Electricity: Class 10 Science Chapter 11 Hinglish mein

Science · Hinglish notes, simulations aur practice · Official NCERT chapter PDF ↗

Published 12 Oct 2026 · Updated 12 Oct 2026

100 min
3 simulations
46 practice questions
27 MCQs

Is chapter mein aap kya seekhenge

Electric current aur potential difference ko define karo, I = Q/t aur V = W/Q use karo, aur batao ki ammeter aur voltmeter kaise jode jaate hain
Standard symbols ke saath circuit diagrams banao aur padho
Ohm's law batao, V–I readings se use check karo aur problems solve karne ke liye V = IR use karo
Samjhao ki resistance length, area of cross-section aur material par kaise depend karta hai, aur R = ρl/A use karo
Series, parallel aur simple combinations mein resistors ka equivalent resistance nikaalo, aur har ek mein current aur voltage bhi
H = I²Rt ke saath current ka heating effect samjhao, aur heaters, filament bulbs aur fuses mein iska use batao
Electric power aur energy calculate karo, kilowatt hours aur joules ke beech convert karo, aur bijli ka kharcha nikaalo

NCERT Science Class 10, Chapter 11, pp. 171–194 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

Garmi ki ek shaam power cut ke baad bijli wapas aane wala pal socho. Fans ghoomne lagte hain, tube lights timtimaakar jal uthti hain, charging par lagaya phone jag jaata hai, aur koi check karta hai ki fridge phir se hum kar raha hai ya nahi. Yeh sab ek hi cheez par chalte hain: electric charge, jise energy ka ek source ek closed path mein ghumata hai.

Is chapter mein tum jaanoge ki electric current kya hai, use kya chalata hai aur uska size kaun control karta hai. Tum Ohm's law se miloge aur dekhoge ki resistors series aur parallel mein jodne par kaisa behave karte hain. Aakhir mein tum dekhoge ki current wire ko garam kyun karta hai, appliances kitni power use karte hain, aur bijli ka bill kaise banta hai.

11.1 Electric Current aur Circuit

Doorbell ka switch dabao aur corridor ke doosre chhor par bell turant baj jaati hai. Wires mein kuch chal pada hai: electric charge. Charge ke steady flow ko electric current (विद्युत धारा) kehte hain. Metal wires mein jo charges chalte hain woh electrons hote hain.

Dhyaan do ki doorbell ko ek nahi, do wires chahiye. Charge supply se nikalta hai, bell se guzarta hai aur wapas supply mein aata hai: use ek poora raasta chahiye. Source ke ek terminal se doosre terminal tak bina toote conducting loop ko electric circuit (विद्युत परिपथ) kehte hain. Switch bas loop mein ek gap hai jise tum khol ya band kar sakte ho. Kholo, to loop ke har hisse mein flow ek saath ruk jaata hai; band karo, to har jagah ek saath shuru ho jaata hai (Fig. 11.1).

3/3Agreement se, cell ke bahar current + se − ki taraf dikhaya jaata hai: electrons ke ulta.

Fig. 11.1Switch loop ko poora karta hai; electrons ek taraf drift karte hain, conventional current doosri taraf dikhaya jaata hai.NCERT Fig. 11.1 · p. 172

Wire ka ek single loop left mein ek cell, upar ek switch, right mein ek bulb aur neeche ek ammeter ko jodta hai. Pehle step mein switch khula hai, bulb andhera hai aur kuch nahi chalta. Doosre step mein switch band hai, bulb jalta hai, aur electrons, jo minus signs wale chhote blue dots ki tarah bane hain, cell ki negative plate se loop mein ghoomkar positive plate tak drift karte hain. Teesre step mein red arrows conventional current ko doosri taraf jaate dikhate hain, positive terminal se bulb aur ammeter se hokar wapas negative terminal tak.

Current kis taraf jaata hai? Jo rule hum aaj bhi use karte hain woh electrical science ke shuruaati dinon mein tay hua tha, 1897 mein electron ki discovery se ek sadi se bhi pehle: maana jaata hai ki current source ke positive terminal se nikalta hai, circuit mein ghoomta hai aur negative terminal par wapas aata hai. Ise conventional current kehte hain. Electrons negative hote hain, isliye wire mein woh asal mein doosri taraf drift karte hain, − se + ki taraf.

Current kitna bada hai? Hum ginte hain ki har second wire ke kisi bhi cross-section se kitna charge guzarta hai. Agar charge $Q$ guzarne mein laga time hai $t,$ to current hai

I = \frac{Q}{t}

(11.1)

jahan

$I$: current, amperes (A) mein
$Q$: guzarne wala charge, coulombs (C) mein
$t$: laga time, seconds (s) mein

Charge coulombs (C) mein naapa jaata hai. Ek electron par sirf itna charge hota hai: $1.6 \times 10^{- 19} C,$ isliye ek coulomb lagbhag $6.25 \times 10^{18}$ electrons ka charge hai. Current ki unit, ampere (A), France ke André-Marie Ampère (1775–1836) ke naam par hai. $1 A$ current ka matlab hai har second ek coulomb charge ka guzarna, isliye $1 A = 1 C / s .$ Chhote currents milliamperes aur microamperes mein diye jaate hain: $1 m A = 10^{- 3} A,$ aur $1 μ A = 10^{- 6} A .$

Definition

Electric current

Kisi conductor ke diye gaye cross-section se har second guzarne wale charge ki amount, $I = Q / t .$ Iski SI unit ampere (A) hai.

