Series and parallel resistance: Question bank — Hinglish deep dive

1.8.17 · D5 · Physics › Electromagnetism › Series and parallel resistance

Definition Teen symbols jo is page par zyada use hote hain

Conductance $G$ simply resistance ka reciprocal hai: $G = \frac{1}{R}$ , siemens mein measure hota hai. Matlab hai "current kitni aasaani se flow karta hai" — resistance ka bilkul ulta. Jahan resistors current se ladte hain, conductance ease measure karta hai. Parallel branches apna $G$ add karte hain kyunki zyada raaste kholne se zyada ease milti hai.
Summation sign $\sum$ "sab ko jodo" ka shorthand hai. $\sum_{k} R_{k}$ ka matlab hai $R_{1} + R_{2} + \dots + R_{n}$ — subscript $k$ bas har resistor se guzarta hai. Ye lambe " $+ \dots +$ " chains likhne se bachata hai.
Sign / polarity convention: har resistor par hum current arrow us direction mein draw karte hain jis mein hum assume karte hain charge flow karta hai, aur jo end current enter karta hai use $+$ mark karte hain aur jo end leave karta hai use $-$ (current hamesha resistor mein higher se lower potential ki taraf drop hoti hai). KVL phir loop mein rise (battery) ke liye $+$ aur har drop ke liye $-$ padhta hai; KCL node mein aate arrows ko positive count karta hai. Neeche figure dekho.

Upar ke teen chote circuits wo mental pictures hain jo is page ke har question ko test kar rahe hain — jab bhi koi "picture" shaky lage, inhe dobara dekho. Notice karo: series sketch mein ek arrow dono resistors se guzarta hai (same $I$ ); parallel sketch mein arrow top node par split hota hai aur bottom par wapas milta hai (same $V$ , currents add hote hain); mixed sketch mein $R_{1}$ ek parallel pair ke saath series mein hai.

True or false — justify karo

Series mein zyada resistors add karne se hamesha total resistance badhti hai.

True. Series matlab ek hi path hai, toh har extra obstacle same current ke liye aur zyada voltage drop pile karta hai —

R_{e q} = \sum_{k} R_{k}

sirf badh sakta hai.

Parallel mein zyada resistors add karne se hamesha total resistance ghatti hai.

True. Har naya branch current ke liye ek extra raasta hai, toh conductance

\sum_{k} G_{k} = \sum_{k} 1/ R_{k}

badhti hai, matlab

R_{e q}

shrink hoti hai — smallest branch se bhi kam.

Parallel combination mein equivalent resistance sabse bade branch resistance ke barabar ho sakti hai.

False.

R_{e q}

hamesha strictly sabse chote branch se bhi chhoti hoti hai, kyunki koi bhi doosra path add karne se current ke liye aur room khuljata hai.

Do identical resistors series ya parallel mein wire karne par same

R_{e q}

dete hain.

False. Series

2 R

deta hai, parallel

R /2

deta hai — 4 ka factor apart. Same parts, ulti wiring, ulta effect.

Series mein, sabse bada resistor sabse bada voltage apne upar rakhta hai.

True. Current

I

shared hai, aur

V_{k} = I R_{k}

, toh voltage directly

R_{k}

ke saath scale karta hai — sabse bada obstacle push ka sabse bada share leta hai.

Parallel mein, sabse bada resistor sabse zyada current carry karta hai.

False. Voltage shared hai, aur

I_{k} = V / R_{k}

, toh current

R

se inversely proportional hai — sabse chota resistor sabse zyada current grab karta hai.

Agar network mein har resistance double karo, toh

R_{e q}

bhi double hoti hai.

True. Dono

\sum_{k} R_{k}

(series) aur

(\sum_{k} 1/ R_{k})^{- 1}

(parallel) common factor ke saath linearly scale karte hain, toh unhe nest karne par bhi

R_{e q}

us factor se scale hoti hai.

Ek wire (zero resistance) ek resistor ke parallel mein rakh do toh pair resistor ki tarah behave karta hai.

False. Ye resistor ko short-circuit kar deta hai:

R_{e q} = R \cdot 0/ (R + 0) = 0

. Saara current free path leta hai aur resistor bypass ho jaata hai.

Parallel pair ke ek branch mein break (infinite resistance) hone par doosra branch normally kaam karta rehta hai.

True. Toota branch simply koi current carry nahi karta; surviving branch ab bhi same nodes par same voltage dekhta hai aur akele ki tarah behave karta hai.

Error dhundo

3 Ω

aur

6 Ω

parallel mein,

R_{e q} = 3 + 6 = 9 Ω

Galat — ye series answer hai. Parallel mein conductances add hote hain (

G = 1/ R

R_{e q} = \frac{3 \cdot 6}{9} = 2 Ω

, jo

3 Ω

se kam hona chahiye.

"Maine

\frac{1}{R _{e q}} = \frac{1}{3} + \frac{1}{6} = \frac{1}{2}

likha, toh

R_{e q} = \frac{1}{2} Ω

Reciprocal lena bhool gaye:

\frac{1}{2}

asal mein

1/ R_{e q}

hai (total conductance), toh usse flip karna zaroori hai —

R_{e q} = 2 Ω

"Dono wires ek seedhi line jaisi dikhti hain, toh wo resistors series mein hain."

Seedha dikhna kaafi nahi hai. Check karo koi node hai jahan current split hoti ho; agar hoti hai, toh ye parallel hain. Sirf unbranched path truly series hai.

"Dono parallel branches same voltage share karti hain, toh unhe equal current carry karni chahiye."

