Series and parallel resistance: Visual walkthrough

1.8.17 · D2 · Physics › Electromagnetism › Series and parallel resistance

Step 0 — Har symbol ka matlab (kuch bhi assume nahi)

Kisi bhi formula se pehle, teen actors se milo. Figure dekho: ek battery charge ko ek loop mein push karti hai ek grey block (resistor) ke through.

Do aur facts, bhi pictures ke saath (Kirchhoff ke do laws):

Baaki sab isi ka consequence hai.

Step 1 — Series: single path dhundho

KYA: Do resistors $R_{1}, R_{2}$ end-to-end ek unbroken line mein jode gaye, voltage $V$ ki battery se fed.

KYUN: Hum woh ek resistor $R_{e q}$ chahte hain jise battery is chain se alag na kar sake — same push, same flow. Pehle samajhna hoga ki "one path" kya force karta hai.

PICTURE: Red arrow follow karo. Yeh $R_{1}$ mein enter karta hai, $R_{1}$ se nikalta hai, aur jaane ki ek hi jagah hai — seedha $R_{2}$ mein. Koi junction nahi, koi fork nahi.

Step 2 — Series: push split hoti hai (voltage add hoti hai)

KYA: Har resistor jo voltage use karta hai woh likho, phir add karo.

KYUN: Battery ki push $V$ ko har choke cross karte waqt "spend" karna padta hai. KVL kehta hai spent amounts ka sum supplied amount ke barabar hona chahiye.

PICTURE: Do "waterfalls" ek ke baad ek — total drop dono drops ka sum hai. Red bracket total $V$ mark karta hai; do black brackets uske pieces hain.

Har resistor par Ohm apply karo (Step 1 se same $I$ ): $V_{1} = I R_{1} V_{2} = I R_{2}$ Yahan $V_{1}$ woh push hai jo $R_{1}$ cross karne mein lagti hai, aur same $I$ dono mein aata hai kyunki Step 1 kehta hai.

Ab KVL — supplied push drops ke sum ke barabar hai: $supplied V = I R_{1} V_{1} + I R_{2} V_{2} = I R_{1} + I R_{2} = I (R_{1} + R_{2})$ Dhyan do ki humne $I$ factor out kiya — yeh tab hi legal hai jab har term mein same $I$ ho. Yahi Step 1 ka payoff hai.

Step 3 — Series: equivalent resistance padh lo

KYA: Definition $R_{e q} = V / I$ se compare karo.

KYUN: $R_{e q}$ define hai (poori combination ke across voltage) ÷ (usme jaane wala current). Humhare paas already $V = I (R_{1} + R_{2})$ hai, toh divide karo.

PICTURE: Do-resistor chain ek mote block mein collapse hoti hai — red equivalent resistor — jise battery identically experience karti hai.

$R_{e q} = \frac{V}{I} = \frac{I ( R _{1} + R _{2} )}{I} = R_{1} + R_{2}$ $I$ upar-niche cancel hota hai, bachta hai: $R_{ser i es} = R_{1} + R_{2} + \dots + R_{n}$ ( $n$ resistors ke liye, Step 2 mein bas zyada terms hain; factor-out phir bhi kaam karta hai.)

Step 4 — Parallel: same do nodes dhundho

KYA: Do resistors dono same do junctions ke beech wire kiye gaye hain, unhe node A (upar) aur node B (niche) kaho.

KYUN: Voltage do points ke beech ka difference hota hai. Agar dono resistors ek taraf A aur doosri taraf B ko touch karte hain, toh dono ko same difference feel hoga.

PICTURE: Dono black resistors do red nodes A aur B ke beech latkh rahe hain — jaise do ladders same do shelves ke beech lete hue.

Step 5 — Parallel: flow split hoti hai (currents add hoti hai)

KYA: Har branch mein current likho, phir junction par add karo.

KYUN: Node A par incoming current $I$ ko fork milta hai — use do branches mein split hona padta hai, aur KCL kehta hai pieces wapas $I$ mein add hote hain.

PICTURE: Red total current A par aata hai aur do black branch arrows mein divide hota hai; woh B par wapas melte hain.

Har branch par Ohm apply karo (Step 4 se same $V$ ), current ke liye solve karo: $I_{1} = \frac{V}{R _{1}} I_{2} = \frac{V}{R _{2}}$ Yahan same $V$ har fraction ke upar baitha hai — woh Step 4 kaam kar raha hai.

Ab KCL — branch currents total mein add hote hain: $total I = V / R_{1} I_{1} + V / R_{2} I_{2} = \frac{V}{R _{1}} + \frac{V}{R _{2}} = V (\frac{1}{R _{1}} + \frac{1}{R _{2}})$ Is baar humne $V$ factor out kiya — legal tab hi jab har branch mein same $V$ ho.