Example $2 A$ ka current har second ek point se $2 C$ charge le jaata hai.

Current ko ammeter (ऐमीटर) se naapa jaata hai. Gine jaane ke liye current ko meter ke andar se guzarna padta hai, isliye ammeter hamesha series mein joda jaata hai: tum circuit kholte ho aur meter se hokar use phir band karte ho. Fig. 11.1 mein ammeter neeche wali wire mein laga hai. Single loop mein tum ise kahin bhi lagao, reading same aayegi.

Example 11.1NCERT Example 11.1 · p. 172

Charging cable se guzarta charge

Ek phone charger cable 40 minutes tak $1.5 A$ ka steady current le jaati hai. (a) Isse kitna charge guzarta hai? (b) Yeh charge kitne electrons le jaate hain? (Electron par charge $= 1.6 \times 10^{- 19} C .$ )

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

Practice · 3 questions

Practice 11.1 (NCERT in-text questions, p. 172): answers ke saath

Short answerEasyNCERT in-text Questions p. 172, Q1
Riya cell ke ek terminal se torch bulb tak ek wire jodti hai, lekin bulb ka doosra contact khula chhod deti hai. Bulb andhera rehta hai. Electric circuit ke idea se samjhao kyun, aur batao ki use kya jodna chahiye.
Solution dikhao
AnswerKoi closed path nahi hai; use bulb ka doosra contact cell ke doosre terminal se jodna hoga.
Charge sirf ek poore, closed conducting path mein behta hai: yaani electric circuit mein. Yahan path bulb par khatam ho jaata hai, isliye koi charge cell tak wapas nahi aa sakta aur koi current nahi behta. Riya ko bulb ke free contact se cell ke doosre terminal tak ek doosri wire jodni hogi (is wire mein switch lagane se woh bulb ko on aur off kar sakegi).
Short answerEasyNCERT in-text Questions p. 172, Q2
Theek-theek samjhao ki one ampere current ka kya matlab hai, aur ampere ko doosri units mein likho.
Solution dikhao
Answer1 A matlab har second ek cross-section se 1 C charge ka flow: 1 A = 1 C/s.
Current charge ke flow ka rate hai, $I = Q / t .$ One ampere woh current hai jab har second conductor ke kisi bhi cross-section se ek coulomb charge guzarta hai, isliye $1 A = 1 C / s .$
Quick answerMediumNCERT in-text Questions p. 172, Q3
Ek wire mein current hai $1 m A .$ Har second wire ke ek point se kitne electrons guzarte hain? (Electron par charge $= 1.6 \times 10^{- 19} C .$ )
electrons
Answer dekho
Answer 6250000000000000 electrons
$1 s$ mein charge hai $Q = I t = 10^{- 3} C .$ Electrons ki sankhya hai $10^{- 3} / (1.6 \times 10^{- 19}) = 6.25 \times 10^{15} .$

Quick check

3 mein se 1

Cell ke bahar ki wires mein electrons kis taraf drift karte hain?

Positive terminal se negative terminal ki tarafNegative terminal se positive terminal ki tarafWoh bilkul nahi chalteAadhe ek taraf, aadhe doosri taraf jaate hain, isliye koi drift nahi hota

Answer dekho

Answer (B) Negative terminal se positive terminal ki taraf

Electrons negative hote hain, isliye woh negative terminal se ghoomkar positive terminal ki taraf drift karte hain. + se − wali direction conventional current hai, jo electrons ke pata chalne se pehle chuni gayi thi.

11.2 Electric Potential aur Potential Difference

Copper wire free electrons se bhari hoti hai, phir bhi apne aap usmein koi current nahi behta. Charges ko aage dhakelne ke liye kuch chahiye. Pipe ke ek closed loop mein paani socho (Fig. 11.2). Agar use koi nahi chalata, to paani bas pada rehta hai. Pump on karo aur woh pressure mein difference bana deta hai: paani high pressure par pump se nikalta hai, loop mein ghoomta hai, aur low pressure par wapas aata hai taaki phir se dhakela jaaye.

3/3Cell = pump, bulb = patli pipe, current = flow, potential difference = pressure ka gap.

Fig. 11.2Pump pressure difference banaye rakhkar paani chalata hai; cell charge ke liye yahi karta hai.

Paani ki pipe ke closed loop mein left side par ek pump aur right side par ek patla hissa hai. Pehle step mein pump band hai aur paani nahi chalta. Doosre step mein pump ghoomta hai, pump ke theek baad wali pipe par high pressure aur theek pehle wali par low pressure likha hai, aur arrows dikhate hain ki paani loop mein clockwise ghoomta hai aur patli pipe se nichudkar guzarta hai. Teesre step mein hisson ke naam circuit ke hisaab se badal diye gaye hain: pump cell hai, patli pipe bulb ya resistor hai, flow current hai aur pressure ka gap potential difference hai.