Same voltage ka matlab same current nahi hota jab tak resistances equal na hon.

I_{k} = V / R_{k}

, toh unequal

R

se unequal current milta hai.

n

equal resistors

R

parallel mein

n R

dete hain."

Ulta hai.

n

equal resistors ka parallel

R / n

deta hai (zyada paths, kam resistance).

n R

series ka result hai.

"Total current battery se nikalne ke liye main un branch currents ko add karta hoon jo maine abhi compute nahi kiye."

Circular hai. Pehle

R_{e q}

mein collapse karo aur Ohm's Law se

I = V / R_{e q}

use karo; branch currents baad mein shared node voltage se milte hain.

"Battery ki internal resistance external circuit ke parallel mein hai."

Nahi — internal resistance external load ke saath series mein hoti hai (dekho EMF and Internal Resistance); same current dono se guzarta hai.

Why questions

Series mein har resistor se current same kyun hoti hai?

Sirf ek hi path hota hai, aur charge kisi bhi point par accumulate nahi ho sakta (KCL), toh jo pehle resistor mein enter karta hai wahi last se exit karta hai.

Parallel mein har resistor ke across voltage same kyun hoti hai?

Voltage do nodes ke beech ka difference hai, aur saare parallel branches exactly same do nodes connect karte hain — toh har ek same difference feel karta hai.

Parallel mein

R

ki jagah conductance

G = 1/ R

kyun add karte hain?

Kyunki currents add hote hain (KCL):

I = \sum_{k} V / R_{k} = V \sum_{k} 1/ R_{k}

. Jo quantity sum hoti hai wo

1/ R

(conductance) hai, aur

I / V = 1/ R_{e q}

"Product over sum" shortcut

\frac{R _{1} R _{2}}{R _{1} + R _{2}}

exactly do resistors ke liye hi kyun kaam karta hai?

Ye bas

(\frac{1}{R _{1}} + \frac{1}{R _{2}})^{- 1} = (\frac{R _{2} + R _{1}}{R _{1} R _{2}})^{- 1}

rewritten hai. Teen ya zyada ke liye ise ek single fraction mein itni cleanly collapse nahi kar sakte.

Ye formulas sirf Ohm's law aur Kirchhoff's laws se hi kyun aane chahiye?

Ohm's law har resistor ka

V

aur

I

relate karta hai; KVL fix karta hai ki voltages series mein kaise combine hote hain; KCL fix karta hai ki currents parallel mein kaise combine hote hain. Kuch aur assume nahi kiya jaata (dekho Kirchhoff's Laws).

Mixed network innermost-first kyun collapse karte hain?

Series ya parallel rule tab hi apply kar sakte ho jab group purely ek ya doosra ho. Innermost pure group reduce karne se ek complicated network simpler ban jaata hai jise tum dobara classify kar sako.

Edge cases

Ek akela resistor

R

"series mein kuch aur nahi" —

R_{e q}

kya hai?

Bas

R

khud. Ek term ki chain usi ek term mein sum hoti hai; koi doosra drop add karne ke liye nahi hai.

Ek akela resistor

R

"parallel mein kuch aur nahi" —

R_{e q}

kya hai?

Woh bhi bas

R

. Reciprocal sum mein ek term hai,

1/ R_{e q} = 1/ R

, toh seedha wapas

R

flip ho jaata hai — series wale ka symmetric trivial case.

Series chain ka ek resistor plain wire (

0 Ω

) se replace ho jaaye toh kya hoga?

R_{e q}

us resistor ki value se girta hai; current badhti hai aur us segment par zero voltage dikhta hai — lekin chain ke baaki hisse ab bhi current limit karte hain.

Series chain ka ek resistor open break (

\infty

) ban jaaye toh current kya hai?

Har jagah zero. Series mein single path hota hai, toh ek break saari flow rok deta hai; har kaam karne wala resistor ab koi voltage drop nahi dikhata.

Do equal resistors

R

parallel mein —

R_{e q}

kya hai, aur current kaise split hoti hai?

R_{e q} = R /2

, aur current exactly half-half split hoti hai kyunki equal resistance matlab equal

I_{k} = V / R

hota hai.

Jaise ek parallel branch ki resistance

R_{1} \to \infty

hoti hai,

R_{e q}

kya approach karta hai?

Ye doosri branch ki resistance

R_{2}

approach karta hai. Infinitely resistive branch koi current carry nahi karti aur drop out ho jaati hai, sirf

R_{2}

bachta hai.

Jaise ek parallel branch ki resistance

R_{1} \to 0

hoti hai,

R_{e q}

kya approach karta hai?

0

approach karta hai. Near-perfect conductor pair ko short-circuit kar deta hai;

\frac{0 \cdot R _{2}}{0 + R _{2}} = 0

, aur almost saara current us branch se guzarta hai.

Resistors ka ek network jo na purely series hai na purely parallel — kya ye do rules hamesha solve kar sakte hain?

Nahi. Balanced ya unbalanced bridge (dekho Wheatstone Bridge) mein koi pure series/parallel groups nahi hote; tumhe full Kirchhoff analysis chahiye.

Connections

Ohm's Law — $V = I R$ , upar har branch mein invoke hota hai.
Kirchhoff's Laws — KVL "same current / voltages add" trap justify karta hai; KCL "same voltage / currents add" trap justify karta hai.
EMF and Internal Resistance — series-vs-parallel internal-resistance trap.
Wheatstone Bridge — wo edge case jahan ye rules khatam ho jaate hain.
Power in Circuits — currents/voltages pata hone ke baad natural follow-up.
Parent: Hinglish topic note.