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

1.8.18 · D4 · Physics › Electromagnetism › Kirchhoff's current law (KCL), Kirchhoff's voltage law (KVL)

Yahan $I$ = current (amperes, A), $R$ = resistance (ohms, $Ω$ ), $ε$ = EMF of a source (volts, V), $V$ = potential difference (volts, V).

Level 1 — Recognition

Exercise 1.1

Teen currents ek node par milte hain. $I_{1} = 4 A$ andar flow karta hai, $I_{2} = 1 A$ andar flow karta hai, aur $I_{3}$ bahar flow karta hai. $I_{3}$ dhundho.

Recall Solution 1.1

KAUN sa law? Junction par currents → yeh charge conservation hai, isliye hum KCL use karenge. Sign rule apply karo (in $+$ , out $-$ ): $I_{1} + I_{2} - I_{3} = 0.$ Solve karo: $I_{3} = I_{1} + I_{2} = 4 + 1 = 5 A .$ Sanity: jo bhi andar aata hai, woh bahar nikalna chahiye — bare junction par koi charge store nahi hota.

Exercise 1.2

Tum ek loop mein apni chosen direction mein travel karte ho aur ek resistor $R = 3 Ω$ se uske current $I = 2 A$ ki same direction mein guzarte ho. Kya potential change ek rise hai ya drop, aur uska signed value kya hai?

Recall Solution 1.2

KYA ho raha hai? Current ek resistor mein high potential se low potential ki taraf flow karta hai (yahi "resistance" karta hai — woh energy burn karta hai). Current ke saath chalna matlab potential mein neeche jaana. To yeh ek drop hai, minus sign ke saath likha jaata hai: $Δ V = - I R = - (2) (3) = - 6 V .$ Agar tum current ke against chalte, to same magnitude $+ 6 V$ (ek rise) ke roop mein aata.

Level 2 — Application

Exercise 2.1

Ek single loop: EMF $ε = 9 V$ in series with $R_{1} = 1 Ω$ aur $R_{2} = 2 Ω$ . Loop current $I$ dhundho.

Recall Solution 2.1

Step 1 — direction assume karo. Maano $I$ clockwise flow karta hai. (Galat guess sirf answer ka sign flip karti hai.) Step 2 — KVL apply karo battery ke $-$ terminal se start karke, clockwise jaate hue: $+ ε - I R_{1} - I R_{2} = 0.$ Ye signs kyun? Battery ke andar $-$ se $+$ ek rise hai ( $+ ε$ ); har resistor jo current ke saath traverse hota hai woh ek drop hai ( $- I R$ ). Step 3 — solve karo: $I = \frac{ε}{R _{1} + R _{2}} = \frac{9}{3} = 3 A .$ Series and Parallel Resistors se cross-check karo: series resistance $= 3 Ω$ , to $I = 9/3 = 3 A$ . ✓

Exercise 2.2

Wohi loop jaise 2.1 mein, lekin ab $R_{2}$ ke across voltage dhundho aur confirm karo ki do resistor voltages $ε$ mein add up hoti hain.

Recall Solution 2.2

Pehle se $I = 3 A$ use karke aur Ohm's Law se: $V_{1} = I R_{1} = 3 \times 1 = 3 V, V_{2} = I R_{2} = 3 \times 2 = 6 V .$ KVL se check karo: $V_{1} + V_{2} = 3 + 6 = 9 V = ε .$ ✓ Yeh bilkul Conservation of Energy ka statement hai: battery jo energy har coulomb ko deti hai ( $9 J/C$ ) woh sab do resistors mein karch ho jaati hai.

Level 3 — Analysis

Exercise 3.1

Battery $ε = 12 V$ do parallel resistors $R_{1} = 6 Ω$ aur $R_{2} = 3 Ω$ ko drive karti hai. Branch currents $I_{1}, I_{2}$ aur total current $I$ dhundho.

Recall Solution 3.1

Har branch loop par KVL — dono resistors seedha battery ke across hain, isliye dono $ε$ dekhte hain: $I_{1} = \frac{ε}{R _{1}} = \frac{12}{6} = 2 A, I_{2} = \frac{ε}{R _{2}} = \frac{12}{3} = 4 A .$ Top node par KCL (total do branches mein split hota hai): $I = I_{1} + I_{2} = 2 + 4 = 6 A .$ Check: $R_{eq} = \frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{6 \cdot 3}{9} = 2 Ω$ , to $I = 12/2 = 6 A$ . ✓ Note karo ki chhota resistor ( $R_{2}$ ) zyada current carry karta hai — current easy path prefer karta hai.