Circuit mein cell pump ka kaam karta hai. Uske andar chemical reactions charge ko ek terminal ki taraf aur doosre se door dhakelte hain, isliye dono terminals ka electrical 'pressure' alag hota hai. Yahi difference potential difference (विभवांतर), ya p.d., hai. Dukaan ki shelf par rakhe naye cell mein bhi yeh pehle se hota hai. Circuit jodo, aur p.d. charge ko ghumata hai: har coulomb cell ke andar energy leta hai aur bulb mein use de deta hai. Energy cell ke chemicals se aati hai, isiliye cells khatam ho jaate hain. Do ya zyada cells ko jodkar battery banti hai.

Isliye p.d. ko energy per unit charge ki tarah naapa jaata hai. Agar charge $Q$ ko do points ke beech le jaane mein laga work hai $W,$ to un points ke beech p.d. hai

V = \frac{W}{Q}

(11.2)

jahan

$V$: potential difference, volts (V) mein
$W$: kiya gaya work, joules (J) mein
$Q$: le jaaya gaya charge, coulombs (C) mein

P.d. ki unit volt (V) hai, jo Italian physicist Alessandro Volta (1745–1827) ke naam par hai. Jab $1 J$ work $1 C$ charge ko do points ke beech le jaata hai, to unke beech p.d. one volt hota hai:

1 V = \frac{1 J}{1 C} = 1 J C^{- 1}

(11.3)

Yaani $1.5 V$ ka cell usse guzarne wale har coulomb ko $1.5 J$ energy deta hai. P.d. ko voltmeter (वोल्टमीटर) se naapa jaata hai, jo hamesha parallel mein joda jaata hai: uski do leads test ho rahe part ke dono taraf ke do points ko chhooti hain, circuit ko tode bina.

Definition

Potential difference

Circuit ke do points ke beech charge chalne par per coulomb transfer hui energy, $V = W / Q .$ Iski SI unit volt hai; $1 V = 1 J / C .$

Example $9 V$ ki battery usse guzarne wale har coulomb ko $9 J$ deti hai.

Example 11.2NCERT Example 11.2 · p. 173

E-rickshaw ki battery dwara kiya gaya work

Ek e-rickshaw ke battery pack ka potential difference hai $48 V .$ Motor circuit mein $5 C$ charge ghumane mein woh kitna work karta hai?

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

Practice · 3 questions

Practice 11.2 (NCERT in-text questions, p. 174): answers ke saath

Quick answerEasyNCERT in-text Questions p. 174, Q1
Torch mein kaun sa part woh push (potential difference) deta hai jo bulb ke filament se charge ko chalata hai?
switchmetal ka caseglass reflectorcells
Answer dekho
Answer (D) cells
Cells chemical action se apne terminals ke beech p.d. banaye rakhte hain. Switch aur case sirf path ko poora karte hain.
Short answerEasyNCERT in-text Questions p. 174, Q2
Ek wire ke do ends ke beech potential difference hai $1 V .$ Isse energy aur charge ke baare mein kya pata chalta hai?
Solution dikhao
AnswerEk end se doosre end tak le jaaye gaye har 1 C charge ke liye 1 J work hota hai.
Yaad karo $V = W / Q,$ yaani $1 V$ p.d. ka matlab hai ki dono ends ke beech chalne wale har coulomb charge ke liye $1 J$ work hota hai (energy transfer hoti hai).
3
NumericalEasyNCERT in-text Questions p. 174, Q3
Ek car battery ka p.d. hai $12 V .$
1. Battery se guzarte hue har coulomb charge ko kitni energy milti hai?
  J
  Answer dekho
  Answer 12 J
  $W = V Q = 12 \times 1 = 12 J .$
2. $3 C$ charge ko kitni energy milti hai?
  J
  Answer dekho
  Answer 36 J
  $W = V Q = 12 \times 3 = 36 J .$

Quick check

3 mein se 1

Circuit ki water-loop wali picture mein potential difference ka kaam kaun karta hai?

Khud paaniPaani ki speedPipe ki widthPump ka banaya hua pressure difference

Answer dekho

Answer (D) Pump ka banaya hua pressure difference

Pump ek pressure difference banaye rakhta hai jo flow ko chalata hai, bilkul waise hi jaise cell ek p.d. banaye rakhta hai jo current ko chalata hai. Paani charge jaisa hai aur patli pipe resistance jaisi.

11.3 Circuit Diagram

Circuit ko bilkul waisa sketch karna jaisa woh dikhta hai, time leta hai aur zaroori cheezein chhupa deta hai. Isliye hum circuit diagram banate hain: har part ek simple standard symbol ban jaata hai, aur wires seedhi lines ban jaati hain. Jo bhi symbols jaanta hai woh circuit padh sakta hai, chahe woh koi bhi bhasha bolta ho. Fig. 11.3 is chapter mein use hone wale symbols dikhata hai.

1/5Is chapter mein tum jo symbols use karoge.

Fig. 11.3Standard symbols se koi bhi ek nazar mein circuit padh sakta hai.NCERT Table 11.1 · p. 174

Do columns mein baarah circuit symbols, har ek apne naam ke saath. Left column mein ek cell, do cells ki battery, khula switch, band switch, dot se judi do wires, aur bina jude cross karti do wires hain, jahan ek wire doosri ke upar se kood jaati hai. Right column mein bulb cross wale circle ki tarah, resistor zigzag ki tarah, rheostat aar-paar arrow wale zigzag ki tarah, fuse wire par chhote rectangle ki tarah, ammeter A wale circle ki tarah aur voltmeter V wale circle ki tarah dikhaye gaye hain. Steps pehle sources, phir switches aur wires, phir components, phir meters ko highlight karte hain.

