7 jaisa logic). Yeh step kyun? "Battery ke across parallel mein" yeh word-cue hai equal branch voltages ke liye.
Branch currents Ohm's Law se: $I_{head} = \frac{12}{3} = 4 A, I_{tail} = \frac{12}{9} \approx 1.33 A .$ Yeh step kyun? Har branch ki current uski apni resistance se set hoti hai shared $12 V$ ke under; Ohm's law $I = V / R$ har known voltage-and-resistance ko current mein convert karta hai.
KCL — total draw: $I_{fuse} = 4 + \frac{4}{3} = \frac{16}{3} \approx 5.33 A .$ Yeh step kyun? Fuse dono branches ka sum carry karta hai.

Verify: $5.33 A < 10 A$ ⇒ fuse holds. ✓ Cross-check: $R_{eq} = \frac{3 \cdot 9}{12} = 2.25 Ω$ , $I = 12/2.25 = 5.33 A$ . ✓

Ex 9 — Cell K: exam twist, open gap ke across voltage

Forecast: agar koi current nahi flow karta, toh reading $0$ hogi ya $10 V$ ?

Pehle current: ek ideal voltmeter ek open circuit hai, isliye KCL force karta hai $I = 0$ (phir se Cell G). Yeh step kyun? Charge ke liye koi closed path nahi ⇒ har jagah zero current.
$I = 0$ ke saath, koi resistor koi voltage drop nahi karta: $I R_{1} = I R_{2} = 0$ . Yeh step kyun? Ohm's Law: drop ke liye current chahiye; $I = 0$ ⇒ zero drop.
Gap voltage $V_{g}$ dhundne ke liye KVL: loop mein chalo, sirf battery ka term non-zero hai: $ε - I R_{1} - V_{g} - I R_{2} = 0 \Rightarrow V_{g} = ε = 10 V .$ Yeh step kyun? Poori EMF us jagah appear hoti hai jahan koi current nahi flow karta — open gap. Yeh exam trap hai: "no current" ka matlab "no voltage" nahi hota.

Verify: poore $10 V$ kahi na kahi appear hone chahiye; zero resistor drops ke saath, yeh sab gap pe aata hai. ✓

Recall Quick self-test

Ex 5 short-circuit current $ε = 12$ , $r = 0.5$ ke saath? ::: $24 A$ Ex 9 open gap ke across voltmeter reading? ::: $10 V$ (poori EMF) Ex 4b current jab do equal EMFs oppose karte hain? ::: $0 A$ (deadlock)

Connections

Ohm's Law — upar har "drop" $V = I R$ hai.
Series and Parallel Resistors — Ex 1, 7, 8 mein sanity check ke roop mein use hua.
Conservation of Charge — isliye KCL ek gap ke across $I = 0$ force karta hai (Ex 6, 9).
Conservation of Energy — Ex 5 mein power balances.
Electric Potential — Ex 9 mein gap voltage ek potential difference hai.
Wheatstone Bridge — inme se kai cells ek saath combine karta hai.
Mesh and Nodal Analysis — Ex 7 ki two-loop bookkeeping automate karta hai.