Current Electricity: Class 12 Physics Chapter 3 Hinglish mein

Physics Part 1 · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

130 min
3 simulations
17 practice questions
43 MCQs

Is chapter mein aap kya seekhenge

Electric current aur current density ko define karo, aur current ko charge carriers ke drift se jodo
Ek simple collision model se metal ki drift velocity, Ohm's law aur conductivity derive karo
R = ρl/A aur resistivity ki temperature par dependence use karke resistances aur temperatures nikaalo
Current–voltage graphs se non-ohmic behaviour pehchaano
Electrical power aur energy calculate karo, aur samjhao ki power high voltage par kyun transmit ki jaati hai
Cells ke emf aur internal resistance ke saath kaam karo, akele cell ke liye bhi aur series ya parallel combination ke liye bhi
Kisi network mein currents nikaalne ke liye Kirchhoff's rules apply karo, Wheatstone bridge samet

NCERT Physics Class 12, Chapter 3, pp. 81–106 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

3.1 Introduction

Phone ko charger mein lagao, to ek ghante ya usse zyada der tak battery mein charge ki ek chhoti, steady dhaara behti hai. Lightning strike bhi charge ko move karti hai, lekin second ke ek chhote se hisse mein, ek zordaar jhatke mein. Pichhle do chapters mein har charge ruka hua tha. Ab hum charges ko chalaate hain aur poochhte hain ki unhe chalta kaun rakhta hai, unhe kya rokta hai aur unki energy kahan jaati hai.

Yeh chapter steady currents ke baare mein hai, jaise woh currents jo TV remote, pocket calculator ya solar lantern chalaate hain. Hum current ko theek-theek define karenge, dekhenge ki metals conduct kyun karte hain, electrons ki motion se Ohm's law derive karenge, aur seekhenge ki resistance material, shape aur temperature par kaise depend karta hai. Phir hum energy ka peechha karenge, cell ke andar jhaankenge, aur do aise rules ke saath khatam karenge jo koi bhi circuit solve kar sakte hain, saath mein resistance naapne ka ek smart circuit bhi.

3.2 Electric current

Current naapne ke liye hume charge ginne ki koi jagah chahiye. Ek aisi surface chuno jisse moving charges guzarte hain, jaise wire ka cross-section, aur tay karo ki uske across kaunsi direction forward ginni hai. Current le jaate salt solution mein positive ions aur negative ions ulti directions mein chalte hain, aur dono mein se kuch galat taraf bhi bhatak jaate hain. Forward le jaaya gaya positive charge aur backward le jaaya gaya negative charge, dono ka effect same hai: dono forward side ko zyada positive bana dete hain. Isliye time $t$ mein hum $q_{+}$ note karte hain, yaani forward gaya positive charge minus wapas aaya positive charge, aur $q_{-}$ , negative charge ke liye yahi count. Forward transfer hua net charge $q = q_{+} - q_{-}$ hai.

Steady flow mein time double karne par $q$ bhi double ho jaata hai, isliye ratio $q / t$ is baat par depend nahi karta ki hum kitni der ginte hain. Yahi ratio surface se guzarne wala electric current (विद्युत धारा) hai:

Steady current

I = \frac{q}{t}

(3.1)

jahan

$q$: forward cross karne wala net charge
$t$: laga hua time

3/3Current $q$ ko $t$ se divide karne par milta hai, aur uski direction wahi hai jidhar positive charge chalta hai.

Fig. 3.1Current kisi surface ko per unit time cross karne wala net charge ginta hai. Backward chalte negative charges bhi forward current mein jud jaate hain.

Ek horizontal band conducting region dikhata hai, uske beech mein ek dashed oval window bani hai aur ek arrow forward direction mark karta hai. Pehle step mein chaar positive aur chaar negative carriers random directions mein chalte hain, isliye har taraf barabar sankhya mein cross karte hain aur net charge zero hai. Doosre step mein saare positive carriers forward aur saare negative carriers backward chalte hain, aur ek label par likha hai q barabar q plus minus q minus, yaani time t mein net forward charge. Teesre step mein band ke neeche ek mota arrow current dikhata hai, positive carriers wali direction mein hi, jis par I barabar q by t likha hai.

Kai currents har pal badalte rehte hain; jaise mobile phone speaker ka current ek second mein hazaaron baar direction badalta hai. Aise current ko kisi ek instant par pakadne ke liye, time $t$ se shuru hone wale ek bahut chhote interval $Δ t$ mein cross karne wala net charge $Δ Q$ gino, aur interval ko chhota hone do:

Instantaneous current

I (t) = \lim_{Δ t \to 0} \frac{Δ Q}{Δ t} = \frac{d Q}{d t}

(3.2)

Agar $I$ negative aaye, to positive charge ka net transfer backward hai. Current ki SI unit ampere (A) hai, jo saat SI base units mein se ek hai. 2019 se ise elementary charge ko exact value $e = 1.602176634 \times 10^{- 19} C$ dekar fix kiya gaya hai, isliye $1 A$ ka current har second kisi point se lagbhag $6.24 \times 10^{18}$ elementary charges guzaarta hai. Chapter 4 us magnetic effect ko describe karta hai jisse pehle ampere define kiya jaata tha.

Table 3.1Roz ki zindagi ke currents das se zyada powers of ten tak phaile hain.

Situation	Typical current
Nerve fibre mein signal	kuch microamperes
Charger par LED indicator	lagbhag 10 mA
Phone ki charging cable	1 se 3 A
230 V supply par room heater	lagbhag 8 A
Car ka starter motor	100 se 300 A
Lightning stroke	microseconds ke liye 10 000 se 100 000 A

Quick check

3 mein se 1

Ek phone charger

30

minutes tak steady

2.0 A

deta hai. Cable se kitna charge behta hai?

Answer dekho

Answer 3600 C

q = I t = 2.0 A \times (30 \times 60) s = 3600 C

. Minutes ko seconds mein badalna yaad rakho.

3.3 Conductors mein electric currents

Field har charge par push lagata hai, lekin current sirf wahi charges le ja sakte hain jo chalne ke liye free hain. Free charges tumhari soch se zyada common hain. Salt water aur car battery ke acid mein mobile ions hote hain, hamare bahut upar ionosphere free electrons aur ions se bhara hai, aur har metal mein free electrons ka ek samundar hota hai.

Metals apni conductivity ek sharing arrangement ki wajah se paate hain. Metal ka har atom apne ek ya zyada outer electrons chhod deta hai, jo phir poore solid ke ho jaate hain aur usmein kahin bhi ghoom sakte hain. Yahi conduction electrons current le jaate hain, jabki peeche chhoote positive ions lattice mein fixed rehte hain. Glass aur rubber mein aise shared electrons nahi hote: har electron apne atom ya bond se bandha rehta hai, isiliye woh insulate karte hain. Is chapter ka baaki hissa metals par focus karta hai.

Saare fields band kar do aur table par padi copper wire ko dhyaan se dekho. Uske conduction electrons bilkul bhi ruke nahi hain: woh bahut tez speed se idhar-udhar bhaagte hain, vibrate karte ions se takraate hain aur har collision ke baad nayi direction mein ud jaate hain. Phir bhi wire mein koi current nahi hai. Har electron jo cross-section ko ek taraf cross karta hai, uske badle doosra electron doosri taraf cross karta hai, isliye net flow zero hai.

Ab sirf stored charge se current chalaane ki koshish karo. $+ Q$ aur $- Q$ charges wali do metal plates lo, aur unhe ek chhote bulb aur switch ke through wire se jodo (Fig. 3.2). Switch band karte hi wire ke along ek field aa jaata hai, jo positive plate se door point karta hai, aur electrons us plate ki taraf kheeche jaate hain. Bulb ek pal ke liye chamakta hai. Lekin har pahunchne wala electron plate ka thoda charge cancel kar deta hai, isliye field kamzor hota jaata hai, aur second ke ek hisse mein plates neutral ho jaati hain aur current ruk jaata hai.

4/4Cell charges ko baar-baar restore karta hai, isliye field aur current steady rehte hain.

Fig. 3.2Charged plates sirf thodi der ka current deti hain. Cell baar-baar charge supply karta rehta hai, isliye field aur current steady rehte hain.NCERT Fig. 3.1 · p. 83

Left mein plus Q wali ek plate aur minus Q wali ek plate upar ki taraf ek wire se judi hain, jis mein switch aur bulb lage hain. Right mein current vs time ka graph hai. Pehle step mein switch khula hai, bulb band hai aur graph khaali hai. Doosre step mein switch band hai, arrows dikhate hain ki current positive plate se bulb hokar negative plate tak beh raha hai, bulb jalta hai, aur graph shuru mein bada current dikhata hai. Teesre step mein plates ke charges fade ho gaye hain, bulb band hai aur graph ka curve zero tak gir gaya hai. Chauthe step mein neeche se ek cell plates ko jodta hai, bulb steady jalta hai aur graph steady current ki flat line dikhata hai.

Isliye lambe samay tak current ke liye ek aisa device chahiye jo ends par charges ko baar-baar restore karta rahe, taaki wire ke andar ka field kabhi khatam na ho. Cell yeh kaam chemistry se karta hai aur generator magnetism se; dono ghanton tak current chala sakte hain. Is chapter ke baaki saare currents isi steady type ke hain.

3.4 Ohm's law

1820s mein, electrons ke baare mein kisi ko pata chalne se bahut pehle, Georg Simon Ohm ne naapa ki metal wires mein current unke across voltage ke saath kaise badalta hai. Fixed temperature par rakhi ek wire ke liye potential difference double karne par current bhi double ho gaya.

LawOhm's law

Constant temperature par kisi metallic conductor se guzarne wala current $I$ uske across potential difference $V$ ke proportional hota hai: $V \propto I$ .

Ohm's law

V = R I

(3.3)

jahan

$V$: conductor ke across potential difference
$I$: usse guzarne wala current
$R$: uska resistance

Definition

Resistance

Conductor ke across potential difference aur usse guzarne wale current ka ratio $R = V / I$ . Iski SI unit ohm hai, $1 Ω = 1 V / A$ .

ExampleJo heater $230 V$ supply se $4.0 A$ leta hai, uska resistance $57.5 Ω$ hai.

Resistance (प्रतिरोध) sirf metal se tay nahi hota: ek lambi patli copper wire aur ek chhoti moti copper bar ke resistances bahut alag hote hain. Simulation mein khud try karo, usmein do wires hain. Har $I$ – $V$ line ka slope tumhe $1 / R$ batata hai.

SimulationOhm's law bench

Voltage slider wala ek cell do resistors mein se ek ke through current chalata hai, jinhe wire A aur wire B kaha gaya hai, saath mein ek ammeter aur ek voltmeter hai. Slider ki har setting circuit ke bagal wale current vs voltage graph mein ek point jodti hai, aur ek readout ratio V by I dikhata hai.

Fig. 3.3Cell voltage badlo aur har wire ke liye

I

–

V

graph par points record karo.

Try karo

Cell slider ko move karke wire A ke kam se kam paanch points record karo. Kya woh origin se guzarti seedhi line par hain?
Wire B par switch karo aur dohrao. Kaunsi line zyada steep hai, aur steepness resistance ke baare mein kya batati hai?
Har point par V ÷ I readout check karo. Kya voltage badalne par woh badalta hai?