Table 11.1Har symbol ko kaise pehchaanein

Component	Symbol kaisa dikhta hai
Cell	ek chhoti moti line (−) ke paas ek lambi patli line (+)
Battery	do ya zyada cell symbols ek line mein jude hue
Switch ya plug key	lever wala ek gap: khula (utha hua) ya band (seedha)
Judi wires	lines jo ek solid dot par milti hain
Bina judi wires	lines jo bina dot ke cross karti hain (ya chhoti si chhalaang ke saath)
Bulb	andar cross wala ek circle
Resistor	ek zigzag line
Rheostat	aar-paar arrow wala zigzag
Fuse	ek chhota rectangle jiske beech se wire guzarti hai
Ammeter	A letter wala ek circle
Voltmeter	V letter wala ek circle

Kuch aadatein diagrams ko padhne mein aasaan bana deti hain. Wires ko square corners wali seedhi lines ki tarah banao. Cell ki lambi plate ko positive terminal banao. Jahan wires judi hon wahan dot lagao, aur jahan sirf cross karti hon wahan mat lagao. School labs mein aksar plug key switch ka kaam karti hai: uska brass plug andar dabane se circuit band hota hai, aur bahar nikaalne se khul jaata hai.

Quick check

2 mein se 1

Har symbol ko us part se match karo jiske liye woh hai.

Left mein koi item tap karo, phir right mein uska match.

Answer dekho

Answer

ek chhoti moti line ke paas ek lambi patli line : cell
ek zigzag line : resistor
aar-paar arrow wala zigzag : rheostat
V letter wala ek circle : voltmeter
ek dot par milti do lines : judi wires

Yeh Fig. 11.3 ke standard symbols hain. Rheostat ke aar-paar arrow dikhata hai ki uska resistance badla ja sakta hai.

11.4 Ohm's Law

Agar tum push double karo, to kya current bhi double hoga? Yeh jaanne ke liye hume aisa circuit chahiye jismein p.d. step by step badhaya ja sake aur har baar current padha ja sake.

Activity 11.1NCERT Activity 11.1 · p. 175

Kya current voltage ke saath-saath badhta hai?

Tumhe chahiye

holder mein chaar $1.5 V$ dry cells
ek $10 Ω$ coil (card par lapeti constantan wire)
ek ammeter (0–1 A) aur ek voltmeter (0–10 V)
ek switch aur connecting wires

Pehle guess karo

Tum ek cell se do cells par jaate ho. Tumhe ammeter mein kya dikhne ki ummeed hai?

Lagbhag aadha currentLagbhag utna hi currentLagbhag double currentLagbhag chaar guna current

Ek cell, switch, ammeter aur coil ko ek single loop mein jodo. Voltmeter ko coil ke dono ends ke across jodo.
Switch band karo, dono meters jaldi se padho, phir switch khol do taaki coil garam na ho.
Holder mein do, teen aur chaar cells ke saath dohrao.
Har row ke liye V ÷ I nikaalo. V (upar ki taraf) ko I (aar-paar) ke against plot karo.

Table 11.2Tumhari readings

Cells	I (A)	V (V)	V ÷ I (Ω)
1
2
3
4

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 11.5

Readings ko plot karne par origin se guzarti seedhi line milti hai (Fig. 11.4). Double p.d. se double current, teen guna p.d. se teen guna current, aur ratio V/I kabhi nahi badalta.

3/4Wire B zyada steep line deta hai: double slope, yaani double resistance.

Fig. 11.4Fixed temperature par resistor ke liye V–I origin se guzarti seedhi line hai; uska slope R hai.NCERT Fig. 11.3 · p. 176

Ek graph jismein aar-paar amperes mein current I aur upar ki taraf volts mein potential difference V hai. Pehle step mein wire A ke chaar blue points 0.3, 0.6, 0.9 aur 1.2 amperes par 1.5, 3, 4.5 aur 6 volts ke saath hain. Doosre step mein origin se guzarti ek seedhi line unhe jodti hai, aur ek slope triangle 3 volts ko 0.6 amperes se divide karke dikhata hai, yaani R 5 ohms hai. Teesre step mein wire B ki orange dashed line, jiske points aadhe current par hain, zyada steep hai aur uspar 10 ohms likha hai. Chauthe step mein filament lamp ka dotted purple curve upar ki taraf mudta hai, kyunki garam hone par lamp ka resistance badhta hai.

1827 mein German physicist Georg Simon Ohm (1787–1854) ne metal wires ke liye yeh pattern report kiya. Ise ab Ohm's law (ओम का नियम) kehte hain.

LawOhm's law

Fixed temperature par rakhe metal conductor ke liye, usse guzarne wala current uske ends ke beech potential difference ke proportional hota hai.