Step 6 — Parallel: equivalent padh lo (aur flip karo!)

KYA: $\frac{1}{R _{e q}} = \frac{I}{V}$ se compare karo.

KYUN: $I = V (\frac{1}{R _{1}} + \frac{1}{R _{2}})$ se, dono sides ko $V$ se divide karo taaki woh combination isolate ho jo $1/ R_{e q}$ ke barabar hai.

PICTURE: Do side-by-side branches collapse hokar A aur B ke beech ek red equivalent resistor ban jaati hain.

$\frac{I}{V} = \frac{1}{R _{1}} + \frac{1}{R _{2}} ⟹ \frac{1}{R _{p a r a l l e l}} = \frac{1}{R _{1}} + \frac{1}{R _{2}} + \dots + \frac{1}{R _{n}}$

Step 7 — Degenerate & edge cases (taaki kuch bhi surprise na kare)

Har derivation ko apni extremes survive karni chahiye. Figure dekho — teen limiting scenarios side by side draw kiye hain.

Step 8 — Ek fully collapsed mixed network (saare moves ek saath)

KYA: $R_{1} = 4 Ω$ series mein ek parallel pair $R_{2} = 6 Ω, R_{3} = 12 Ω$ ke saath, $V = 10$ V se driven.

KYUN: Dikhana ki do rules composable hain — pehle parallel island collapse karo, phir result ko ek ordinary series link treat karo.

PICTURE: Network ko shrink hote dekho: pair → single 4 Ω block (red) → series sum → single 8 Ω block.

Parallel island collapse karo (Step 6 shortcut): $\frac{6 \cdot 12}{6 + 12} = \frac{72}{18} = 4 Ω$ .
Ab series (Step 3): $4 Ω + 4 Ω = 8 Ω = R_{e q}$ .
Battery current: $I = \frac{V}{R _{e q}} = \frac{10}{8} = 1.25$ A.
Island ke across voltage: $V_{i s l an d} = I \cdot 4 = 5$ V, isliye $I_{2} = \frac{5}{6} \approx 0.83$ A, $I_{3} = \frac{5}{12} \approx 0.42$ A, aur $0.83 + 0.42 = 1.25$ ✓ (KCL sanity check).

Ek-picture summary

Dono derivations, ek doosre ki mirror images, ek canvas par: series = same current, voltages add; parallel = same voltage, currents add.

Recall Feynman: poori kahani ek 12-saal ke bachhe ko batao

Charges ko logon ki bheed sochon. Resistor ek darwaza hai — narrow wala mushkil se nikalta hai. Series darwaze hain ek ke peeche ek ek single hallway mein: sabko sab se guzarna padta hai baari-baari, isliye same crowd-per-second har ek se squeeze hoti hai, aur difficulties pile up hoti jaati hain — hallway mushkil hoti jaati hai. Parallel darwaze hain side by side same wall mein: bheed wall tak pahunchti hai aur split hoti hai, zyada darwaze matlab bheed zyada tez guzarti hai overall, isliye total difficulty akele sabse aasaan darwaze se bhi niche gir jaati hai. Har side-by-side darwaze ke across push (voltage) same wall-pressure hai; har in-line darwaze se guzarne wali crowd-per-second (current) same hai. Ek line mein pushes add karo; fork par flows add karo. Ohm ( $V = I R$ ) pushes ko flows mein aur wapas badalta hai; Kirchhoff bas books rakhta hai — kuch create nahi, kuch lost nahi. Yahi poora chapter hai.

Recall Quick self-check

Series derivation mein $I$ factor out kyun kar sakte hain? ::: Kyunki single path same $I$ ko har resistor se force karta hai (Step 1). Parallel derivation mein $V$ factor out kyun kar sakte hain? ::: Kyunki har branch ke dono ends same do nodes ko touch karte hain, isliye har ek same $V$ feel karta hai (Step 4). Plain wire ke parallel mein ek resistor $R_{e q} =$ ? deta hai ::: $0$ (wire use short-circuit karta hai, Case B). Broken branch ke parallel mein ek resistor $R_{e q} =$ ? deta hai ::: $R_{1}$ (open branch koi current carry nahi karta, Case C). Series mein open circuit kya karta hai? ::: Saara current rok deta hai, $R_{e q} = \infty$ — ek hi path toot jaata hai.

Connections

Ohm's Law — $V = I R$ ne har step ki picture ko equation mein badla.
Kirchhoff's Laws — KCL ne "same current" (series) aur "currents add" (parallel) diya; KVL ne "voltages add" diya.
Resistivity and Resistance — jahan se ek single $R$ value aati hai.
EMF and Internal Resistance — internal resistance cell ke saath series mein hoti hai.
Wheatstone Bridge — ek network jo na pure series hai na pure parallel.
Power in Circuits — $P = I^{2} R = V^{2} / R$ combination mein share hoti hai.