$R$ size par kaise depend karta hai, yeh jaanne ke liye ek uniform wire se shuru karo jiska resistance $R$ , length $l$ aur cross-section ka area $A$ hai. Ab socho ki aise identical pieces ko do tarah se joda jaaye (Fig. 3.4).

4/4Isliye $R \propto l$ aur $R \propto 1 / A$ , jisse $R = ρ l / A$ milta hai.

Fig. 3.4Length double karne par resistance double ho jaata hai. Area double karne par woh aadha ho jaata hai. Dono milakar

R = ρ l / A

.NCERT Fig. 3.2 · p. 83

Rectangular conductor pieces ki rows; har piece mein left se ek current arrow aata hai aur neeche ek double arrow potential difference dikhata hai. Pehle step mein ek piece current I le jaata hai aur potential difference V hai, isliye R barabar V by I. Doosre step mein ek row aati hai jismein do pieces ek ke baad ek jude hain: same current I ke liye 2V chahiye, isliye resistance 2R hai, label par likha hai dugni length. Teesre step mein ek row aati hai jismein do pieces agal-bagal rakhe hain: same V par current 2I hai, isliye resistance R by 2 hai, label par likha hai dugna area. Chauthe step mein ek box combined result dikhata hai, R barabar rho l by A.

DerivationResistance length aur area par kaise depend karta hai

Do pieces ek ke baad ek. Same current $I$ har piece se guzarta hai, aur har piece ko potential difference $V = I R$ chahiye. Total potential difference $2 V$ hai, isliye combination ka resistance $2 V / I = 2 R$ hai. Dugni lambi wire ka resistance dugna hota hai, isliye $R \propto l$ .

Do pieces agal-bagal. Pieces ko dono ends par jodo taaki dono par same potential difference $V$ ho. Har ek current $I$ le jaata hai, isliye total $2 I$ hai aur combination ka resistance $V / (2 I) = R / 2$ hai. Yeh dugne cross-section wala ek hi conductor hai, isliye $R \propto 1 / A$ .

Dono results ko milakar $R \propto l / A$ . Constant of proportionality sirf material aur uske temperature par depend karta hai:

Uniform conductor ka resistance

R = ρ \frac{l}{A}

(3.9)

jahan

$ρ$: material ki resistivity
$l$: current ki direction mein length
$A$: cross-sectional area

Definition

Resistivity

Material ki property $ρ$ jo $R = ρ l / A$ se define hoti hai. Yeh unit length aur unit cross-section wale piece ka resistance hai, jo ohm metre ( $Ω m$ ) mein naapa jaata hai. Yeh material aur uske temperature par depend karti hai, shape par nahi.

ExampleRoom temperature par copper ka $ρ \approx 1.7 \times 10^{- 8} Ω m$ hota hai.

Resistivity (प्रतिरोधकता) se hum Ohm's law ko aise form mein likh sakte hain jo material ke andar har point par sahi ho. Current density $j = I / A$ define karo, yaani cross-section ke per unit area current (unit $A / m^{2}$ ). Uniform wire mein field $E = V / l$ hota hai. $V = I R$ mein $V = E l$ aur $R = ρ l / A$ rakhne par $E l = (j A) (ρ l / A)$ milta hai, yaani $E = ρ j$ .

Current density ek vector hai jo ordinary conductor mein $𝐄$ ke along point karta hai, isliye hum result ko vector form mein likh sakte hain. Resistivity ka reciprocal, $σ = 1 / ρ$ , conductivity (चालकता) hai, jo siemens per metre ( $1 S = 1 Ω^{- 1}$ ) mein naapi jaati hai.

Ek point par Ohm's law

𝐄 = ρ 𝐣

(3.12)

Conductivity ke terms mein Ohm's law

𝐣 = σ 𝐄

(3.13)

Example

Copper wire mein resistance, current density aur field

Ek copper wire $50 m$ lambi hai aur uska diameter $1.0 m m$ hai. Copper ki resistivity $1.7 \times 10^{- 8} Ω m$ hai. (a) Wire ka resistance nikaalo. (b) Agar ismein $5.0 A$ behta hai, to iske andar current density aur electric field nikaalo. (c) Wire ko kheench kar dugni length ka kar diya jaata hai, volume fixed rehta hai. Iska naya resistance kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Try karke dekho

Ek resistor jo tum draw kar sakte ho

Tumhe chahiye

ek soft pencil (6B ya 8B)
moti paper ki ek sheet
ek ruler
resistance naapne par set ek digital multimeter
do metal paper clips

Pehle guess karo

Tum

2 c m

lambi pencil strip ka resistance naapte ho, phir usi strip ke

8 c m

ka. Tum kya expect karte ho?

Lagbhag chaar guna resistanceLagbhag utna hi resistanceLagbhag ek chauthai resistanceLagbhag solah guna resistance

Lagbhag $1 c m$ chaudi aur $10 c m$ lambi ek strip shade karo, us par kai baar pencil chalao taaki graphite layer moti aur ek jaisi ho.
Ek probe ko strip ke shuru mein clip karo. Doosre probe ko strip par $2$ , $4$ , $6$ aur $8 c m$ par dabao aur har baar resistance note karo.
Dugni chaudi ek doosri strip shade karo, utne hi pressure aur utni hi baar pencil chalakar, aur uska $8 c m$ naapo.

Tumhari readings

Length (cm)	Resistance (kΩ)
2
4
6
8
8 (dugni width)

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 3.3

Quick check

4 mein se 1

Ek metal wire ko kheench kar uski original length ka teen guna kar diya jaata hai. Uska volume same rehta hai. Uska resistance kitne guna badalta hai?

Teen guna ho jaata haiSame rehta haiEk tihaai ho jaata haiNau guna ho jaata hai

Answer dekho

Answer (D) Nau guna ho jaata hai

Length

3 l

ho jaati hai aur constant volume par area

A / 3

ho jaata hai. Isliye

R = ρ l / A

3 \times 3 = 9

guna badhta hai. Sirf teen guna kehna patle cross-section ko ignore karta hai.

3.5 Electrons ka drift aur resistivity ka origin

Current voltage ke proportional hona hi kyun chahiye? Ions ki lattice mein takraate electrons ka ek simple model iska jawab deta hai aur batata hai ki resistivity kis par depend karti hai.

Bina field ke shuru karo. Har electron seedhi line mein udta hai jab tak kisi ion se nahi takraata, phir lagbhag usi speed se ek random nayi direction mein nikal jaata hai. Isliye kisi bhi instant par electrons ki velocities har taraf point karti hain. $N$ conduction electrons ko $i = 1, 2, \dots, N$ number do aur us instant par unki velocities ko $𝐯_{i}$ kaho; random directions ki wajah se average zero ho jaata hai:

Field nahi: average velocity zero

\frac{1}{N} \sum_{i = 1}^{N} 𝐯_{i} = 0

(3.14)

Ab ek field $𝐄$ on karo. Charge $- e$ aur mass $m$ wale har electron ko collisions ke beech same acceleration milta hai:

Electron ka acceleration

𝐚 = - \frac{e 𝐄}{m}

(3.15)

DerivationCollisions se drift velocity

Har electron ko ek imaginary stopwatch do jo har collision par zero par reset ho jaati hai. Kisi instant par electron $i$ ki stopwatch $t_{i}$ dikhati hai, aur woh apne pichhle collision se velocity $𝐯_{i}$ ke saath nikla tha. Is poore time mein usne constant acceleration (3.15) mehsoos kiya, isliye ab uski velocity hai

𝐕_{i} = 𝐯_{i} - \frac{e 𝐄}{m} t_{i}

(3.16)

Drift nikaalne ke liye ise same instant par saare $N$ electrons par average karo. Pehla term average mein zero ho jaata hai: electron har collision se random direction mein nikalta hai, isliye (3.14) in velocities par bhi laagu hota hai. Stopwatches alag-alag readings dikhati hain, kyunki collisions random moments par hote hain, lekin unka average ek fixed time $τ$ hai, relaxation time, chahe hum koi bhi instant chunein. Isliye average velocity hai

Drift velocity

𝐯_{d} = ⟨ 𝐕_{i} ⟩ = - \frac{e 𝐄}{m} τ

(3.17)

Dhyaan do yeh kya kehta hai. Har electron accelerate ho raha hai, phir bhi average velocity time ke saath nahi badalti. Har collision woh mita deta hai jo electron ne abhi field se kamaya tha, isliye poori bheed $𝐄$ ke opposite ek steady, dheemi drift (अपवाह वेग) mein aa jaati hai.

3/3Kai collisions ke baad electron field ke against $v_{d} t$ distance drift kar chuka hai.

Fig. 3.5Collisions ke beech field har flight ko ek hi taraf mod deta hai. Random hissa average out ho jaata hai aur

𝐄

ke opposite ek dheemi drift bachti hai.NCERT Fig. 3.3 · p. 85

Positive ions ka ek regular grid poori picture mein bhara hai. Pehle step mein, bina field ke, ek electron collisions ke beech seedhi flights ka zigzag path leta hai, ek open circle se shuru hokar ek filled dot par khatam hota hai jo shuruaat ke paas hi hai. Doosre step mein E label wala field arrow right ki taraf point karta hai; purana path dashed dikhaya gaya hai aur ek naya blue path har flight mein thoda left ki taraf mudta hai, isliye woh start se kaafi left mein khatam hota hai. Teesre step mein start se naye end point tak ek mota arrow, field ke opposite point karta hua, v d t label ke saath net drift dikhata hai.

Drift se current tak. Wire ke across, $𝐄$ ke right angle par, area $A$ ki ek flat surface socho (Fig. 3.6). Electrons speed $∣ 𝐯_{d} ∣$ se ek dheemi line mein uski taraf aate hain. Agle chhote time $Δ t$ mein jo usse pahunchenge, woh abhi uske bagal mein, jidhar se woh aa rahe hain us taraf, $∣ 𝐯_{d} ∣ Δ t$ length ki ek patli slice mein hain. Slice ka volume $A ∣ 𝐯_{d} ∣ Δ t$ hai, isliye per unit volume $n$ conduction electrons ke saath usmein $n A ∣ 𝐯_{d} ∣ Δ t$ electrons hain, har ek ka charge size $e$ ka. $Δ t$ mein cross karne wala charge $I Δ t$ hai, isliye

3/3 $Δ t$ mein $n A v_{d} Δ t$ electrons cross karte hain, har ek charge $e$ le jaata hai.

Fig. 3.6Time

Δ t

mein shaded slice ka har electron surface cross karta hai, jisse

I = n e A v_{d}

milta hai.NCERT Fig. 3.4 · p. 86

Ek horizontal cylindrical wire electrons ke liye blue dots se bhari hai, beech mein area A ka ek dashed oval cross-section hai. Wire ke upar field arrow E right point karta hai aur neeche drift arrow v d left point karta hai. Doosre step mein cross-section ke theek right mein v d delta t length ki wire ki slice shaded hai aur uske andar ke electrons par gola bana hai. Teesre step mein saare electrons left khisak gaye hain, gole wale electrons ab cross-section ke theek paar hain, ek current arrow I right point karta hai, aur result I barabar n e A v d dikhaya gaya hai.