Ohm's law

V = I R

(11.5)

jahan

$R$: conductor ka resistance, ohms (Ω) mein

Constant $R$ batata hai ki conductor se har ampere dhakelne mein kitne volts lagte hain. Ise resistance (प्रतिरोध) kehte hain: yeh jitna bada hoga, conductor charge ke flow ko utna zyada rokega. Ise ohms mein naapa jaata hai, aur Ω (Greek capital omega) likha jaata hai. Jis wire mein har volt theek ek ampere chalata hai uska resistance hai $R = 1 Ω,$ isliye $1 Ω = 1 V / 1 A .$ Ohm's law ko do kaam ke tareekon se ulta-palta kiya ja sakta hai:

R = \frac{V}{I}

(11.6)

I = \frac{V}{R}

(11.7)

Definition

Resistance

Conductor charge ke flow ko jitna rokta hai, $R = V / I .$ Iski SI unit ohm (Ω) hai.

ExampleJo bulb $6 V$ par $0.5 A$ leta hai uska resistance hai $R = 12 Ω .$

Stage light ka dimmer, toy car ka speed control aur purane fan ka regulator, sab ek hi kaam karte hain: supply wahi rehti hai par current badal jaata hai. Form (11.7) dikhata hai kaise. $V$ fixed ho to current aur resistance inversely proportional hote hain, isliye teen guna resistance ek-tihaai current hi guzarne deta hai. Jis part ka resistance adjust kiya ja sake use variable resistor kehte hain. Iska lab version, rheostat (धारा नियंत्रक), mein ek sliding contact hota hai jo tay karta hai ki circuit mein kitni resistance wire hai.

Neeche wala simulation activity ko dohraata hai. Ek fark dhyaan do: yeh current ko p.d. ke against upar ki taraf plot karta hai, isliye wahan zyada steep line chhote resistance ki hoti hai (uska slope $1 / R$ hai).

SimulationOhm's law bench

Voltage slider wala ek cell do resistance coils mein se ek se current chalata hai, 10 ohms ki coil A ya 20 ohms ki coil B, jismein ammeter series mein aur voltmeter coil ke across laga hai. Slider ki har setting current vs voltage graph par ek point jodti hai, aur ek readout V ko I se divide karke deta hai. Ek switch circuit ko kholta aur band karta hai.

Fig. 11.5Activity 11.1 ki copy karne ke liye Cell slider ko

1.5 V

ke steps mein chalao, phir coil B par switch karo.NCERT Fig. 11.2 · p. 175

Try karo

Coil A ke liye cell ko 1.5, 3, 4.5 aur 6 V par set karo. Kya V ÷ I readout badalta hai?
Coil B par switch karke dohrao. Is graph par kaun si line zyada steep hai, aur iska matlab chhota resistance kyun hai?
Switch kholo. Meters kya padhte hain?

Kya ek hi source se circuit ka har part same current guzaarta hai? Agli activity kai parts ko compare karti hai.

Activity 11.2NCERT Activity 11.2 · p. 176

Same cells, alag parts

Tumhe chahiye

holder mein do $1.5 V$ cells
ek ammeter (0–3 A), ek switch aur do crocodile-clip leads
test karne ke liye: pencil lead, rubber band, torch bulb aur 10 cm patli nichrome wire

Pehle guess karo

Tumhe kis test item se sabse kam current guzarne ki ummeed hai?

nichrome wiretorch bulbpencil leadrubber band

Cells, switch aur ammeter ka ek loop banao, do crocodile clips ke beech ek gap chhodkar.
Ek test item ko gap ke across clip karo, switch band karo, ammeter padho aur turant switch khol do.
Har item ke liye dohrao, phir unhe sabse zyada current se sabse kam current tak rank karo.

Table 11.3Tumhari readings

Gap mein item	Current (A)	Rank
Nichrome wire
Torch bulb
Pencil lead
Rubber band

Tumhari entries apne aap save hoti hain.

Parts alag kyun hote hain? Metal ke andar free electrons aaraam se nahi fisal sakte: woh vibrate karte atoms se baar-baar takraate hain, jisse unka drift dheema ho jaata hai. Koi part drift ko kitna dheema karta hai, wahi uska resistance hai. Rubber band ek chhor par hai: ek insulator, jiska resistance itna zyada hai ki koi current naapa hi nahi ja sakta. Doosre chhor par copper jaisa good conductor hai, jiska resistance uske size ke hisaab se bahut chhota hai. Beech mein woh parts hain jo jaan-boojhkar ek known, kaam ke resistance ke saath banaye jaate hain, jaise nichrome wire aur tumhari use ki hui coil: yeh resistors hain.

Quick check

3 mein se 1

Ek geyser element

220 V

supply se

10 A

leta hai. Uska resistance kya hai?

Answer dekho

Answer 22 Ω

R = V / I = 220 / 10 = 22 Ω .

11.5 Conductor ka Resistance kin Factors par Depend karta hai

Electrician air conditioner ke liye moti wire use karta hai lekin bedside lamp ke liye patli wire, aur heater coils kabhi copper ki nahi banti. Yeh choices is baat par depend karti hain ki wire ka resistance uske size aur uske material ke saath kaise badalta hai.

Activity 11.3NCERT Activity 11.3 · p. 177

Length, motaai aur metal

Tumhe chahiye

ek $1.5 V$ cell, ek ammeter (0–5 A) aur ek switch
same motaai ki constantan wire ke teen 50 cm tukde
usi motaai ki 50 cm copper wire
ek metre rule aur do crocodile clips

Pehle guess karo

Tum clips ke beech constantan wire ki length double karte ho. Current ka kya hota hai?

