Series and parallel resistance: Worked examples — Hinglish deep dive

1.8.17 · D3 · Physics › Electromagnetism › Series and parallel resistance

Yeh page Series and parallel resistance ke liye exhaustive worked-example lab hai. Parent note ne tumhe chaar clean examples diye. Yahan hum deliberately har case class ko dhundh ke nikalte hain jisse ek problem banai ja sakti hai — weird, degenerate, aur exam-trap wale cases bhi — taaki tum koi bhi scenario pehli baar na dekho jab exam mein aaye.

Yahan sab kuch sirf do tools par tika hua hai, dono parent aur uske links mein define kiye gaye hain:

Ohm's Law: $V = I R$ — ek resistor ke across voltage uske current aur resistance ka product hota hai.
Kirchhoff's Laws: ek node par current in = current out (KCL); ek loop mein voltages zero mein add hote hain (KVL).

Agar koi symbol neeche unfamiliar lage, woh parent note mein build kiya gaya tha — pehle woh padho.

The scenario matrix

Kuch bhi work karne se pehle, chalte hain cases ka pura space enumerate karein. Har resistor-network problem jo tumhe kabhi milegi woh in cells mein se ek hai (ya kuch ka combination):

#	Case class	Kya special hai isme	Covered by
A	Pure series	ek hi path, current shared	Ex 1
B	Pure parallel (two)	do nodes, voltage shared	Ex 2
C	Parallel, unequal branches	current unequally split hoti hai	Ex 2
D	Mixed (series + parallel block)	stages mein collapse karo	Ex 3
E	Degenerate: short circuit ( $R \to 0$ branch)	ek $0 Ω$ path sab kuch le jaati hai	Ex 4
F	Degenerate: open circuit ( $R \to \infty$ branch)	ek broken branch kuch nahi carry karti	Ex 5
G	Equal resistors, $n$ of them	fast $R / n$ aur $n R$ rules	Ex 6
H	Limiting behaviour: ek $R$ ≫ doosre se	kaun sa "wins"?	Ex 7
I	Real-world word problem (household wiring)	words → circuit mein translate karo	Ex 8
J	Exam twist: symmetric bridge / hidden equal split	trick pakdo, poora algebra avoid karo	Ex 9

Case A — Pure series

Worked example Ex 1 · Cell A — same current, resistances sum

Teen resistors ek line mein: $R_{1} = 2 Ω$ , $R_{2} = 3 Ω$ , $R_{3} = 5 Ω$ , ek $V = 20$ V battery ke across. $R_{e q}$ , current $I$ , aur har voltage drop nikalo.

Forecast: Ek hi path, toh current har jagah same hai. Guess karo: kya $R_{e q}$ $5 Ω$ se bada hoga ya chhota? (Bada — series hamesha pile up karta hai.)

Step 1. $R_{e q} = R_{1} + R_{2} + R_{3} = 2 + 3 + 5 = 10 Ω$ . Yeh step kyon? Ek unbranched path → sab mein same $I$ → KVL se drops add hote hain → resistances add hoti hain.

Step 2. $I = V / R_{e q} = 20/10 = 2$ A. Yeh step kyon? Ohm's Law poore collapsed chain par apply kiya.

Step 3. Drops: $V_{1} = I R_{1} = 2 \cdot 2 = 4$ V, $V_{2} = 2 \cdot 3 = 6$ V, $V_{3} = 2 \cdot 5 = 10$ V. Yeh step kyon? Same $I$ hai, toh har drop bas $I R_{k}$ hai.

Verify: $V_{1} + V_{2} + V_{3} = 4 + 6 + 10 = 20$ V $= V$ ✓ (KVL). Aur $10 Ω > 5 Ω$ , jaise forecast kiya tha. Units: $A \cdot Ω = V$ ✓.

Cases B & C — Pure parallel, unequal split

Worked example Ex 2 · Cells B + C — same voltage, currents unequally split hoti hain

$R_{1} = 4 Ω$ aur $R_{2} = 12 Ω$ same do nodes ke beech, $V = 24$ V. $R_{e q}$ , branch currents, aur total nikalo.

Forecast: Dono ko poora $24$ V milta hai. Kaun sa branch zyada current carry karta hai — $4 Ω$ wala ya $12 Ω$ wala? (Chhota wala, $4 Ω$ : kam obstacle → zyada flow.)

Step 1. Product over sum: $R_{e q} = \frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{4 \cdot 12}{4 + 12} = \frac{48}{16} = 3 Ω$ . Yeh step kyon? Same do nodes → same $V$ → conductances add hoti hain. Figure dekho: dono green paths identical node pair (blue dots) ko bridge karte hain.

