Series and parallel resistance: Exercises — Hinglish deep dive

1.8.17 · D4 · Physics › Electromagnetism › Series and parallel resistance

Shuru karne se pehle, yeh teen hi tools hain jo hum kabhi bhi use karte hain:

Level 1 — Recognition

Tum sirf decide karte ho ki connection kis type ka hai aur rule padh lete ho. Abhi zyada arithmetic nahi hai.

Recall Solution 1.1

Kya dekhna hai: ek single path, koi junction nahi jahan current split ho sake → yeh series hai. Kyun: same current pehle $R_{1}$ phir $R_{2}$ se guzarna chahiye (charge ikatta nahi ho sakta, KCL). Calculate: resistances add hote hain, $R_{e q} = 10 + 15 = 25 Ω$ . Sanity: $25 Ω$ dono resistors se bada hai — series ke liye sahi hai.

Recall Solution 1.2

Kya hai: dono ends same do nodes se jude hain → current do branches mein split ho sakti hai → parallel. Kyun: har resistor ko identical node-to-node voltage feel hoti hai. Calculate (product over sum): $R_{e q} = \frac{10 \times 15}{10 + 15} = \frac{150}{25} = 6 Ω$ . Sanity: $6 Ω$ sabse chhote ( $10 Ω$ ) se bhi chhota hai — parallel ke liye sahi hai.

Recall Solution 1.3

Top diagram: ek single line, current kahi aur ja hi nahi sakti → series. Bottom diagram: node A par current do raston mein split hoti hai aur node B par wapas milti hai → parallel. Trick yeh hai ki drawing ki shape nahi dekni — yeh dekhna hai ki koi junction current ko split karta hai ya nahi.

Level 2 — Application

Ab numbers ko Ohm's law se guzaro taaki currents aur voltage drops nikal sako.

Recall Solution 2.1

(a) Series → add karo: $R_{e q} = 4 + 8 = 12 Ω$ . (b) Poori chain par Ohm: $I = \frac{V}{R _{e q}} = \frac{12}{12} = 1 A$ . Kyun: equivalent resistor wohi current draw karta hai jo battery actually deliver karti hai. (c) Har drop $I R_{k}$ hai: $V_{1} = 1 \times 4 = 4 V$ , $V_{2} = 1 \times 8 = 8 V$ . KVL check: $4 + 8 = 12 V$ ✓ — drops applied voltage mein add hote hain.

Recall Solution 2.2

(a) Product over sum: $R_{e q} = \frac{4 \times 8}{4 + 8} = \frac{32}{12} = \frac{8}{3} \approx 2.67 Ω$ . (b) Har branch ko poora $12 V$ feel hota hai (same nodes): $I_{1} = \frac{12}{4} = 3 A$ , $I_{2} = \frac{12}{8} = 1.5 A$ . (c) KCL — branches add hote hain: $I = 3 + 1.5 = 4.5 A$ . Cross-check: $I = \frac{V}{R _{e q}} = \frac{12}{8/3} = 4.5 A$ ✓. Dhyan do: chhota resistor ( $4 Ω$ ) bada current leta hai.

Recall Solution 2.3

(a) $n$ equal series mein → $n R = 5 \times 20 = 100 Ω$ . (b) $n$ equal parallel mein → $R / n = 20/5 = 4 Ω$ . Kyun: paanch equal raaste load share karte hain, resistance paanchwa hissa kar dete hain.

Level 3 — Analysis

Mixed networks ko ek block at a time collapse karo, phir numbers wapas bahar nikalo.

Recall Solution 3.1

Step A — parallel pair collapse karo. Do equal $10 Ω$ → $R / n = 10/2 = 5 Ω$ . Kyun: woh same do nodes share karte hain. Step B — $R_{1}$ ke saath series. Battery se poora current pehle $R_{1}$ se phir collapsed block se guzarta hai: $R_{e q} = 5 + 5 = 10 Ω$ . Battery current: $I = \frac{15}{10} = 1.5 A$ . Parallel block ke across voltage: $V_{b l oc k} = I \times 5 = 1.5 \times 5 = 7.5 V$ (baaki $7.5 V$ $R_{1}$ par drop hote hain; KVL: $7.5 + 7.5 = 15$ ✓). Branch currents: har ek $= \frac{7.5}{10} = 0.75 A$ . Check: $0.75 + 0.75 = 1.5 A$ ✓.

Recall Solution 3.2

Step A: $6$ aur $3$ ka parallel: $\frac{6 \times 3}{6 + 3} = \frac{18}{9} = 2 Ω$ . Step B: series: $R_{e q} = 2 + 8 = 10 Ω$ . Current: $I = \frac{20}{10} = 2 A$ . Parallel block ke across voltage: $V = I \times 2 = 2 \times 2 = 4 V$ . ( $R_{3}$ par drop $2 \times 8 = 16 V$ hai; $4 + 16 = 20$ ✓.)

Recall Solution 3.3

Step A — inner series: $R_{3} + R_{4} = 1 + 3 = 4 Ω$ (dono se same current guzarta hai). Step B — parallel: woh $4 Ω$ branch, $R_{2} = 6 Ω$ ke saath parallel mein: $\frac{4 \times 6}{4 + 6} = \frac{24}{10} = 2.4 Ω$ . Step C — outer series: $R_{e q} = 2 + 2.4 = 4.4 Ω$ . Current: $I = \frac{12}{4.4} = \frac{30}{11} \approx 2.73 A$ .