Drift speed se current

I = n e A ∣ 𝐯_{d} ∣

(3.18)

DerivationDrift model se Ohm's law

Current density paane ke liye (3.18) ko $A$ se divide karo, $j = n e ∣ 𝐯_{d} ∣$ , aur (3.17) se drift speed $∣ 𝐯_{d} ∣ = e τ E / m$ rakho. Current density $𝐄$ ke along point karti hai (electrons ulti taraf drift karte hain, lekin woh negative hain), isliye

𝐣 = \frac{n e^{2} τ}{m} 𝐄

(3.22)

Iska form bilkul equation (3.13) ke $𝐣 = σ 𝐄$ jaisa hai. Isliye collision model Ohm's law ko predict karta hai aur conductivity ko microscopic quantities ke terms mein deta hai:

Metal ki conductivity

σ = \frac{n e^{2} τ}{m}

(3.23)

Prediction tab tak sahi hai jab tak $n$ aur $τ$ field par depend nahi karte. Metals ke liye yeh fields ki bahut badi range par sahi hai, isiliye metal wires Ohm's law ko itni achhi tarah follow karti hain.

Chalo copper ke numbers daalte hain: $n \approx 8.5 \times 10^{28} m^{- 3}$ aur $ρ = 1 / σ \approx 1.7 \times 10^{- 8} Ω m$ . (3.23) ko $τ$ ke liye solve karne par $τ = m / (n e^{2} ρ) \approx 2.5 \times 10^{- 14} s$ milta hai. Copper mein ek electron lattice se ek second mein lagbhag chaalis million million baar takraata hai.

SimulationCopper wire mein drift karte electrons

Stage ke upar ek circuit hai jismein cell, switch aur copper wire hai. Uske neeche wire ka ek magnified hissa electrons ko idhar-udhar bhaagte aur takraate dikhata hai, aur do tracer electrons trails chhodte hain. Sliders cell voltage, load resistance aur wire ka cross-sectional area set karte hain, aur readouts current, millimetres per second mein drift speed aur ek metre drift karne ka time dikhate hain.

Fig. 3.7Electron cloud tezi se hilta hai lekin dheere drift karta hai. Readout copper ke liye

v_{d} = I / (n e A)

use karta hai.

Try karo

Current badhao aur tracer trails dekho. Drift speed readout kaise badalta hai?
Current fixed rakho aur wire ka cross-sectional area chaar guna kar do. Drift speed ka kya hota hai, aur kyun?
Switch kholo. Kya electrons chalna band kar dete hain, ya sirf drift karna band karte hain?

Example 3.1NCERT Example 3.1 · p. 87

Drift kitni dheemi hai?

Geyser ke liye ghar ka wiring circuit $4.0 m m^{2}$ cross-section wali aluminium wire use karta hai aur $16 A$ le jaata hai. Aluminium ki density $2.70 \times 10^{3} k g / m^{3}$ aur molar mass $27.0 g / m o l$ hai, aur har atom teen conduction electrons deta hai. (a) Electrons ki drift speed nikaalo. (b) Ise compare karo (i) $300 K$ par electron ki typical random thermal speed se, jo classically $\sqrt{3 k_{B} T / m}$ se estimate hoti hai, aur (ii) us speed se jisse electric signal wire ke along travel karta hai, jo lagbhag $3 \times 10^{8} m / s$ hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Random motion mein electron ki speed bahut zyada hai; quantum theory ise aur bhi bada bana deti hai, metals mein lagbhag $10^{6} m / s$ . Uske upar judi drift bahut chhoti hai. Phir bhi current bada ho sakta hai kyunki $n$ bahut bada hai: har cubic metre mein lagbhag $10^{29}$ carriers.

Example 3.2NCERT Example 3.2 · p. 88

Drift karte electrons ke baare mein paanch claims

Ek classmate paanch claims karta hai. Har ek ke liye tay karo ki woh sahi hai ya nahi, aur samjhao. (a) “Kyunki electrons ek second mein millimetre ka sirf ek hissa drift karte hain, switch se 5 m door bulb ghanton baad jalna chahiye.” (b) “Electron par steady force use lagaataar badhti speed dega, isliye drift speed badhti rehni chahiye.” (c) “Itni kam drift speed aur har electron par itne chhote charge ke saath bada current possible hi nahi hai.” (d) “Jab electrons higher potential ki taraf drift karte hain, to wire ka har free electron usi taraf chalta hai.” (e) “Do collisions ke beech electron ka path hamesha seedhi line hota hai.”

Pehle khud try karo. Phir solution ek-ek step karke kholo.

(a) Galat. Wire hamesha free electrons se bhari hoti hai, jaise paani se pehle se bhara pipe. Tap kholo aur door wale end se turant paani girne lagta hai, tap par ghusa paani pahunchne se bahut pehle. Switch band karne se field ka change circuit ke along lagbhag light ki speed se jaata hai, isliye $5 m$ door filament ke electrons lagbhag $2 \times 10^{- 8} s$ mein push ho jaate hain. Phir current ko apni steady value tak aane mein bahut thoda time lagta hai.
(b) Galat, kyunki yeh collisions ko ignore karta hai. Electron $- 𝐄$ ke along velocity sirf apni chhoti free flight ke dauraan kamaata hai; agla collision uski direction bigaad deta hai aur kamaayi hui velocity mita deta hai. Kai flights par average karne se extra velocity ek fixed amount hai, $e E τ / m$ . Pegs lage board par neeche ludhakti marble bhi aise hi behave karti hai: woh baar-baar tez hoti hai aur takraati hai, aur aakhir mein neeche ki taraf ek steady average speed paati hai.

Socho zara

Drift kaunsi motion se hoti hai?

Har electron ki velocity ko do hisson mein baanto. Jo hissa uske pichhle collision ke theek baad tha, woh kahin bhi point kar sakta hai, isliye poori bheed par woh jud kar zero ho jaata hai. Jo hissa usne tab se $𝐄$ se kamaya hai, woh har electron ke liye ek hi taraf point karta hai. Averaging ke baad sirf yeh doosra hissa bachta hai, isliye drift poori tarah field ki wajah se hai.

Ek puzzle: $5 A$ le jaati copper wire ki net charge density zero hai, kyunki ions ka positive charge electrons ke negative charge ko bilkul cancel kar deta hai. To kya $𝐣 = ρ_{q} 𝐯$ (charge density times velocity) koi current nahi predict karta? Nahi. Yeh relation ek baar mein ek hi type ke carrier par lagana chahiye. Ions apni jagah rehte hain aur kuch contribute nahi karte, jabki electrons, charge density $ρ_{-}$ ke saath, $𝐣 = ρ_{-} 𝐯_{d}$ dete hain.

Quick check

2 mein se 1

Metal wire mein electric field double kar diya jaata hai, temperature same rehta hai. Electrons ki drift speed ka kya hota hai?

Same rehti haiDouble ho jaati haiChaar guna ho jaati haiTime ke saath badhti rehti hai

Answer dekho

Answer (B) Double ho jaati hai

v_{d} = e τ E / m

aur

τ

nahi badla, isliye

v_{d} \propto E

. Yeh time ke saath nahi badhti, kyunki collisions kamaayi hui velocity ko baar-baar reset karte rehte hain.

3.5.1 Mobility

Har conductor ko mobile charge carriers chahiye, lekin carriers alag-alag hote hain. Metal mein electrons hote hain; gas discharge, jaise neon sign, mein electrons aur positive ions hote hain; car battery ke acid mein positive aur negative ions hote hain. Alag carriers field par kitni aasaani se respond karte hain, yeh compare karne ke liye hum unki mobility (गतिशीलता) use karte hain, yaani unit field se banne wali drift speed:

Mobility

μ = \frac{∣ 𝐯_{d} ∣}{E}

(3.24)

Mobility $m^{2} V^{- 1} s^{- 1}$ mein naapi jaati hai aur hamesha positive number ke roop mein di jaati hai, carrier ka sign kuch bhi ho. Semiconductor data tables aksar ise $c m^{2} V^{- 1} s^{- 1}$ mein dete hain, jo SI unit ka $10^{- 4}$ hai. Electrons ke liye (3.17) se milta hai

Mobility aur relaxation time

μ = \frac{e τ}{m}

(3.25)

(3.23) ke saath milane par ek compact result milta hai, $σ = n e μ$ : conductivity carrier density times charge times mobility hai. Copper ke liye $μ = e τ / m \approx 4.3 \times 10^{- 3} m^{2} V^{- 1} s^{- 1}$ . Pure silicon mein electrons lagbhag tees guna zyada mobile hain, phir bhi silicon bahut kharab conduct karta hai kyunki usmein carriers bahut hi kam hain.

Example

Same drift speed tak do raaste

Pichhle example ki copper wire par wapas aao, jo $5.0 A$ le jaati hai aur jiske andar $0.108 V / m$ ka field hai. Copper ke liye $n = 8.5 \times 10^{28} m^{- 3}$ aur $ρ = 1.7 \times 10^{- 8} Ω m$ lo. Relaxation time aur mobility nikaalo, aur mobility se drift speed nikaalo. Answer ko $v_{d} = I / (n e A)$ se check karo.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

1.5 m m^{2}

cross-section wali copper wire

3.0 A

le jaati hai.

n = 8.5 \times 10^{28} m^{- 3}

ke saath drift speed millimetres per second mein kitni hai?

mm/s

Answer dekho

Answer 0.147 mm/s

v_{d} = I / (n e A) = 3.0 / [(8.5 \times 10^{28}) (1.6 \times 10^{- 19}) (1.5 \times 10^{- 6})] = 1.47 \times 10^{- 4} m / s = 0.147 m m / s

3.6 Ohm's law ki limitations

Ohm's law metals aur kai doosre materials ko achhi tarah describe karta hai, lekin bahut se useful devices ise todte hain. Yeh departures teen tarah ke hote hain, aur har ek current vs voltage graph par aasaani se dikh jaata hai (Fig. 3.8).

$I$ , $V$ ke proportional nahi hai. Graph mud jaata hai. Filament lamp roz ka example hai: current badhne par filament garam hota hai, uska resistance badhta hai, aur graph voltage axis ki taraf mud jaata hai.
Response $V$ ke sign par depend karta hai. Voltage ulatne se current bas ulat nahi jaata. Diode ek taraf bada current jaane deta hai aur doosri taraf lagbhag kuch nahi; yeh one-way behaviour rectifiers mein use hota hai, jinse tum Chapter 14 mein miloge.
Ek current, kai voltages. Kuch materials, jaise gallium arsenide (GaAs), ka graph badhta hai, girta hai aur phir badhta hai, isliye same current ek se zyada voltage par aata hai.

4/4GaAs-type curve: same current teen alag voltages par aata hai.

Fig. 3.8Ohmic conductor origin se guzarti seedhi line deta hai. Baaki curves woh teen tareeke dikhate hain jinse koi device Ohm's law tod sakta hai.NCERT Figs. 3.5–3.7 · p. 89

Axes par voltage V horizontal aur current I vertical hai, origin beech-left ke paas hai taaki negative values dikh sakein. Pehle step mein ek green seedhi line origin se guzarti hai, jis par ohmic likha hai. Doosre step mein seedhi line dashed reference ban jaati hai aur ek orange curve zyada voltages par voltage axis ki taraf mudta hai, jis par filament lamp likha hai. Teesre step mein ek vermillion diode curve negative voltages par zero ke paas rehta hai aur positive voltages par tezi se upar jaata hai. Chauthe step mein ek purple curve badhta hai, girta hai aur phir badhta hai; ek dashed horizontal line use teen points par milti hai, jis par likha hai ek current, teen voltages.