Exercise 3.2 (the negative-answer case)

Ek node par, $I_{1} = 5 A$ andar flow karta hai. Do currents bahar flow karne ki assume ki gayi hain: $I_{2} = 8 A$ (known) aur $I_{3}$ (unknown, bahar assume kiya). $I_{3}$ dhundho aur uske sign ko interpret karo.

Recall Solution 3.2

KCL (in $+$ , out $-$ ): $I_{1} - I_{2} - I_{3} = 0 \Rightarrow I_{3} = I_{1} - I_{2} = 5 - 8 = - 3 A .$ Minus sign interpret karo: humne assume kiya tha ki $I_{3}$ bahar flow karta hai, lekin woh negative nikla. Matlab real current node mein andar flow karta hai magnitude $3 A$ ke saath. Sanity: to in-flow $= 5 + 3 = 8 A$ , out-flow $= 8 A$ . Balanced. ✓

Level 4 — Synthesis

Exercise 4.1

Figure dekho: ek $ε = 14 V$ battery, ek series resistor $R_{1} = 2 Ω$ , phir current do parallel resistors $R_{2} = 6 Ω$ aur $R_{3} = 3 Ω$ mein split hoti hai, phir recombine hoti hai aur return karti hai. Total current $I$ , split currents $I_{2}, I_{3}$ , aur parallel pair ke across voltage dhundho.

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

Recall Solution 4.1

Step 1 — parallel pair collapse karo (Series and Parallel Resistors se): $R_{p} = \frac{R _{2} R _{3}}{R _{2} + R _{3}} = \frac{6 \cdot 3}{9} = 2 Ω.$ Step 2 — outer loop par KVL ( $I$ = $R_{1}$ aur $R_{p}$ se guzarne wala total current): $ε - I R_{1} - I R_{p} = 0 \Rightarrow I = \frac{ε}{R _{1} + R _{p}} = \frac{14}{2 + 2} = 3.5 A .$ Step 3 — parallel pair ke across voltage (figure mein "mint node"): $V_{p} = I R_{p} = 3.5 \times 2 = 7 V .$ Step 4 — KCL + Ohm se split karo (dono branches $V_{p} = 7 V$ dekhti hain): $I_{2} = \frac{V _{p}}{R _{2}} = \frac{7}{6} \approx 1.1667 A, I_{3} = \frac{V _{p}}{R _{3}} = \frac{7}{3} \approx 2.3333 A .$ Split node par KCL check: $I_{2} + I_{3} = \frac{7}{6} + \frac{7}{3} = \frac{7}{6} + \frac{14}{6} = \frac{21}{6} = 3.5 A = I .$ ✓

Exercise 4.2

Do batteries ek loop mein ek doosre ka oppose karti hain: $ε_{1} = 10 V$ aur $ε_{2} = 4 V$ (opposite direction mein push karti hai), ek single resistor $R = 3 Ω$ ke saath. Current dhundho.

Recall Solution 4.2

$I$ clockwise assume karo us direction mein jisme $ε_{1}$ push karta hai. Clockwise travel karte hue:

$ε_{1}$ se guzarte hue ( $-$ se $+$ ): rise $+ ε_{1}$ ,
$ε_{2}$ se guzarte hue ( $+$ se $-$ , kyunki woh oppose karta hai): drop $- ε_{2}$ ,
$R$ se current ke saath guzarte hue: drop $- I R$ . KVL: $ε_{1} - ε_{2} - I R = 0 \Rightarrow I = \frac{ε _{1} - ε _{2}}{R} = \frac{10 - 4}{3} = 2 A .$ Zyada strong battery jeet jaati hai; net driving EMF $10 - 4 = 6 V$ hai.

Level 5 — Mastery

Exercise 5.1 (balanced Wheatstone bridge)

Figure ke Wheatstone Bridge mein, chaar arms hain $R_{1} = 2 Ω$ , $R_{2} = 4 Ω$ (top pair) aur $R_{3} = 3 Ω$ , $R_{4} = 6 Ω$ (bottom pair). Ek galvanometer middle ko bridge karta hai. Dikhao ki bridge balanced hai (galvanometer se koi current nahi), aur $ε = 15 V$ source se total current dhundho.