Double ho jaata haiAadha ho jaata haiWahi rehta haiZero ho jaata hai

Constantan ka ek 50 cm tukda metre rule ke saath circuit mein clip karo. Switch thodi der band karo aur current note karo.
Do tukdon ko end to end jodkar 100 cm banao aur uski jagah use clip karo.
Ab do 50 cm tukdon ko usi clips ke beech aas-paas (side by side) clip karo, taaki current ko guzarne ke liye double cross-section mile.
Aakhir mein 50 cm copper wire clip karo. Switch sirf ek second ke liye band karo: current bada hota hai.

Table 11.4Tumhari readings

Clips ke beech wire	Current (A)
Constantan, 50 cm
Constantan, 100 cm
Do 50 cm tukde side by side
Copper, 50 cm

Tumhari entries apne aap save hoti hain.

3/4Double area: aadha resistance, double current.

Fig. 11.6Har baar same cell: length, area aur material, teeno current badalte hain.NCERT Fig. 11.5 · p. 178

Do clips ke beech pakdi ek wire ek horizontal bar ke upar bani hai, jiski length same cell se aane wala current dikhati hai. Pehle step mein length l ki constantan wire 0.40 amperes deti hai. Doosre step mein wire double lambi hai aur bar aadha hokar 0.20 amperes ho jaata hai. Teesre step mein wire moti hai, double area ke saath, aur bar double hokar 0.80 amperes ho jaata hai. Chauthe step mein wire original size ki copper ki hai, aur bar scale se bahar chala jaata hai, uspar far bigger likha hai.

Uniform metal wires par dhyaan se kiye gaye measurements do rules dete hain. Resistance length ke saath-saath badhta hai, $R \propto l,$ aur cross-section badhne par ghatta hai, $R \propto 1 / A .$ Lambi wire electrons ke liye lamba obstacle course hai; moti wire aas-paas zyada lanes deti hai. Dono ko saath rakhne par milta hai

R = ρ \frac{l}{A}

(11.10)

jahan

$ρ$: material ki resistivity, ohm metres (Ω m) mein
$l$: conductor ki length, metres mein
$A$: area of cross-section, square metres mein

Constant $ρ$ (Greek rho) material ki resistivity (प्रतिरोधकता) hai. Resistance ek khaas wire ka hota hai, lekin resistivity substance ki hoti hai. Copper wire ko aadha kaato to uska resistance aadha ho jaata hai, phir bhi uski resistivity wahi rehti hai. Rearrange karne par milta hai $ρ = R A / l,$ isliye resistivity ki unit $Ω \times m^{2} / m,$ yaani ohm metre (Ω m) hai.

Definition

Resistivity

Us material ke unit length aur unit area of cross-section wale conductor ka resistance, $ρ = R A / l .$ Yeh material aur uske temperature par depend karti hai, wire ki shape par nahi. SI unit: ohm metre (Ω m).

Table 11.520 °C par kuch metals, alloys aur insulators ki resistivity (problems ke liye handbook values; inhe yaad karne ki zaroorat nahi)NCERT Table 11.2 · p. 179

Material	Resistivity (Ω m)
Silver	$1.59 \times 10^{- 8}$
Copper	$1.68 \times 10^{- 8}$
Gold	$2.44 \times 10^{- 8}$
Aluminium	$2.65 \times 10^{- 8}$
Tungsten	$5.60 \times 10^{- 8}$
Nickel	$6.99 \times 10^{- 8}$
Iron	$9.7 \times 10^{- 8}$
Manganin (Cu, Mn, Ni ka alloy)	$4.8 \times 10^{- 7}$
Constantan (Cu, Ni ka alloy)	$4.9 \times 10^{- 7}$
Nichrome (Ni, Cr, Fe ka alloy)	$1.10 \times 10^{- 6}$
Glass (insulator)	$10^{11}$ se $10^{15}$
Hard rubber (insulator)	lagbhag $10^{13}$
Fused quartz (insulator)	lagbhag $7.5 \times 10^{17}$

Table 11.5 mein values ka phailaav dekho. Metals aur alloys ki range hai lagbhag $10^{- 8}$ se $10^{- 6} Ω m;$ glass, rubber aur quartz jaise insulators bahut upar hain, lagbhag $10^{11}$ se shuru hokar, aur kuch to paar kar jaate hain $10^{17} Ω m .$ Resistivity temperature ke saath bhi badalti hai: metals ke liye yeh garam hone par badhti hai.

Yeh numbers batate hain ki har kaam ke liye alag material kyun chuna jaata hai:

Wires aur power lines ko kam se kam energy waste karni chahiye, isliye unmein sabse kam resistivity wale affordable metals use hote hain: copper, aur aluminium, jo halka bhi hai.
Irons, toasters aur room heaters ke heating elements ko chhoti coil mein bada resistance chahiye aur unhe ghanton tak red-hot rehna padta hai. Nichrome jaise alloys inke liye sahi hain: alloy ki resistivity aam taur par uske metals se kahin zyada hoti hai, aur nichrome aasaani se oxidise (burn) nahi hota.
Bulb filaments ko bina pighle white-hot chamakna padta hai, isliye woh tungsten se bante hain. Tungsten tab bhi tika rehta hai jab temperature paar kar jaata hai $3000^{\circ} C .$

Example 11.3NCERT Example 11.3 · p. 179

Soldering iron ya immersion rod?