Key idea

R = V/I ek definition hai, Ohm's law ek claim hai

Tum kisi bhi device ke liye kisi bhi operating point par $V$ ko $I$ se divide karke result ko wahan ka resistance keh sakte ho. Yeh bas ek definition hai. Ohm's law aage ka claim hai ki yeh ratio $V$ ke size ya sign badalne par bhi same rehta hai, taaki $I$ – $V$ graph origin se guzarti seedhi line ho.

Local form $𝐄 = ρ 𝐣$ ek saaf test deta hai: material ohmic hai agar uska $ρ$ field kitna bhi strong ho aur kisi bhi taraf point kare, same rahe. Copper wire, ya circuit board par lage resistor ka doped silicon, roz ke use mein yeh test pass karte hain. Lekin koi bhi material har field par yeh test pass nahi karta. Bahut strong field mein carriers collisions ke beech itni energy kamaa lete hain ki $τ$ (aur kuch materials mein $n$ ) badal jaata hai, aur seedhi $I$ – $V$ line mud jaati hai.

Ohm's law todne wale devices koi kharabi nahi hain jinse bachna ho. Diodes, transistors, LEDs aur solar cells sab non-ohmic behaviour par hi depend karte hain. Lekin is chapter ke baaki hisse mein, aur iske baad ke kai chapters mein, har resistor ko ohmic maana gaya hai.

3.7 Alag-alag materials ki resistivity

Resistivity matter ki lagbhag kisi bhi doosri property se zyada powers of ten tak phaili hai: silver aur best insulators ke beech lagbhag 24. Logarithmic scale par materials teen bade groups mein aate hain jinke beech chaude gaps hain (Fig. 3.9).

Conductors. Metals aur unke alloys ki resistivities lagbhag $10^{- 8}$ aur $10^{- 6} Ω m$ ke beech hoti hain.
Semiconductors. Germanium, silicon aur aise materials beech mein aate hain. Temperature badhne par inki resistivity girti hai, aur kisi suitable impurity ki bahut thodi si matra milaane (doping) se yeh kai powers of ten kam ho sakti hai. Tumhare phone ki har chip isi par tiki hai: uske transistors silicon crystal ke chune hue regions ko dope karke banaye jaate hain.
Insulators. Glass, rubber, ceramics aur plastics ki resistivities lagbhag $10^{10}$ se $10^{16} Ω m$ se bhi aage tak hoti hain: metal ki lagbhag $10^{18}$ se $10^{24}$ guna.

3/3Glass aur rubber: $10^{10} Ω m$ aur usse bahut aage.

Fig. 3.9Logarithmic scale par teeno groups ek doosre se bahut door hain: rubber band same shape ke copper se lagbhag

10^{23}

guna zyada current ko rokta hai.

Ohm metre mein resistivity ka ek horizontal logarithmic axis ten to the minus eight se ten to the sixteen tak jaata hai, har chaar powers of ten par tick labels hain. Pehle step mein bilkul left mein ek patla orange band metals label ke saath hai, copper aur nichrome ke markers ke saath. Doosre step mein beech mein ek chauda green band semiconductors label ke saath hai, germanium ka marker lagbhag ek ohm metre ke paas aur silicon ka lagbhag do hazaar ke paas. Teesre step mein right mein ek blue band insulators label ke saath hai, glass ka marker lagbhag ten to the twelve ke paas aur rubber ka lagbhag ten to the fifteen ke paas.

Table 3.2Room temperature ke aas-paas approximate values. Alloys kaafi high resistivity ke saath bahut chhota

α

rakhte hain.

Material	20 °C par resistivity ρ (Ω m)	Temperature coefficient α (har °C par)	Typical use
Silver	$1.6 \times 10^{- 8}$	$3.8 \times 10^{- 3}$	contacts aur high-end connectors
Copper	$1.7 \times 10^{- 8}$	$3.9 \times 10^{- 3}$	ghar ki wiring, motor windings
Aluminium	$2.7 \times 10^{- 8}$	$3.9 \times 10^{- 3}$	overhead power lines
Tungsten	$5.6 \times 10^{- 8}$	$4.5 \times 10^{- 3}$	lamp filaments
Manganin (alloy)	$4.8 \times 10^{- 7}$	lagbhag $2 \times 10^{- 6}$	standard resistors
Nichrome (alloy)	$1.1 \times 10^{- 6}$	lagbhag $1.7 \times 10^{- 4}$	heating elements
Germanium (pure)	$0.46$	negative, bada	shuruaati transistors, detectors
Silicon (pure)	lagbhag $2 \times 10^{3}$	negative, bada	chips aur solar cells
Glass	$10^{10}$ se $10^{14}$	—	insulating supports
Hard rubber	$10^{13}$ se $10^{16}$	—	cable insulation aur gloves

Quick check

3 mein se 1

In materials ko badhti resistivity ke order mein lagao.

nichrome
germanium
glass
copper

Answer dekho

Answer

copper
nichrome
germanium
glass

Copper (

\sim 10^{- 8} Ω m

) < nichrome (

\sim 10^{- 6}

) < germanium (

\sim 0.5

) < glass (

10^{10}

ya zyada). Alloy phir bhi conductor hai, kisi bhi semiconductor se bahut neeche.

3.8 Resistivity ki temperature par dependence

Torch bulb ke filament ka resistance multimeter se naapo to chhoti value milti hai. Bulb ki rating se jalte samay ka resistance calculate karo to kahin badi value milti hai. Iski wajah temperature hai, aur materials mein bahut fark hota hai ki temperature unki resistivity ko kitna, aur kis direction mein, badalta hai.

Metals sabse simple case hain. Metal wire ko ek moderate range mein garam karo, maan lo kuch sau degrees, to uski resistivity temperature rise ke lagbhag proportional badhti hai:

Resistivity aur temperature

ρ_{T} = ρ_{0} [1 + α (T - T_{0})]

(3.26)

jahan

$ρ_{0}$: reference temperature $T_{0}$ par resistivity
$ρ_{T}$: temperature $T$ par resistivity
$α$: temperature coefficient of resistivity

Definition

Temperature coefficient of resistivity

Temperature mein har degree rise par resistivity ka fractional change, $α = \frac{ρ_{T} - ρ_{0}}{ρ_{0} (T - T_{0})}$ . Iski unit $K^{- 1}$ hai (ya $^{\circ} C^{- 1}$ , same size). Metals ke liye yeh positive hota hai.

ExampleRoom temperature ke paas copper ke liye $α \approx 3.9 \times 10^{- 3}^{\circ} C^{- 1}$ : uski resistivity har degree par lagbhag $0.4 %$ badhti hai.

Fig. 3.10 mein copper curve dekho. Lagbhag $100 K$ ke upar yeh lagbhag seedha hai, jo (3.26) maan kar chalta hai. Kam temperatures par yeh mud jaata hai aur ek chhoti residual resistivity par flat ho jaata hai, jo vibrate karte ions se nahi balki crystal ki impurities aur defects se tay hoti hai. Isliye (3.26) ko ek local fit samjho: $T_{0}$ ko un temperatures ke paas chuno jo tumhare kaam ke hain, aur seedhi line achha kaam karegi.

4/4Temperature badhne par semiconductor ki resistivity girti hai.

Fig. 3.10Metals:

T

ke saath

ρ

badhta hai. Nichrome jaise alloys: lagbhag koi change nahi. Semiconductors:

T

badhne par

ρ

tezi se girta hai.NCERT Figs. 3.8–3.10 · p. 90

Vertical axis resistivity ko uski 300 kelvin wali value se divide karke dikhata hai, 0 se lagbhag 2.5 tak, aur horizontal axis temperature 0 se 600 kelvin tak dikhata hai. Pehle step mein ek orange copper curve zero ke paas lagbhag flat shuru hota hai, phir lagbhag linearly badhta hai aur 600 kelvin par lagbhag 2 tak pahunchta hai. Doosre step mein ek dashed seedhi line copper curve ko 300 kelvin par chhooti hai, jahan rho zero point mark hai, aur formula rho T barabar rho zero times one plus alpha times T minus T zero dikhaya gaya hai; lagbhag 100 kelvin ke neeche line curve se alag ho jaati hai. Teesre step mein nichrome ke liye ek blue lagbhag flat line judti hai, jo 1 ke paas rehti hai. Chauthe step mein ek purple semiconductor curve kam temperature par high values se tezi se girkar 600 kelvin par zero ke paas aa jaata hai.

Figure ke baaki do curves bilkul alag behave karte hain. Nichrome, manganin ya constantan jaisa alloy mushkil se badalta hai: uska $α$ copper se das se hazaaron guna chhota hai, isiliye precision resistors in alloys se lapete jaate hain. Semiconductor ulta chalta hai, garam hone par uski resistivity tezi se girti hai. Digital fever thermometer ka thermistor temperature sense karne ke liye bilkul yahi effect use karta hai.

Itna alag behaviour kyun? (3.23) ko ulta karo:

Drift model se resistivity

ρ = \frac{1}{σ} = \frac{m}{n e^{2} τ}

(3.27)

Denominator mein do quantities hain: carrier density $n$ aur relaxation time $τ$ . Garam karne se har solid mein $τ$ chhota ho jaata hai, kyunki ions jitna zyada vibrate karte hain, electron utni jaldi raaste se bhatak jaata hai. Result $n$ tay karta hai. Metal mein conduction electrons ka poora set pehle se hai, isliye $n$ mushkil se badalta hai aur chhota $τ$ , $ρ$ ko badha deta hai. Semiconductor ya insulator mein garmi extra carriers ko azaad kar deti hai, aur $n$ itni tezi se badhta hai ki $τ$ ki kami ko dabaa deta hai: $ρ$ girta hai.

Jis wire ke dimensions mushkil se badalte hain, uska resistance bhi isi rule ko follow karta hai, $R_{T} = R_{0} [1 + α (T - T_{0})]$ . Isliye resistance naapna temperature naapna hai. Platinum resistance thermometer ko ice point aur steam point par calibrate kiya jaata hai, aur beech ki reading $R_{t}$ temperature $t = \frac{R_{t} - R_{0}}{R_{100} - R_{0}} \times 100^{\circ} C$ deti hai.

Example 3.3NCERT Example 3.3 · p. 91

Heater coil kitni garam hai?

Room heater ki nichrome coil $25.0^{\circ} C$ ke room mein multimeter par $40.0 Ω$ padhti hai; meter ka chhota current use garam nahi karta. $230 V$ par on karne par heater pehle bada current leta hai, jo kuch seconds mein girta hai aur coil chamakne par $5.10 A$ par tik jaata hai. Nichrome ke liye $α = 1.70 \times 10^{- 4}^{\circ} C^{- 1}$ lo. Chamakti coil kitni garam hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Example 3.4NCERT Example 3.4 · p. 92

Garam khaane ke liye ek probe

Ek school canteen platinum probe (Pt100 sensor) se check karti hai ki dal garam serve ho rahi hai. Pighalti barf mein iska resistance $100.00 Ω$ aur standard pressure par ubalte paani mein $138.50 Ω$ hai. Dal ke bartan mein dubaane par yeh $127.08 Ω$ padhta hai. Dal ka temperature kya hai? Platinum ka temperature coefficient kya hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Try karke dekho

Thanda filament, garam filament

Tumhe chahiye

ek 12 V, 21 W bike ya car bulb (ya ek 6 V torch bulb)
ek digital multimeter
ek 12 V battery ya adapter (ya torch bulb ke liye 6 V battery)
connecting wires

Pehle guess karo

Bulb ki rating tumhe uska garam resistance batati hai,

R = V^{2} / P

. Multimeter se naapa gaya thanda resistance isse kaisa hoga?