Recall Solution 5.1

Balance condition. Bridge balanced hota hai jab do midpoints (figure mein B aur D) same potential par hote hain, to galvanometer se koi current cross nahi karta. Yeh tab hota hai jab $\frac{R _{1}}{R _{2}} = \frac{R _{3}}{R _{4}} .$ Check karo: $\frac{R _{1}}{R _{2}} = \frac{2}{4} = 0.5$ aur $\frac{R _{3}}{R _{4}} = \frac{3}{6} = 0.5$ . Equal → balanced. ✓ YEH ratio kyun? Galvanometer zero current carry kar raha hai (KCL force karti hai ki $R_{1}$ se guzarne wala current $R_{2}$ se guzarne wale ke equal ho, aur isi tarah $R_{3}$ - $R_{4}$ pair), to har side ek simple series voltage divider hai. Equal ratios dono midpoints ko $ε$ ke same fraction par rakhte hain → equal potential → koi bridge current nahi (yeh Electric Potential single-valuedness use karta hai). Total current (galvanometer open, to do branches independent hain, phir parallel mein): $Left branch: R_{1} + R_{2} = 6 Ω, Right branch: R_{3} + R_{4} = 9 Ω.$ $R_{eq} = \frac{6 \cdot 9}{6 + 9} = \frac{54}{15} = 3.6 Ω, I = \frac{ε}{R _{eq}} = \frac{15}{3.6} = 4.1 \overline{6} A .$

Exercise 5.2 (full two-mesh system)

Ek circuit mein do meshes hain jo ek middle resistor share karte hain. Source $ε = 20 V$ ; left mesh resistor $R_{a} = 4 Ω$ , shared middle resistor $R_{m} = 2 Ω$ , right mesh resistor $R_{b} = 6 Ω$ . Mesh currents $I_{a}$ (left loop) aur $I_{b}$ (right loop) set up karo, dono clockwise, source sirf left mesh mein aur $R_{b}$ sirf right mesh mein. $I_{a}, I_{b}$ solve karo.

Recall Solution 5.2

Yeh Mesh and Nodal Analysis hai — ek systematic KVL application. Shared resistor $R_{m}$ difference $(I_{a} - I_{b})$ carry karta hai kyunki do clockwise loops usse opposite directions mein push karte hain. KVL, left loop (source + $R_{a}$ + shared $R_{m}$ ): $ε - I_{a} R_{a} - (I_{a} - I_{b}) R_{m} = 0.$ Numbers plug karo: $20 - 4 I_{a} - 2 (I_{a} - I_{b}) = 0 \Rightarrow 6 I_{a} - 2 I_{b} = 20.$ KVL, right loop ( $R_{b}$ + shared $R_{m}$ , koi source nahi): $- (I_{b} - I_{a}) R_{m} - I_{b} R_{b} = 0 \Rightarrow - 2 (I_{b} - I_{a}) - 6 I_{b} = 0 \Rightarrow 2 I_{a} - 8 I_{b} = 0 \Rightarrow I_{a} = 4 I_{b} .$ Substitute $I_{a} = 4 I_{b}$ into $6 I_{a} - 2 I_{b} = 20$ : $6 (4 I_{b}) - 2 I_{b} = 20 \Rightarrow 22 I_{b} = 20 \Rightarrow I_{b} = \frac{10}{11} \approx 0.9091 A .$ $I_{a} = 4 I_{b} = \frac{40}{11} \approx 3.6364 A .$ Shared resistor mein current: $I_{a} - I_{b} = \frac{40}{11} - \frac{10}{11} = \frac{30}{11} \approx 2.7273 A$ (left loop ke clockwise sense mein flow karta hai). ✓

Recall Master checklist (finish karne ke baad kholo)

Ek ek line — is set se jo reflexes train ho rahe hain. KCL sahi tool kab hai? ::: Jab charge kisi node par milta ya split hota hai — currents ka sum $= 0$ . KVL sahi tool kab hai? ::: Jab tum ek closed loop trace kar sako — signed potential changes ka sum $= 0$ . Negative current ka matlab kya hai? ::: Assumed arrow backward tha; magnitude rakho, direction flip karo. Current ke saath traverse kiye gaye resistor ka sign? ::: Ek drop, $- I R$ . − terminal par enter ki gayi battery ka sign? ::: Ek rise, $+ ε$ . Do clockwise mesh loops dwara shared resistor mein current? ::: Do mesh currents ka difference. Wheatstone balance condition? ::: $R_{1} / R_{2} = R_{3} / R_{4}$ → equal midpoint potentials → zero galvanometer current.

Connections

Ohm's Law — upar har solution ne $V = I R$ ko do laws ke beech bridge ki tarah use kiya.
Series and Parallel Resistors — L3–L5 mein networks collapse karne ke liye use kiya.
Conservation of Charge — har KCL step ke peeche "kyun" hai.
Conservation of Energy — har KVL loop ke zero sum hone ke peeche "kyun" hai.
Electric Potential — single-valuedness Wheatstone balance argument ko justify karti hai.
Wheatstone Bridge — Exercise 5.1.
Mesh and Nodal Analysis — Exercise 5.2 systematic method ka entry point hai.