$220 V$ supply par in dono mein se har ek kitna current leta hai: (a) soldering iron jiske element ka resistance hai $880 Ω,$ aur (b) immersion water-heater rod jiska resistance $44 Ω$ hai?

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

Example 11.4NCERT Example 11.4 · p. 180

Voltage badhana

Toy car ka ek resistor $0.30 A$ leta hai jab uske across p.d. hai $4.5 V .$ Agar p.d. $7.5 V$ tak badha diya jaaye to woh kitna current lega? Maan lo uska temperature nahi badalta.

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

Example 11.5NCERT Example 11.5 · p. 180

Yeh kaun sa alloy hai?

Rohan ek purane hair dryer se heating wire nikaalta hai. Iski $2.0 m$ length ka diameter $0.40 m m$ hai aur 20 °C par resistance $17.5 Ω$ hai. Wire ki resistivity nikaalo aur Table 11.5 se material ka naam batao.

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

Example 11.6NCERT Example 11.6 · p. 180

Lambi aur patli

Ek alloy ki wire ka resistance hai $2 Ω .$ Usi alloy ki doosri wire teen guna lambi hai aur uska area of cross-section aadha hai. Uska resistance kya hai?

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

Practice · 5 questions

Practice 11.3 (NCERT in-text questions, p. 181): answers ke saath

Short answerEasyNCERT in-text Questions p. 181, Q1
Fixed temperature par metal wire ka resistance tay karne wali teen cheezon ke naam batao, aur batao ki har ek use kaise affect karti hai.
Solution dikhao
AnswerLength (R ∝ l), area of cross-section (R ∝ 1/A) aur material (uski resistivity ρ ke through).
Dekho $R = ρ l / A :$ lambi wire ka resistance zyada hota hai ( $R \propto l$ ); moti wire, jiska area of cross-section bada hai, ka kam ( $R \propto 1 / A$ ); aur material apni resistivity $ρ$ ke through matter karta hai (nichrome ka copper se kahin zyada). Temperature bhi badal deta hai $ρ .$
Quick answerEasyNCERT in-text Questions p. 181, Q2
Barabar length ki do copper wires, ek moti aur ek patli, baari-baari se same cell se jodi jaati hain. Kismein current zyada aasaani se behta hai?
Patli wire mein, kyunki raaste mein kam metal aata haiMoti wire mein, kyunki uska area of cross-section bada haiDono mein barabar, kyunki dono same metal ki hainPatli wire mein, kyunki usmein electrons tez chalte hain
Answer dekho
Answer (B) Moti wire mein, kyunki uska area of cross-section bada hai
$R \propto 1 / A :$ moti wire ka resistance chhota hai, isliye same cell se usmein zyada current behta hai.
Quick answerEasyNCERT in-text Questions p. 181, Q3
Ek resistor mein current hai $0.6 A .$ Uske across p.d. pehle ki value ka ek-tihaai kar diya jaata hai; uska resistance nahi badalta. Naya current kya hai?
A
Answer dekho
Answer 0.2 A
$I = V / R,$ isliye ek-tihaai p.d. se ek-tihaai current milta hai: $0.6 / 3 = 0.2 A .$
Short answerEasyNCERT in-text Questions p. 181, Q4
Electric room heater ki coil copper ki nahi, nichrome ki kyun banti hai?
Solution dikhao
AnswerNichrome ki resistivity zyada hai, isliye chhoti coil bhi khoob garam hoti hai, aur yeh red heat par oxidise nahi hota ya pighalta nahi; copper ki resistivity kam hai aur garam hone par yeh oxidise ho jaata hai.
Heating coil ko chhoti jagah mein bada resistance chahiye, isliye use zyada resistivity wala material chahiye: nichrome ki resistivity copper se lagbhag 65 guna hai. Use ghanton tak red-hot bhi rehna padta hai: nichrome aasaani se oxidise (burn) nahi hota aur uska melting point high hai, jabki copper oxidise ho jaata aur khatarnaak had tak bada current guzaarta.
5
Quick answerEasyNCERT in-text Questions p. 181, Q5
Inke jawab dene ke liye Table 11.5 use karo.
1. Behtar conductor kaun hai, nickel ya iron?
  nickeliron
  Answer dekho
  Answer (A) nickel
  Nickel ki resistivity ( $6.99 \times 10^{- 8} Ω m$ ) iron ki resistivity ( $9.7 \times 10^{- 8} Ω m$ ) se kam hai.
2. Table mein kaun sa alloy sabse kharaab conductor hai?
  manganinconstantannichrome
  Answer dekho
  Answer (C) nichrome
  Diye gaye alloys mein nichrome ki resistivity sabse zyada hai, $1.10 \times 10^{- 6} Ω m .$

Quick check

3 mein se 1

Ek wire ki jagah usi metal ki aisi wire lagayi jaati hai jo double lambi hai aur jiska area of cross-section double hai. Uska resistance:

double ho jaata haiaadha ho jaata haiwahi rehta haichaar guna ho jaata hai

Answer dekho

Answer (C) wahi rehta hai

R = ρ l / A :

l

double karne se double hota hai

R,

A

double karne se aadha, isliye dono cancel ho jaate hain. Chaar guna chunne ka matlab hai ki dono changes ko galti se aisa maana gaya ki woh badhate hain

R .