Level 4 — Synthesis

Ulta sochna hai: tumhe target diya gaya hai aur tumhe design karna hai ya infer karna hai.

Recall Solution 4.1

Goal-driven reasoning: $9 Ω$ , $6 Ω$ (ek) aur $18 Ω$ (teen series mein) ke beech mein hai, toh kuch aisa chahiye jo add up kare. Do ko parallel mein rakho: $6/2 = 3 Ω$ . Phir usे teesre $6 Ω$ ke series mein rakho: $3 + 6 = 9 Ω$ . ✓ Wiring: teesra resistor ek line mein, jo ek node par jaata hai jahan se do parallel $6 Ω$ resistors mein split hota hai.

Recall Solution 4.2

Parallel equation set up karo: $\frac{1}{4} = \frac{1}{R} + \frac{1}{12}$ . Isolate karo: $\frac{1}{R} = \frac{1}{4} - \frac{1}{12} = \frac{3}{12} - \frac{1}{12} = \frac{2}{12} = \frac{1}{6}$ . Flip karo: $R = 6 Ω$ . Check (product over sum): $\frac{6 \times 12}{6 + 12} = \frac{72}{18} = 4 Ω$ ✓.

Recall Solution 4.3

Words ko translate karo: $R_{1} + R_{2} = 18$ aur $\frac{R _{1} R _{2}}{R _{1} + R _{2}} = 4$ . Doosre se: $R_{1} R_{2} = 4 \times 18 = 72$ . Toh hume do numbers chahiye jinki sum 18 aur product 72 ho. $x^{2} - 18 x + 72 = 0$ solve karo → $x = \frac{18 \pm 324 - 288}{2} = \frac{18 \pm 6}{2}$ . $R_{1} = 12 Ω, R_{2} = 6 Ω$ . Check: series $12 + 6 = 18$ ✓; parallel $\frac{72}{18} = 4$ ✓.

Level 5 — Mastery

Boundary cases, limits, aur full case-coverage — jahan naive plugging break ho jaata hai.

Recall Solution 5.1

Parallel formula: $\frac{1}{R _{e q}} = \frac{1}{10} + \frac{1}{0}$ . $\frac{1}{0}$ wala term infinite conductance hai. Iska matlab: $\frac{1}{R _{e q}} \to \infty$ , toh $R_{e q} \to 0 Ω$ . Physically: $0 Ω$ wire ek free road hai; essentially poora current ussi se jaata hai aur $10 Ω$ resistor bypass ho jaata hai (approximately $0$ current carry karta hai). Kisi bhi resistor ke across short uska effect khatam kar deta hai.

Recall Solution 5.2

Parallel formula: $\frac{1}{R _{e q}} = \frac{1}{10} + \frac{1}{\infty} = \frac{1}{10} + 0$ . Toh $R_{e q} = 10 Ω$ — toota hua branch aise hai jaise woh tha hi nahi. Kyun: infinite resistance matlab zero current us taraf flow kar sakti hai, toh woh kuch contribute nahi karta. Series se contrast: ek series branch mein break hone par $R_{e q}$ infinite hota hai (single path cut ho jaata hai, current bilkul band ho jaati hai). Wohi toota component, ulta consequence — location sab decide karta hai.

Recall Solution 5.3

$R_{e q} = \frac{10 R}{10 + R}$ use karo aur har regime check karo:

$R \to 0$ (short): $R_{e q} \to \frac{0}{10} = 0 Ω$ . 5.1 se match karta hai.
$R = 10 Ω$ (equal): $R_{e q} = \frac{100}{20} = 5 Ω = R_{0} /2$ . Do equal resistors → aadha.
$R \to \infty$ (open): $R_{e q} \to \frac{10 R}{R} = 10 Ω = R_{0}$ . 5.2 se match karta hai. Conclusion: $R_{e q}$ monotonically $0$ se upar $10 Ω$ ki taraf badhta hai lekin kabhi nahi pahunchta. Toh parallel combination hamesha strictly chhote wale resistor se chhoti hoti hai — $10 Ω$ boundary ek ceiling hai jo tabhi approach hoti hai jab doosri branch vanish ho jaaye.

Recall Feynman recap — woh ek aadat jo sab solve kar deti hai

Kisi bhi network ke liye: (1) ek pure series ya pure parallel block dhundo, (2) use collapse karo, (3) redraw karo, (4) tab tak repeat karo jab tak ek resistor na bache, phir Ohm's Law use karo aur current aur voltage wapas bahar nikalo. Series resistance add karta hai aur voltage split karta hai; parallel conductance add karta hai aur current split karta hai. Bas yahi pura game hai.

Connections

Ohm's Law — har collapsed $R_{e q}$ ko currents aur drops mein convert karne ke liye use hota hai.
Kirchhoff's Laws — KVL checks (voltages add hote hain) aur KCL checks (currents add hote hain) har solution verify karte hain.
Power in Circuits — agla step: inhi networks mein $P = I^{2} R$ distribute karo.
Wheatstone Bridge — un networks ke liye jo na pure series hain na pure parallel.
EMF and Internal Resistance — internal resistance sirf ek aur resistor hai battery ke series mein.