Step 2. Branch currents (har ek ko poora $V$ milta hai): $I_{1} = V / R_{1} = 24/4 = 6$ A, $I_{2} = V / R_{2} = 24/12 = 2$ A. Yeh step kyon? Ohm's Law har branch par — voltage shared hai, current nahi.

Step 3. Total: $I = I_{1} + I_{2} = 6 + 2 = 8$ A. Yeh step kyon? KCL — branch currents wapas milti hain.

Verify: Equivalent se check: $I = V / R_{e q} = 24/3 = 8$ A ✓. Note karo $I_{1} = 3 I_{2}$ exactly, aur $R_{2} = 3 R_{1}$ — chhote resistor ne teen guna current pakad li (Cell C, unequal split). Aur $R_{e q} = 3 Ω < 4 Ω$ (sabse chhote se bhi chhota) ✓.

Case D — Mixed network, stages mein collapse karo

Worked example Ex 3 · Cell D — ek resistor aur ek parallel block in series

$R_{1} = 6 Ω$ in series ek parallel pair $R_{2} = 10 Ω$ aur $R_{3} = 15 Ω$ ke saath. Supply $V = 30$ V.

Forecast: Pehle kaun collapse karoge — series part ya parallel block? (Parallel block: use pehle ek number mein reduce karna hoga before tum use series mein add kar sako.)

Step 1. Parallel block: $R_{23} = \frac{R _{2} R _{3}}{R _{2} + R _{3}} = \frac{10 \cdot 15}{25} = \frac{150}{25} = 6 Ω$ . Yeh step kyon? $R_{2}, R_{3}$ do nodes share karte hain (figure mein yellow node pair). Inhe ek single equivalent block mein collapse karo.

Step 2. Ab yeh ek plain series chain hai: $R_{e q} = R_{1} + R_{23} = 6 + 6 = 12 Ω$ . Yeh step kyon? Same current $R_{1}$ mein se aur phir poore block mein se jaata hai — ek path → add karo.

Step 3. Battery current: $I = V / R_{e q} = 30/12 = 2.5$ A. Yeh step kyon? Fully collapsed circuit par Ohm's Law.

Step 4. Parallel block ke across voltage: $V_{23} = I \cdot R_{23} = 2.5 \cdot 6 = 15$ V. Phir $I_{2} = 15/10 = 1.5$ A, $I_{3} = 15/15 = 1$ A. Yeh step kyon? Block ka apna voltage dono branches share karti hain; har branch Ohm's Law se split hoti hai.

Verify: $I_{2} + I_{3} = 1.5 + 1 = 2.5$ A $= I$ ✓ (KCL). $R_{1}$ par drop $= 2.5 \cdot 6 = 15$ V, plus block $15$ V $= 30$ V ✓ (KVL).

Case E — Degenerate: a short circuit ( $R \to 0$ )

Worked example Ex 4 · Cell E — ek wire ek resistor ko short kar rahi hai

$R_{1} = 8 Ω$ ek ideal wire $R_{2} = 0 Ω$ ke parallel, $V = 12$ V ke across.

Forecast: $R_{e q}$ kya hai? Current kahan jaayegi? (Guess karo $R_{e q} = 0$ , aur saari current $R_{1}$ ko dodge karegi.)

Step 1. $\frac{1}{R _{e q}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} = \frac{1}{8} + \frac{1}{0}$ . Term $1/0 \to \infty$ hai, toh $1/ R_{e q} \to \infty$ , isliye $R_{e q} = 0$ . Yeh step kyon? Parallel formula ko literally uske limit tak le jaao. Infinite conductance dominate karti hai.

Step 2. Pair ke across voltage $= 0$ (yeh $I \cdot R_{e q} = I \cdot 0$ hai). Toh $I_{1} = V_{b l oc k} / R_{1} = 0/8 = 0$ A $R_{1}$ se. Yeh step kyon? Dono branches block voltage share karti hain; agar block $0 Ω$ ka hai toh $0$ V drop karta hai, toh $R_{1}$ ko kuch nahi milta.

Step 3. Saari current wire se jaati hai: $I_{2} = I_{t o t a l}$ (sirf baaki circuit / source ki limit se limited). Yeh step kyon? KCL — total current phir bhi flow karni chahiye, aur sirf wire hi ek aise path offer karta hai jiska voltage-independent capacity hai.

Verify: Sanity: $R_{e q} = 0 < 8 Ω$ (sabse chhote se bhi chhota — shortcut rule limit par bhi survive karta hai) ✓. Two-resistor shortcut bhi agree karta hai: $\frac{R _{1} R _{2}}{R _{1} + R _{2}} = \frac{8 \cdot 0}{8 + 0} = \frac{0}{8} = 0$ ✓.