Lagbhag utna hiKahin zyadaKahin kamBilkul zero

Rating se garam resistance nikaalo: $12 V$ , $21 W$ bulb ke liye $R = 12^{2} / 21 \approx 6.9 Ω$ .
Bulb ko disconnect aur thanda rakhkar multimeter se uska resistance naapo.
Agar current range wala meter ho, to bulb ko low-voltage source se jodo aur jalne do. Current (meter series mein) aur potential difference (meter bulb ke across) record karo, phir $V / I$ nikaalo.

Filament resistance

Condition	Resistance (Ω)
Thanda (multimeter)
Garam (rating se)
Garam (V ÷ I se)

Tumhari entries apne aap save hoti hain.

Quick check

3 mein se 1

Ek copper coil ka resistance

20^{\circ} C

par

10.0 Ω

hai.

α = 3.9 \times 10^{- 3}^{\circ} C^{- 1}

ke saath

70^{\circ} C

par uska resistance kya hai?

Answer dekho

Answer 11.95 Ω

R = 10.0 [1 + (3.9 \times 10^{- 3}) (50)] = 10.0 \times 1.195 = 11.95 Ω

3.9 Electrical energy, power

Current jis bhi wire se guzarta hai use garam karta hai, kabhi kaam ka (kettle), kabhi bekaar (garam hota charger). Yeh heat aati kahan se hai?

DerivationEnergy kahan jaati hai

Ek resistor se steady current $I$ guzarne do. Jis end se woh ghusta hai woh, jis end se nikalta hai usse $V$ jitne higher potential par hai. Ek chhote time $Δ t$ mein charge $Δ Q = I Δ t$ high end par ghusta hai aur utna hi charge low end se nikalta hai; andar kuch jama nahi hota. Net effect wahi hai jo $Δ Q$ ko seedha potential drop $V$ ke neeche le jaane ka, isliye uski electric potential energy badalti hai

Δ U_{pot} = - Δ Q V = - I V Δ t < 0

(3.28)

Agar carriers ko kuch na rokta, to yeh khoyi potential energy kinetic energy $Δ K = I V Δ t$ ban kar wapas aati, aur carriers wire ke along speed pakad lete. Aisa nahi hota: woh steady $v_{d}$ se drift karte hain. Har collision par electron abhi kamaayi energy us ion ko de deta hai jisse woh takraata hai, isliye lattice zyada vibrate karti hai aur conductor garam hota hai. Ise Joule heating kehte hain, aur $Δ t$ mein paida heat $Δ W = I V Δ t$ hai.

$Δ t$ se divide karne par heat mein badalne wali power milti hai. $V = I R$ use karne se do aur useful forms milte hain:

Electrical power

P = I V

(3.32)

Resistor mein power

P = I^{2} R = \frac{V^{2}}{R}

(3.33)

Yahi power toaster coil ko garam karti hai ya filament ko safed-garam chamkaati hai. Resistor sirf energy convert karta hai, isliye koi aur use supply karta hai: cell ke chemicals, ya power station ka generator. (3.33) ka kaunsa form zyada kaam ka hai, yeh is par depend karta hai ki kya fixed hai. Series mein jude resistors (same $I$ ) ke liye $P = I^{2} R$ dikhata hai ki sabse bada resistance sabse garam hota hai. Same mains supply par appliances (same $V$ ) ke liye $P = V^{2} / R$ dikhata hai ki sabse chhota resistance sabse zyada power leta hai.

Power ki SI unit watt hai ( $1 W = 1 J / s$ ). Electricity boards energy ka bill kilowatt hours mein banate hain, jo tumhare bill ke “units” hain: $1 k W h = 1000 W \times 3600 s = 3.6 \times 10^{6} J$ .

Example

Geyser aur bijli ka bill

Ek $1.5 k W$ , $230 V$ geyser roz $40$ minutes chalta hai. (a) Woh kitna current leta hai, aur garam hone par uske element ka resistance kya hai? (b) $30$ din ke mahine mein woh kitni energy use karta hai, aur ₹7 per unit par iska kharcha kitna hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Lambi doori tak power bhejna. Kisi shehar ki bijli saikdon kilometre door ke coal plant ya hydroelectric dam se aa sakti hai, aur beech ki cables mein resistance hota hai. Maan lo door wale end par load ko voltage $V$ par power $P$ milni chahiye, aur cables ka total resistance $R_{c}$ hai. Cables mein load jitna hi current behta hai, $I = P / V$ , isliye jo power woh bekaar heat mein badalti hain woh hai

Transmission cables mein loss

P_{c} = I^{2} R_{c} = \frac{P^{2} R_{c}}{V^{2}}

(3.35)

Same power deliver karne ke liye loss $1 / V^{2}$ ki tarah badalta hai: voltage double karo to loss ek chauthai ho jaata hai; das guna karo to loss sau guna kam ho jaata hai. Isliye India ka long-distance grid $220$ , $400$ aur $765 k V$ par chalta hai. Itne voltages ghar ke andar jaanleva honge, isliye transformers power station se nikalte waqt voltage ko step up karte hain aur tumhari wiring tak pahunchne se pehle wapas $230 V$ tak step down karte hain (Fig. 3.11). Chapter 7 batata hai ki transformers kaise kaam karte hain.

3/3Town ke paas gharon ke liye voltage ko 230 V tak step down kiya jaata hai.

Fig. 3.11Higher voltage par same power ke liye chhota current chahiye, isliye lines mein heat ke roop mein bahut kam power khoti hai.

Left mein ek power plant hai aur right mein do ghar, jo teen lattice towers par tangi line se jude hain; line resistance par R c likha hai aur upar formula P loss barabar I squared R c dikhaya gaya hai. Pehle step mein koi transformer nahi hai, line moti aur garam dikhayi gayi hai, uske upar heat ki lehrein hain, aur caption kehta hai low voltage, bada current, garam wires. Doosre step mein plant ke bagal mein ek step-up transformer aata hai, line patli normal wire ban jaati hai aur heat ki lehrein gayab ho jaati hain: high voltage, chhota current, bahut kam loss. Teesre step mein gharon ke paas ek step-down transformer aata hai aur unki khidkiyaan roshan ho jaati hain, gharon ke liye 230 volts tak step down.

Example

Feeder line 11 kV par kyun chalti hai

Ek gaon ke substation ko $5.0 Ω$ total resistance wali feeder line se $100 k W$ deliver karna hai. Line mein khoyi power compare karo agar power (a) $440 V$ aur (b) $11 k V$ par bheji jaaye.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Resistor se guzarne wala current double kar diya jaata hai. Heating ka rate kitne guna badalta hai?

42½Nahi badalta

Answer dekho

Answer (A) 4

R

fixed ke saath

P = I^{2} R

, isliye

I

double karne se

P

chaar guna ho jaata hai. Sirf double kehna

P

I

ke proportional maan lena hai.

3.10 Cells, emf, internal resistance

Cell ek chemical device hai jo circuit ke ends par charges ko bhar kar rakhta hai. Do electrodes, positive (P) aur negative (N) marked, ek electrolyte mein khade hain. Har electrode par ek chemical reaction metal aur liquid ke beech charge move karti hai, jab tak electrode apne se chhoote electrolyte ke relative ek fixed potential par nahi tik jaata.

Maan lo positive electrode apne bagal ke electrolyte se $V_{+}$ potential upar tikta hai, aur negative electrode apne bagal ke electrolyte se $V_{-}$ potential neeche (dono numbers positive). Current na hone par electrolyte ke andar koi drop nahi hota, isliye N se P tak raaste mein sirf do jumps hote hain: $V_{-}$ upar, phir $V_{+}$ upar. Isliye

Cell ka emf

ε = V_{+} + V_{-}

(3.36)

Definition

Electromotive force (emf)

Jab koi current na behe, tab source ke terminals ke beech ka potential difference $ε$ . Yeh us work ke barabar hai jo source ka non-electrostatic (cell ke liye chemical) action unit positive charge ko uske negative terminal se positive terminal tak le jaane mein karta hai.

ExampleNaye alkaline AA cell ka emf lagbhag $1.5 V$ hota hai.

Emf (विद्युत वाहक बल) naam, jo electromotive force ka short form hai, 1800s ke shuru se chala aa raha hai. Emf asal mein ek potential difference hai, volts mein naapa jaata hai, aur iska newton se koi lena-dena nahi hai.

Circuit ko ek resistor $R$ se band karo. Cell ke bahar current P se $R$ hokar N tak behta hai. Loop poora karne ke liye use electrolyte ko bhi cross karna padta hai, N se wapas P tak, aur electrolyte kisi bhi doosre conductor ki tarah ise rokta hai. Iska resistance $r$ cell ka internal resistance (आंतरिक प्रतिरोध) hai. Current ki direction mein ise cross karne par potential drop $I r$ hota hai (Fig. 3.12), isliye terminals ke beech bacha potential difference, yaani terminal voltage, hai

2/3Load $R$ ke saath electrolyte ke across potential $I r$ girta hai, isliye $V = ε - I r$ .

Fig. 3.12Electrodes par chemical jumps milkar

ε

banate hain. Current behne par electrolyte

I r

le leta hai, aur terminals par

V = ε - I r

bachta hai.NCERT Fig. 3.12 · p. 94

Upar electrolyte ke beaker mein left mein blue negative electrode N aur right mein orange positive electrode P hai. Uske neeche ek graph cell ke andar position ke saath potential dikhata hai, electrodes ke neeche line up karke. Pehle step mein koi external circuit nahi hai: potential N par V minus upar jump karta hai, electrolyte mein flat rehta hai, phir P par V plus upar jump karta hai, isliye P, N se epsilon upar hai. Doosre step mein ek resistor R electrodes ko jodta hai, arrows dikhate hain ki current P se nikal kar R se hokar electrolyte ko N se P tak cross karta hai; flat hissa ab I r neeche jhuk jaata hai, isliye P, N se sirf V barabar epsilon minus I r upar hai, jo epsilon par dashed line ke saamne dikhaya gaya hai. Teesre step mein resistor ki jagah sirf ek wire hai; electrolyte ke across drop epsilon ke barabar hai aur terminal potential difference zero hai.

Current dete waqt terminal voltage

V = ε - I r

(3.38)

Yahi $V$ , $R$ ke across potential difference hai, isliye $V = I R$ bhi hai. $I R = ε - I r$ rakhne par current milta hai:

Cell se current

I = \frac{ε}{R + r}

(3.40)

Isse teen baatein nikalti hain. Pehli, lagbhag zero current lene wala voltmeter khud emf padhta hai. Doosri, load jitna zyada current leta hai, terminal voltage utna gir jaata hai. Teesri, terminals ko moti wire ( $R \approx 0$ ) se jodne par cell sabse bada current deta hai, $I_{\max} = ε / r$ . Aisa short circuit cell ko zyada garam karke kharab kar sakta hai, aur badi batteries isse aag bhi laga sakti hain, isliye manufacturers $ε / r$ se bahut kam ek safe maximum current specify karte hain.

Agar koi external source current ko positive terminal ke andar dhakele, jaise battery charge karte waqt, to $r$ ke across drop ulat jaata hai aur terminal voltage emf se zyada hota hai:

Charge karte waqt terminal voltage

V = ε + I r

(3.60)

Internal resistance bahut chhota bhi ho sakta hai aur bada bhi. Lead-acid car battery ka lagbhag $0.01 Ω$ hota hai, isliye woh starter motor mein sau ampere se bhi zyada push kar sakti hai. Chhote dry cell ka kahin zyada hota hai, aur purana hone par uska $r$ badhta jaata hai. Isiliye purana torch cell voltmeter par ab bhi $1.5 V$ padh sakta hai lekin bulb ko sirf dheema jalata hai: current behte hi uska zyadatar emf uske apne $r$ ke across kho jaata hai.

SimulationInternal resistance wala cell

6 volts emf aur 1 ohm internal resistance wala cell ek network chalata hai jismein R1, parallel mein lage R2 aur R3 ke saath series mein hai. Ek ammeter total current dikhata hai, aur voltmeter ko cell ke across ya network ke kisi bhi hisse ke across lagaya ja sakta hai. Chalte markers har branch ka current dikhate hain, aur sliders emf aur har resistance badalte hain.

Fig. 3.13

r = 1 Ω

wala

6 V

cell

R_{1}

ko chalata hai, jo parallel mein lage

R_{2}

aur

R_{3}

ke saath series mein hai.

Try karo

Voltmeter ko cell ke across lagao. Switch kholo aur emf padho, phir switch band karo. Terminal voltage kitna girta hai?
R1 ko step by step kam karo. Current aur terminal voltage ka kya hota hai? $V = ε - I r$ se compare karo.
R1 ke baad junction par markers dekho. Kya R1 ka current, R2 aur R3 ke currents ke jod ke barabar hai?

Example

Do readings se emf aur internal resistance

Cell ke across laga voltmeter $1.20 V$ padhta hai jab cell $4.0 Ω$ resistor chalata hai, aur $1.35 V$ jab woh $9.0 Ω$ resistor chalata hai. Cell ka emf aur internal resistance nikaalo. Maan lo voltmeter koi current nahi leta.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Try karke dekho

Cell ka emf aur internal resistance nikaalo

Tumhe chahiye

holder mein ek purana AA cell
ek digital multimeter
lagbhag 10 Ω, 4.7 Ω aur 2.2 Ω ke resistors (2 W ya zyada rating wale)
ek push switch aur connecting wires

Pehle guess karo

Open circuit wali reading ke mukaable, jab cell

2.2 Ω

resistor chalata hai to voltmeter kya dikhayega?

Zyada voltageUtna hi voltageZeroKam voltage

Bina kuch aur jode cell ke across voltage naapo. Yeh lagbhag uska emf hai.
Push switch ke through cell ke across ek resistor jodo. Switch dabao, voltage jaldi se padho aur chhod do.
Baaki do resistors ke saath dohrao. Har current $I = V / R$ se nikaalo.
$V$ vs $I$ plot karo. $V$ axis par intercept $ε$ hai; slope $- r$ hai.

Load par cell

Load R (Ω)	Voltage V (V)	Current I = V/R (A)
kuch nahi (open)		0
10
4.7
2.2

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 3.13

3.11 Series aur parallel mein cells

Torch mein do cells ek ke baad ek lagte hain; solar lantern mein kai cells agal-bagal lag sakte hain. Jaise resistors milkar ek equivalent resistor ban jaate hain, waise hi cells ke group ko, baaki circuit ke hisaab se, ek equivalent cell se replace kiya ja sakta hai jiska emf $ε_{e q}$ aur internal resistance $r_{e q}$ ho (Fig. 3.14).

DerivationSeries mein do cells

Cell 1 (emf $ε_{1}$ , internal resistance $r_{1}$ ) ke negative terminal ko cell 2 ( $ε_{2}$ , $r_{2}$ ) ke positive terminal se jodo. Khaali positive terminal ko A, joint ko B aur khaali negative terminal ko C kaho. A se external circuit mein current $I$ behne par, har cell par (3.38) lagaane se $V (A) - V (B) = ε_{1} - I r_{1}$ aur $V (B) - V (C) = ε_{2} - I r_{2}$ milta hai. Jodne par,

V_{A C} = (ε_{1} + ε_{2}) - I (r_{1} + r_{2})

(3.43)

A aur C ke beech ek single cell $V_{A C} = ε_{e q} - I r_{e q}$ deta. $I$ ki har value ke liye dono expressions compare karne par:

Series mein cells

ε_{e q} = ε_{1} + ε_{2}, r_{e q} = r_{1} + r_{2}

(3.45)

Ab cell 2 ko ulta kar do, taaki dono negative terminals B par milein. Ab current cell 2 mein uske positive terminal se negative terminal ki taraf jaata hai, jisse $V (B) - V (C) = - ε_{2} - I r_{2}$ milta hai aur isliye $V_{A C} = (ε_{1} - ε_{2}) - I (r_{1} + r_{2})$ . Pehle ki tarah ek equivalent cell se compare karne par:

ε_{e q} = ε_{1} - ε_{2} (ε_{1} > ε_{2}), r_{e q} = r_{1} + r_{2}

(3.47)

Yahi hisaab-kitaab kitni bhi lambi line ke liye kaam karta hai. Current ki direction mein uske along chalo: jis cell ko tum − se + cross karte ho woh apna emf jodta hai, jise + se − cross karte ho woh ghatata hai, aur har internal resistance judta hai, cell kisi bhi taraf face kare.

3/3Positive terminals jude: $1 / r_{e q} = 1 / r_{1} + 1 / r_{2}$ aur $ε_{e q} / r_{e q} = ε_{1} / r_{1} + ε_{2} / r_{2}$ .

Fig. 3.14Series mein emfs (signs ke saath) judte hain aur internal resistances judte hain. Parallel mein conductances

1 / r

judte hain aur ratios

ε / r

bhi.NCERT Figs. 3.13–3.14 · p. 95

Teen shaded rows mein cells ke pairs hain, lambi plate positive terminal hai. Pehle step mein upar wali row mein points A, B aur C ke beech series mein jude cells hain, dono ek hi taraf face karte hain, aur results epsilon eq barabar epsilon 1 plus epsilon 2 aur r eq barabar r 1 plus r 2 hain. Doosre step mein doosri row aati hai jismein doosra cell ulta hai, jisse epsilon eq barabar epsilon 1 minus epsilon 2 milta hai aur r eq ab bhi r 1 plus r 2 hai. Teesre step mein teesri row do cells agal-bagal dikhati hai, unke positive terminals left mein point X par aur negative terminals right mein point Y par jude hain, aur results one over r eq barabar one over r 1 plus one over r 2 aur epsilon eq over r eq barabar epsilon 1 over r 1 plus epsilon 2 over r 2 hain.

DerivationParallel mein do cells

Ab dono positive terminals ko ek common point X se aur dono negative terminals ko ek common point Y se jodo (Fig. 3.14, aakhri step). Maan lo $I_{1}$ aur $I_{2}$ cells ke positive terminals se nikalne wale currents hain, aur $I$ woh current hai jo X se external circuit mein behta hai. X par charge jama nahi hota, isliye

I = I_{1} + I_{2}

(3.48)

Dono cells ke across same potential difference $V = V (X) - V (Y)$ hai, isliye $V = ε_{1} - I_{1} r_{1} = ε_{2} - I_{2} r_{2}$ . Har ek ko uske current ke liye solve karo aur jodo:

I = \frac{ε_{1} - V}{r_{1}} + \frac{ε_{2} - V}{r_{2}} \Rightarrow V = \frac{ε_{1} r_{2} + ε_{2} r_{1}}{r_{1} + r_{2}} - I \frac{r_{1} r_{2}}{r_{1} + r_{2}}

(3.52)

Ek equivalent cell ke $V = ε_{e q} - I r_{e q}$ se compare karo:

ε_{e q} = \frac{ε_{1} r_{2} + ε_{2} r_{1}}{r_{1} + r_{2}}, r_{e q} = \frac{r_{1} r_{2}}{r_{1} + r_{2}}

(3.54)

Inhe is form mein yaad rakhna aasaan hai

Parallel mein cells

\frac{1}{r_{e q}} = \frac{1}{r_{1}} + \frac{1}{r_{2}}, \frac{ε_{e q}}{r_{e q}} = \frac{ε_{1}}{r_{1}} + \frac{ε_{2}}{r_{2}}

(3.57)

Agal-bagal zyada cells hon to har sum mein bas aur terms jud jaate hain. Ulta joda gaya cell (jiska negative terminal positive rail par ho) apne emf ko ulta karke aata hai, $ε \to - ε$ .

Do special cases yaad rakhne laayak hain. Series mein $n$ identical cells $n ε$ aur $n r$ dete hain: zyada voltage. Parallel mein $m$ identical cells $ε$ aur $r / m$ dete hain: same voltage aur kam internal resistance, isliye woh zyada current de sakte hain aur zyada der chalte hain.

Example

Parallel mein ek strong cell aur ek weak cell

$1.5 V$ emf aur $0.50 Ω$ internal resistance wala cell, $1.2 V$ emf aur $0.30 Ω$ internal resistance wale cell ke saath parallel (positive se positive) mein joda gaya hai. Yeh jodi $2.0 Ω$ resistor chalati hai. (a) Equivalent emf aur internal resistance nikaalo, aur resistor mein current. (b) Har cell kitna current deta hai? (c) Agar resistor hata diya jaaye to kya hota hai?

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

3 mein se 1

Chaar cells, har ek ka emf

1.5 V

aur internal resistance

0.25 Ω

5.0 Ω

lamp ke saath series mein jude hain. Kitna current behta hai?

Answer dekho

Answer 1 A

ε_{e q} = 6.0 V

aur

r_{e q} = 1.0 Ω

, isliye

I = 6.0 / (5.0 + 1.0) = 1.0 A

3.12 Kirchhoff's rules

Series aur parallel rules ek had tak hi kaam aate hain. Do alag branches mein cells lagao, ya resistors ko bridge mein jodo, to kitna bhi combine karo network simple nahi hoga. Gustav Kirchhoff ke diye do rules, Kirchhoff's rules (किरखोफ़ के नियम), aise har case ko handle karte hain.

Inhe use karne se pehle apne labels set karo. Har cell ke + aur − terminals mark karo. Phir har branch ko arrow ke saath ek current symbol do. Arrow sirf direction ka ek guess hai, aur algebra use theek kar dega: jo current negative aaye, woh bas apne arrow ke against beh raha hai. Labels lag jaane ke baad yeh sign rules har element ko cross karte waqt potential change batate hain:

Resistor ko uske current ke arrow ke along cross karne par potential $I R$ girta hai; arrow ke against chalne par $I R$ badhta hai.
Cell ko uske negative se positive terminal ki taraf cross karne par potential $ε$ badhta hai; positive se negative ki taraf, $ε$ girta hai.
Cell ka internal resistance kisi bhi doosre resistor ki tarah treat hota hai. Agar cell mein current $I$ , N se P ki taraf behta hai, to $V (P) - V (N) = ε - I r$ , jaisa (3.38) mein. Agar P se N ki taraf behta hai, to $V (P) - V (N) = ε + I r$ , jaisa (3.60) mein.

LawJunction rule

Kisi bhi junction par andar aane wala total current, bahar jaane wale total current ke barabar hota hai.

Maan lo kisi junction mein har second jitna charge nikalta hai, usse zyada ghusta hai. Tab wahan charge jama hota jaata, junction ka potential badhta rehta, aur currents badal jaate: circuit steady nahi rehta. Isliye steady state mein har junction se charge theek utni hi tezi se nikalta hai jitni tezi se aata hai, isliye junction rule conservation of charge hai. Yeh ek single wire ke kisi bhi point par bhi sahi hai, wire kaise bhi mude: point mein aane wala current bahar jaane wale current ke barabar hota hai.