Case F — Degenerate: an open circuit ( $R \to \infty$ )

Worked example Ex 5 · Cell F — ek broken branch series mein aur parallel mein

Do sub-questions same broken resistor $R_{2} = \infty$ (open) ke saath. (a) In series: $R_{1} = 5 Ω$ ek open $R_{2}$ ke series mein, $V = 12$ V ke across. (b) In parallel: $R_{1} = 5 Ω$ ek open $R_{2}$ ke parallel mein, $V = 12$ V ke across.

Forecast: Kaun se arrangement mein circuit abhi bhi kaam karta hai — series ya parallel? (Parallel: open branch simply ignore ho jaati hai. Series: open sirf ek hi path tod deta hai → bilkul bhi current nahi.)

Step 1 (a). Series: $R_{e q} = R_{1} + R_{2} = 5 + \infty = \infty$ . Current $I = V / R_{e q} = 12/\infty = 0$ A. Yeh step kyon? Ek hi path hai, aur woh cut hai. Charge gap cross nahi kar sakta → poori line dead hai.

Step 2 (b). Parallel: $\frac{1}{R _{e q}} = \frac{1}{5} + \frac{1}{\infty} = \frac{1}{5} + 0 = \frac{1}{5}$ , toh $R_{e q} = 5 Ω$ . Yeh step kyon? Infinite resistance wale resistor ki zero conductance hoti hai — yeh parallel sum mein kuch nahi add karta. Sirf good branch survive karti hai.

Step 3 (b). Current: $I = V / R_{e q} = 12/5 = 2.4$ A, poori $R_{1}$ se; $I_{2} = 12/\infty = 0$ A. Yeh step kyon? Surviving branch par Ohm's Law; open branch kuch carry nahi karti.

Verify: (a) mein $0$ A milta hai — ek broken series chain dead hai, intuition se match karta hai ✓. (b) mein $2.4$ A milta hai — open branch cleanly vanish ho gayi, $R_{e q}$ akele resistor ke barabar hai ✓. Ex 4 se contrast: short in parallel = $R_{e q} \to 0$ ; open in parallel = branch ignore. Opposite extremes.

Case G — $n$ equal resistors (fast rules)

Case H — Limiting behaviour: ek resistor dominate karta hai

Worked example Ex 7 · Cell H — kya hota hai jab ek

$R$ doosre ko swamp kar deta hai? Do parallel mein: $R_{1} = 2 Ω$ fixed, aur $R_{2}$ ko hum badhate hain: try karo $R_{2} = 2$ , phir $20$ , phir $2000 Ω$ . $R_{e q}$ track karo.

Forecast: Jaise $R_{2} \to \infty$ , kaun si value $R_{e q}$ approach karta hai? (Yeh $R_{1} = 2 Ω$ approach karta hai — mota resistor open ban jaata hai, toh sirf chhota wala bachta hai.)

Step 1. $R_{2} = 2$ : $R_{e q} = \frac{2 \cdot 2}{2 + 2} = \frac{4}{4} = 1 Ω$ (equal resistors → half). Step 2. $R_{2} = 20$ : $R_{e q} = \frac{2 \cdot 20}{22} = \frac{40}{22} \approx 1.818 Ω$ . Step 3. $R_{2} = 2000$ : $R_{e q} = \frac{2 \cdot 2000}{2002} = \frac{4000}{2002} \approx 1.998 Ω$ . Yeh steps kyon? Hum limit $lim_{R_{2} \to \infty} \frac{R _{1} R _{2}}{R _{1} + R _{2}}$ dekh rahe hain. Top aur bottom ko $R_{2}$ se divide karo: $\frac{R _{1}}{R _{1} / R _{2} + 1} \to \frac{R _{1}}{0 + 1} = R_{1}$ . Figure mein curve (red) dashed line $R_{e q} = 2$ ki taraf flatten ho raha hai.

Verify: Sequence $1 \to 1.818 \to 1.998$ monotonically $2$ ki taraf climb karta hai lekin kabhi nahi pahunchta — exactly horizontal asymptote $R_{e q} = R_{1}$ ✓. Physical reading: ek bahut bada parallel resistor almost ek open hai (Cell F), toh yeh $R_{e q}$ ko barely change karta hai.

Case I — Real-world word problem

Worked example Ex 8 · Cell I — household appliances ek outlet par

Ek wall outlet $V = 120$ V supply karta hai. Tum ek lamp $240 Ω$ ka aur ek heater $24 Ω$ ka same outlet mein lagaate ho (toh woh outlet ke do terminals share karte hain). Har appliance ki current aur outlet se total current nikalo.

Forecast: Kya yeh series hain ya parallel? (Parallel — dono same do outlet terminals ke across wired hain, toh dono ko poora $120$ V milta hai. Isliye appliances independently kaam karte hain.) Zyada current kaun kheenchta hai — lamp ya heater?

Step 1. Parallel pehchano: dono same node pair ke beech → same $V = 120$ V. Yeh step kyon? Pehle case classify karo, hamesha — parent ki mistake "same-looking wires" dekho.

Step 2. Lamp: $I_{l am p} = 120/240 = 0.5$ A. Heater: $I_{h e a t er} = 120/24 = 5$ A. Yeh step kyon? Ohm's Law har branch par; har ek ko poora mains voltage milta hai.

Step 3. Outlet se total: $I = 0.5 + 5 = 5.5$ A (KCL). Step 4. Equivalent: $R_{e q} = \frac{240 \cdot 24}{240 + 24} = \frac{5760}{264} \approx 21.82 Ω$ .

Verify: $I = V / R_{e q} = 120/21.82 \approx 5.5$ A ✓. Low-resistance heater lamp ki current ka $10 \times$ pakad leta hai — isliye heaters breakers trip karte hain. Note karo $R_{e q} \approx 21.8 Ω < 24 Ω$ (sabse chhote se bhi chhota) ✓.

Case J — Exam twist: symmetry algebra se zyada kaam aati hai

Worked example Ex 9 · Cell J — ek balanced bridge (trick pakdo)

Char resistors input node $P$ aur output node $Q$ ke beech ek diamond banate hain: top path $P \to M \to Q$ mein $R_{a} = 10 Ω$ phir $R_{b} = 20 Ω$ hai; bottom path $P \to N \to Q$ mein $R_{c} = 10 Ω$ phir $R_{d} = 20 Ω$ hai. Ek paanchwa resistor $R_{g} = 50 Ω$ $M$ se $N$ ko bridge karta hai. $P$ – $Q$ ke across supply $V = 30$ V hai. Bridge $R_{g}$ se current nikalo, phir $R_{e q}$ .

Forecast: Dono paths identical hain ( $10$ phir $20$ dono par). Kya tumhe messy bridge algebra ki zaroorat hai? (Nahi — symmetry se $M$ aur $N$ same potential par hain, toh $R_{g}$ se koi current nahi jaati. Use delete kiya ja sakta hai!)

Step 1. Dono arms compare karo. Top: $R_{a} : R_{b} = 10 : 20$ . Bottom: $R_{c} : R_{d} = 10 : 20$ . Ratios match karte hain → balanced bridge (Wheatstone Bridge condition). Yeh step kyon? Jab $R_{a} / R_{b} = R_{c} / R_{d}$ , toh $M$ par voltage $N$ par voltage ke barabar hoti hai. Figure dekho: dono midpoints same dashed equipotential line par aate hain.

Step 2. $M$ , $N$ par equal potentials → $R_{g}$ ke across voltage $0$ → $I_{g} = 0/ R_{g} = 0$ A. Yeh step kyon? Ohm's Law: koi potential difference nahi, koi current nahi. Bridge dead weight hai.

Step 3. $R_{g}$ delete karo. Ab parallel mein do independent series arms hain. Har arm: $10 + 20 = 30 Ω$ . Parallel mein do equal $30 Ω$ arms: $R_{e q} = 30/2 = 15 Ω$ . Yeh step kyon? Bridge hatane ke baad network plain series-then-parallel hai (Cell D + Cell G).

Verify: Battery current $I = V / R_{e q} = 30/15 = 2$ A, har arm mein $1$ A split hoti hai. $M$ par potential ( $P$ se): $R_{a}$ ke across drop $= 1 \cdot 10 = 10$ V, toh $M$ $P$ se $10$ V neeche hai. $N$ ke liye bhi same: $1 \cdot 10 = 10$ V $P$ se neeche. $V_{M} = V_{N}$ ✓ → wakai $I_{g} = 0$ , shortcut legitimate tha confirm karta hai.

Wrap-up: matrix ab fully covered hai

Recall Kaun sa example kaun se cell se tha?

A→Ex1 · B,C→Ex2 · D→Ex3 · E (short)→Ex4 · F (open)→Ex5 · G (n equal)→Ex6 · H (limit)→Ex7 · I (word)→Ex8 · J (bridge twist)→Ex9. Scenario matrix ki har row ka ek worked, verified example hai.

Connections

Parent: Series and parallel resistance — woh do formulas jo yeh examples exercise karte hain.
Ohm's Law — literally har step mein use hota hai.
Kirchhoff's Laws — KCL current splits ke liye, KVL voltage checks ke liye.
Wheatstone Bridge — Ex 9 mein balanced-bridge trick.
Power in Circuits — Ex 8 mein heater $5$ A kheenchta hai: follow-up ke liye $P = V I = 600$ W try karo.
EMF and Internal Resistance — ek real battery inme se kisi bhi circuit mein ek series resistor add kar deti hai.