2/3 $3 + 2 = 4 + I$ , isliye unknown current $1 A$ hai.

Fig. 3.15Junction par charge jama nahi hota, isliye andar aane wale currents ka jod bahar jaane wale currents ke barabar hona chahiye. Wire ke kisi bhi point par bhi yahi sahi hai.NCERT Fig. 3.15 · p. 98

Picture ke beech mein ek node par chaar wires milti hain. Arrows dikhate hain ki 3 amperes left se aur 2 amperes upar se andar aate hain, 4 amperes right ki taraf bahar jaate hain, aur ek unknown current I neeche-right wali wire se bahar jaata hai. Doosre step mein node highlight hota hai, equation 3 plus 2 barabar 4 plus I dikhayi jaati hai, aur unknown current 1 ampere nikalta hai. Teesre step mein junction fade ho jaata hai aur ek single mudi hui wire dikhayi jaati hai jis par ek point mark hai; same current I point mein aata hai aur bahar jaata hai.

LawLoop rule

Circuit ke kisi bhi closed loop mein potential ke badhne aur girne ka jod zero hota hai.

5/5d se a tak wire ka koi resistance nahi hai. Changes ka jod zero hai: $+ 10 - 6 - 4 = 0$ .

Fig. 3.16Loop ka ek chakkar: cell ke across

10 V

upar (uska

12 V

emf minus andar khoye

2 V

), resistors ke across

6 V

aur

4 V

neeche. Total change zero hai.NCERT Fig. 3.15 · p. 98

Left mein ek rectangular loop hai jiske corners a, b, c aur d hain. Left side mein ek dashed box ke andar 1 ohm internal resistance wala 12 volt cell hai, upar wali side mein 3 ohm resistor aur right side mein 2 ohm resistor; arrows 2 amperes ka clockwise current dikhate hain. Right mein points a, b, c, d aur wapas a par potential ka graph hai. Pehle step mein ek dot a par hai aur graph sirf a par zero volts dikhata hai. Doosre step mein dot b par jaata hai aur graph 12 volts tak badhta hai, phir internal resistance ke across 10 volts tak girta hai. Teesre step mein dot c par pahunchta hai aur graph 3 ohm resistor ke across 4 volts tak girta hai. Chauthe step mein dot d par pahunchta hai aur graph 2 ohm resistor ke across zero tak girta hai. Paanchven step mein dot wapas a par hai aur sum plus 10 minus 6 minus 4 barabar 0 dikhaya gaya hai.

Quick check

2 mein se 1

Ek junction par

5 A

aur

2 A

ke currents andar aate hain, aur ek wire se

4 A

bahar jaata hai. Bachi hui ek hi wire se kitna current bahar jaata hai?

Answer dekho

Answer 3 A

Andar = bahar:

5 + 2 = 4 + I

, isliye

I = 3 A

Ek method jo hamesha kaam karta hai.

Circuit draw karo, har branch current ko arrow ke saath label karo, aur turant junction rule use karo taaki kam se kam unknown currents aayein.
Kaafi independent loops chuno: bache hue har unknown ke liye ek. Loop independent hai agar woh kam se kam ek aisi branch se guzarta hai jo pichhle loops mein use nahi hui.
Har loop ke chaaron taraf apni pasand ki direction mein chalo, sign rules se potential changes jodo, aur sum ko zero rakho.
Simultaneous equations solve karo. Negative current apne arrow ke against behta hai. Apne answers ko aise loop se check karo jo tumne use nahi kiya.

Example 3.5NCERT Example 3.5 · p. 98

Chhah resistors, ek smart idea

Chhah resistors, har ek $6.0 Ω$ ka, ek square ke chaar corners P, Q, R aur S ke har pair ko jodte hain: har side ke along ek aur har diagonal ke along ek (diagonals bina chhuye ek doosre ko cross karte hain). Network ko negligible internal resistance wali $9.0 V$ battery chalati hai, jo dikhaye gaye tareeke se padosi corners P aur Q se judi hai. P aur Q ke beech equivalent resistance aur har resistor mein current nikaalo.

Fig. 3.17Har resistor

6.0 Ω

ka hai. Diagonals bina chhuye cross karte hain.NCERT Fig. 3.16 · p. 98

Chaar corners, upar S aur R aur neeche P aur Q, chaaron sides aur dono diagonals ke along resistors se jude hain; dono diagonals ek chhote hop par cross karte hain taaki chhuein nahi. Neeche P aur Q ke beech 9.0 volt battery judi hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Yeh shortcut sirf isliye kaam aaya kyunki network itna symmetric hai: chaaron corners ek jaise hain, isliye kisi bhi do ke across same $3.0 Ω$ aata hai. Agle network mein do cells hain aur koi symmetry nahi, isliye hum equations ka poora set likh kar solve karte hain.

Example 3.6NCERT Example 3.6 · p. 99

Do cells wala network

Neeche diye network mein saare branch currents nikaalo. $12 V$ cell ka internal resistance negligible hai; $6 V$ cell ka internal resistance $1 Ω$ hai. Neeche ki poori rail E ek hi node hai.

Fig. 3.18Maani gayi directions:

I_{1}

A se B,

I_{2}

B se C,

I_{3}

C se D.NCERT Fig. 3.17 · p. 99

Upar ke nodes A, B, C aur D 3, 2 aur 1 ohm ke resistors se jude hain, aur current arrows I 1, I 2 aur I 3 right ki taraf point karte hain. A se ek 12 volt cell, upar positive, neeche ki rail E tak jaata hai. B se ek 6 ohm resistor aur C se ek 2 ohm resistor rail tak neeche jaate hain. D se ek 6 volt cell, upar positive, dashed box mein 1 ohm internal resistance ke saath, rail tak neeche jaata hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Try karke dekho

Simulation mein Kirchhoff's rules check karo

Circuit simulation kholo aur ammeter reading, yaani total current, note karo.
$R_{2}$ aur $R_{3}$ se guzarne wala current nikaalo. Wide screen par yeh har resistor ke bagal mein likha hota hai; phone par kisi resistor ko tap karo taaki voltmeter uske across aa jaaye, aur reading ko uske resistance se divide karo. Kya dono currents ka jod $R_{1}$ ke current ke barabar hai?
Voltmeter ko pehle $R_{1}$ ke across, phir parallel pair ke across, phir cell ke across lagao. Kya terminal voltage baaki do readings ke jod ke barabar hai?

Abhi nahi kar sakte? Simulation try karo · Fig. 3.13

Quick check

3 mein se 1

Loop ke chaaron taraf uske current ki direction mein chalte hue tum ek cell ko uske positive terminal se negative terminal ki taraf cross karte ho. Cell ke emf ke across potential kaise badalta hai?

ε

badhta haiSirf

I r

badalta hai

ε

girta haiNahi badalta

Answer dekho

Answer (C) $ε$ girta hai

+ se − tak potential

ε

girta hai, current kuch bhi kare. Current sirf internal resistance ke

I r

term ka sign tay karta hai.

3.13 Wheatstone bridge

Kirchhoff's rules physics ke sabse useful circuits mein se ek ko samjhate hain, Wheatstone bridge (व्हीटस्टोन सेतु). Yeh ek unknown resistance ko known resistances se compare karke nikaalta hai (Fig. 3.19). Chaar resistors, A se B tak $P$ , B se C tak $Q$ , A se D tak $R$ aur D se C tak $S$ , ek band diamond banate hain. Ek cell opposite corners A aur C ko jodta hai; yeh branch battery arm hai. Ek galvanometer G, jo ek sensitive current detector hai, baaki do corners B aur D ko jodta hai; yeh branch galvanometer arm hai.NCERT textbook arms ko $R_{1}$ (AD), $R_{2}$ (AB), $R_{3}$ (DC) aur $R_{4}$ (BC) naam deti hai aur balance condition ko $R_{2} / R_{1} = R_{4} / R_{3}$ likhti hai. Yeh neeche di gayi $P / Q = R / S$ wali hi condition hai.

3/3Resistors ko tab tak adjust karo jab tak $V_{B} = V_{D}$ na ho jaaye: G mein koi current nahi, aur $P / Q = R / S$ .

Fig. 3.19Jab

P / Q = R / S

ho, to points B aur D same potential par hote hain aur galvanometer zero padhta hai: bridge balanced hai.NCERT Fig. 3.18 · p. 101

Chaar resistors ek diamond banate hain: left mein A se upar B tak P, B se right mein C tak Q, A se neeche D tak R aur D se C tak S. Needle wala ek galvanometer B se seedha neeche D tak juda hai, aur diamond ke neeche ek cell A aur C ko jodta hai. Pehle step mein battery arm aur galvanometer arm highlight aur label kiye gaye hain. Doosre step mein arrows dikhate hain ki current B se D tak galvanometer se hokar behta hai, needle right ki taraf mudi hai aur label par likha hai V B barabar nahi hai V D ke, G mein current. Teesre step mein needle seedhi upar zero par point karti hai aur labels par likha hai V B barabar V D aur P over Q barabar R over S, koi deflection nahi.

Koi bhi resistors lene par needle hilti hai, kyunki kuch current B aur D ke beech G se hokar short cut leta hai. Lekin resistors ko adjust karke hum galvanometer current $I_{g}$ ko bilkul zero kar sakte hain. Yahi special setting, balanced bridge, circuit ko useful banati hai.

DerivationBalance condition

Maan lo $I_{g} = 0$ . B par junction rule kehta hai ki $P$ se aane wala current $I_{1}$ bina badle $Q$ mein aage jaata hai. D par, $R$ se aane wala current $I_{2}$ , $S$ mein aage jaata hai. Is argument mein kuch bhi cell par depend nahi karta, isliye uska emf aur internal resistance kuch bhi ho sakta hai.

Loop ABDA par loop rule lagao ( $P$ se, phir G se, phir $R$ se wapas). G ke across koi drop nahi hai kyunki usmein koi current nahi behta:

- I_{1} P + 0 + I_{2} R = 0 \Rightarrow I_{1} P = I_{2} R

(3.62)

Ab loop BCDB ( $Q$ se, phir $S$ aur G se wapas):

- I_{1} Q + I_{2} S + 0 = 0 \Rightarrow I_{1} Q = I_{2} S

(3.63)

Pehle result ko doosre se divide karne par dono currents hat jaate hain:

Balance condition

\frac{P}{Q} = \frac{R}{S}

(3.64(a))

Shabdon mein: $P$ aur $R$ ke across potential drops barabar hain, isliye B aur D same potential par hain, aur G mein current chalane wala kuch nahi hai.

Practice mein unknown resistance ko $S$ ki jagah lagaya jaata hai. $P$ aur $Q$ , yaani ratio arms, fixed known resistors hain, aur ek variable known resistance $R$ ko tab tak adjust kiya jaata hai jab tak galvanometer needle zero par na ruk jaaye, yaani null point. Tab

Unknown resistance

S = R \frac{Q}{P}

(3.64(b))

Null method ke asli fayde hain. Needle bilkul zero par hai ya nahi, yeh tay karna kisi value ko accurately padhne se kahin aasaan hai. Balance condition mein cell ka emf ya galvanometer ka resistance nahi aata, isliye inmein se kisi ko jaanna ya steady rakhna zaroori nahi. Bridge sabse sensitive tab hota hai jab chaaron arms ke resistances milte-julte hon. School laboratory mein meter bridge yahi idea use karta hai: ek metre lambi uniform wire $P$ aur $Q$ ki jagah leti hai, aur uske along contact khiskane se unka ratio badalta hai.

InteractiveBridge ko balance karo

Fig. 3.20Ek mystery resistor aur ratio arms ka ek pair chuno, phir

R

ko tab tak adjust karo jab tak galvanometer zero na padhe.

Example 3.7NCERT Example 3.7 · p. 101

Unbalanced bridge se current

Dikhaye gaye bridge mein $P = 10 Ω$ (AB), $Q = 20 Ω$ (BC), $R = 15 Ω$ (AD) aur $S = 20 Ω$ (DC) hain. $40 Ω$ resistance wala galvanometer B aur D ke beech juda hai, aur negligible internal resistance wala $6.0 V$ cell A aur C ke across. Galvanometer se guzarne wala current nikaalo. $S$ ki kaunsi value bridge ko balance karegi?

Fig. 3.21Unbalanced bridge:

10 / 20 \neq 15 / 20

.NCERT Fig. 3.19 · p. 101

Chaar resistors ka ek diamond: A se B tak 10 ohm, B se C tak 20 ohm, A se D tak 15 ohm aur D se C tak 20 ohm. 40 ohm ka galvanometer B aur D ko jodta hai, aur neeche 6.0 volt ka cell A aur C ko jodta hai.

Pehle khud try karo. Phir solution ek-ek step karke kholo.

Quick check

4 mein se 1

Ek bridge

P = 10 Ω

Q = 40 Ω

aur

R = 25 Ω

ke saath balance hota hai. Unknown resistance

S

kya hai?

Answer dekho

Answer 100 Ω

S = R Q / P = 25 \times 40 / 10 = 100 Ω

Table 3.3Ek nazar mein quantities

Quantity	Symbol	SI unit	Dimensions	Definition
Electric current	$I$	A	$[A]$	SI base quantity
Charge	$q$ , $Q$	C	$[T A]$	$q = I t$
Potential difference, emf	$V$ , $ε$	V	$[M L^{2} T^{- 3} A^{- 1}]$	per unit charge work
Resistance	$R$	Ω	$[M L^{2} T^{- 3} A^{- 2}]$	$R = V / I$
Resistivity	$ρ$	Ω m	$[M L^{3} T^{- 3} A^{- 2}]$	$R = ρ l / A$
Conductivity	$σ$	S/m	$[M^{- 1} L^{- 3} T^{3} A^{2}]$	$σ = 1 / ρ$
Electric field	$E$	V/m	$[M L T^{- 3} A^{- 1}]$	per unit charge force
Current density	$j$	A/m²	$[L^{- 2} A]$	$j = I / A$
Drift speed	$v_{d}$	m/s	$[L T^{- 1}]$	$v_{d} = e E τ / m$
Relaxation time	$τ$	s	$[T]$	collisions ke beech ka average time
Mobility	$μ$	m² V⁻¹ s⁻¹	$[M^{- 1} T^{2} A]$	$μ = v_{d} / E$
Power	$P$	W	$[M L^{2} T^{- 3}]$	$P = I V$

Flash cards · 10

Yaad rakhne wale key relations

1 / 10

Recap

Summary aur key points

Current kisi surface ko per unit time cross karne wala net charge hai, $I = d Q / d t$ ; yeh ek scalar hai jo amperes mein naapa jaata hai (Section 3.2).
Steady current ke liye closed circuit aur cell jaisa source chahiye, jo conductor ke andar steady field bana kar rakhe (Section 3.3).
Ohm's law: constant temperature par metal ke liye $V = I R$ . Resistance shape par $R = ρ l / A$ ke through depend karta hai, aur locally $𝐣 = σ 𝐄$ (Section 3.4).
Collisions field ke push ko ek steady drift mein badal dete hain, $𝐯_{d} = - e τ 𝐄 / m$ , jisse $I = n e A v_{d}$ aur $ρ = m / (n e^{2} τ)$ milte hain (Section 3.5).
Drift speeds bahut chhoti hain (lagbhag ek millimetre per second ya kam) lekin $n$ bahut bada hai; mobility $μ = v_{d} / E = e τ / m$ (Section 3.5.1).
$V = I R$ kisi bhi device ke liye resistance define karta hai; Ohm's law tabhi sahi hai jab $R$ , $V$ par depend na kare. Lamps, diodes aur GaAs devices ise todte hain (Section 3.6).
Metals ka $ρ \sim 10^{- 8}$ se $10^{- 6} Ω m$ hota hai, insulators ka $10^{18}$ se $10^{24}$ guna zyada, aur semiconductors beech mein aate hain (Section 3.7).
Metals ke liye $ρ_{T} = ρ_{0} [1 + α (T - T_{0})]$ , jahan $α > 0$ ; alloys ka $α$ bahut chhota hota hai; semiconductors ka $ρ$ temperature ke saath girta hai (Section 3.8).
Heat mein badalne wali power $P = I V = I^{2} R = V^{2} / R$ hai; high voltage par transmit karne se line losses kam hote hain, jo $1 / V^{2}$ ki tarah badalte hain (Section 3.9).
Cell ka emf uska open-circuit terminal voltage hai; internal resistance $r$ ke saath $V = ε - I r$ aur $I = ε / (R + r)$ (Section 3.10).
Series mein cells: emfs sign ke saath judte hain aur internal resistances judte hain. Parallel mein: $1 / r_{e q} = \sum 1 / r_{i}$ aur $ε_{e q} / r_{e q} = \sum ε_{i} / r_{i}$ (Section 3.11).
Junction rule (charge conservation) aur loop rule (single-valued potential) koi bhi circuit solve kar dete hain (Section 3.12).
Wheatstone bridge tab balance hota hai jab $P / Q = R / S$ ho, aur null method se unknown resistance deta hai (Section 3.13).

Apna NCERT exercise question dhoondo

Hamare practice questions ncert.ai ke liye likhe gaye hain aur NCERT exercises ko follow karte hain. Apni textbook ke question se match karne wala question dhoondo; uska answer uske saath hai.

Tumhari NCERT textbook mein	Yahan practice question
NCERT Exercise 3.1	Practice Q1
NCERT Exercise 3.2	Practice Q2
NCERT Exercise 3.3	Practice Q3
NCERT Exercise 3.4	Practice Q4
NCERT Exercise 3.5	Practice Q5
NCERT Exercise 3.6	Practice Q6
NCERT Exercise 3.7	Practice Q7
NCERT Exercise 3.8	Practice Q8
NCERT Exercise 3.9	Practice Q9

Quick check: MCQs

3.2 Electric current3 questions
3.3 Conductors mein electric currents3 questions
3.4 Ohm's law4 questions
3.5 Electrons ka drift aur resistivity ka origin2 questions
3.5.1 Mobility3 questions
3.6 Ohm's law ki limitations3 questions
3.7 Alag-alag materials ki resistivity3 questions
3.8 Resistivity ki temperature par dependence3 questions
3.9 Electrical energy, power3 questions
3.10 Cells, emf, internal resistance4 questions
3.11 Series aur parallel mein cells3 questions
3.12 Kirchhoff's rules2 questions
3.12 Kirchhoff's rules3 questions
3.13 Wheatstone bridge4 questions

Glossary · 14

Key terms (English + Hindi)

conductivityचालकताchaalakta: Resistivity ka reciprocal, $σ = 1 / ρ$ , siemens per metre mein, jahan $𝐣 = σ 𝐄$ .Chapter mein pehli baar yahan aaya →
current densityधारा घनत्वdhaara ghanatva: Flow ke normal surface ke per unit area current, $j = I / A$ , $A / m^{2}$ mein. Yeh ek vector hai.Chapter mein pehli baar yahan aaya →

drift velocityअपवाह वेगapavaah veg: Field mein charge carriers ki chhoti average velocity, jo unki random motion ke upar judti hai; electrons ke liye $𝐯_{d} = - e τ 𝐄 / m$ .Chapter mein pehli baar yahan aaya →

electric currentविद्युत धाराvidyut dhaara: Kisi surface ko per unit time cross karne wala net charge, $I = d Q / d t$ , jo amperes mein naapa jaata hai.Chapter mein pehli baar yahan aaya →
emfविद्युत वाहक बलvidyut vaahak bal: Open circuit par source ka terminal potential difference: uska chemical (ya koi aur) action per unit charge jitna work karta hai. Volts mein naapa jaata hai.Chapter mein pehli baar yahan aaya →

internal resistanceआंतरिक प्रतिरोधaantarik pratirodh: Cell ke andar ka resistance $r$ , jaise uske electrolyte ka, jo current behne par $I r$ drop paida karta hai.Chapter mein pehli baar yahan aaya →

Kirchhoff's rules: Do circuit rules: junction mein aane wale currents bahar jaane walon ke barabar hote hain, aur kisi bhi closed loop mein potential changes ka jod zero hota hai.Chapter mein pehli baar yahan aaya →

mobilityगतिशीलताgatisheelta: Per unit field drift speed, $μ = v_{d} / E = e τ / m$ , $m^{2} V^{- 1} s^{- 1}$ mein.Chapter mein pehli baar yahan aaya →

relaxation time: Conduction electrons ke pichhle collision se ab tak ka average time $τ$ ; metals mein lagbhag $10^{- 14} s$ .Chapter mein pehli baar yahan aaya →
resistanceप्रतिरोधpratirodh: Kisi conductor ke liye ratio $R = V / I$ , jo ohms ( $Ω$ ) mein naapa jaata hai.Chapter mein pehli baar yahan aaya →
resistivityप्रतिरोधकताpratirodhakta: $R = ρ l / A$ mein material ki property $ρ$ , $Ω m$ mein. Yeh material aur temperature par depend karti hai, shape par nahi.Chapter mein pehli baar yahan aaya →

temperature coefficient of resistivityप्रतिरोधकता ताप गुणांकpratirodhakta taap gunaank: Har degree par resistivity ka fractional change, $ρ_{T} = ρ_{0} [1 + α (T - T_{0})]$ mein $α$ . Metals ke liye positive.
terminal voltage: Jab source current $I$ de raha ho, tab uske across potential difference: $V = ε - I r$ .Chapter mein pehli baar yahan aaya →

Wheatstone bridgeव्हीटस्टोन सेतुwheatstone setu: Diamond mein lage chaar resistors, ek diagonal ke across cell aur doosre ke across galvanometer. Yeh tab balance hota hai jab $P / Q = R / S$ .Chapter mein pehli baar yahan aaya →

Current Electricity: Class 12 Physics Chapter 3 Hinglish mein

Is chapter mein aap kya seekhenge

3.1 Introduction

3.2 Electric current

3.3 Conductors mein electric currents

3.4 Ohm's law

3.5 Electrons ka drift aur resistivity ka origin

3.5.1 Mobility

3.6 Ohm's law ki limitations

3.7 Alag-alag materials ki resistivity

3.8 Resistivity ki temperature par dependence

3.9 Electrical energy, power

3.10 Cells, emf, internal resistance

3.11 Series aur parallel mein cells

3.12 Kirchhoff's rules

3.13 Wheatstone bridge

Summary aur key points

Practice questions answers ke saath (NCERT “Exercises”)

Apna NCERT exercise question dhoondo

Quick check: MCQs

Key terms (English + Hindi)

Tum chapter ke end tak pahunch gaye