11.6 Resistors ke System ka Resistance

Asli gadgets mein shayad hi kabhi sirf ek resistor hota hai. Radio, phone charger ya ghar ki wiring kai parts ko saath jodti hai, aur aisa karne ke do basic tareeke hain. Series (श्रेणीक्रम) mein resistors ek chain mein end to end jude hote hain, isliye charge ko baari-baari se har ek se guzarna padta hai. Parallel (पार्श्वक्रम) mein har resistor same do points ke across juda hota hai, isliye charge ke paas raaston ka choice hota hai. Dono cases mein Ohm's law ki madad se hum poore group ki jagah ek single equivalent resistance rakh sakte hain.

11.6.1 Series mein Resistors

Kya ek resistor kuch current 'kha jaata' hai, aur agle ke liye kam chhodta hai? Series chain ke saath-saath ammeter ko khiskaakar iska jawab milta hai.

Activity 11.4NCERT Activity 11.4 · p. 182

Kya chain mein kahin current khatam hota hai?

Tumhe chahiye

holder mein teen $1.5 V$ cells ( $4.5 V$ )
$5 Ω,$ $10 Ω$ aur $15 Ω$ ke resistors
ek ammeter (0–1 A), ek switch aur connecting wires

Pehle guess karo

Tum ammeter ko pehle resistor se theek pehle padhte ho aur phir aakhri resistor ke theek baad. Doosri reading hogi:

kam, kyunki har resistor kuch current kha jaata haipehli jitni hizyada, kyunki charges tez ho jaate hainzero

Teeno resistors ko end to end jodo, phir cells, switch aur ammeter se loop poora karo.
Switch band karo, ammeter padho aur switch khol do.
Ammeter ko pehle aur doosre resistor ke beech, phir doosre aur teesre ke beech, phir teesre ke baad lagao. Har baar padho.

Table 11.6Tumhari readings

Ammeter ki position	Current (A)
5 Ω se pehle
5 Ω aur 10 Ω ke beech
10 Ω aur 15 Ω ke beech
15 Ω ke baad

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 11.8

Activity 11.5NCERT Activity 11.5 · p. 183

Voltage ka bantwaara

Tumhe chahiye

Activity 11.4 wala series circuit
ek voltmeter (0–5 V)

Pehle guess karo

Teeno resistors ke across p.d.s poori chain ke across p.d. se kaise compare honge?

Har ek poore ke barabar haiSabse bada poore ke barabar haiSab milkar poore ke barabar hainSab zero hain

Voltmeter ko poori chain ke across jodo, $5 Ω$ resistor ki shuruaat se $15 Ω$ wale ke end tak. Switch band karo aur padho $V .$ Cells ke across p.d. bhi padho.
Voltmeter ko sirf $5 Ω$ resistor ke across lagao, phir $10 Ω,$ phir $15 Ω .$ In readings ke naam rakho $V_{1},$ $V_{2}$ aur $V_{3} .$
$V_{1},$ $V_{2}$ aur $V_{3}$ ko jodo aur total ko compare karo is reading se: $V .$

Table 11.7Tumhari readings

Voltmeter kiske across	Reading (V)
Poori chain
5 Ω
10 Ω
15 Ω

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 11.8

2/3 $0.75 + 1.5 + 2.25 = 4.5 V :$ parts ke across p.d.s milkar total banate hain.

Fig. 11.7Series mein: ek current, bata hua voltage, aur resistances jud jaate hain.NCERT Fig. 11.8 · p. 183

Ek rectangular circuit ki neeche wali wire mein 4.5 volt ka cell hai aur upar wali wire par 5, 10 aur 15 ohms ke teen zigzag resistors hain. Pehle step mein loop ki kai jagahon par red labels sab 0.15 amperes padhte hain, yaani har jagah same current. Doosre step mein resistors ke upar brackets 0.75, 1.5 aur 2.25 volts dikhate hain, aur teeno ke upar ek bada bracket total 4.5 volts dikhata hai. Teesre step mein teeno resistors ki jagah 30 ohms wala ek lamba resistor aa jaata hai, aur current ab bhi 0.15 amperes hai.

DerivationSeries mein equivalent resistance

Maan lo same current $I$ in teeno se behta hai: $R_{1},$ $R_{2}$ aur $R_{3},$ aur poori chain ke across p.d. hai $V .$ Activity ne dikhaya ki alag-alag p.d.s jud jaate hain:

V = V_{1} + V_{2} + V_{3}

(11.11)

Ohm's law us single resistor $R_{s}$ par laagu hota hai jo chain ki jagah le sakta hai, aur jismein same rehte hain $V$ aur $I :$

V = I R_{s}

(11.12)

Yeh har resistor par alag se bhi laagu hota hai:

\begin{array}{c} V_{1} = I R_{1} \\ V_{2} = I R_{2} \\ V_{3} = I R_{3} \end{array}

(11.13)

In sabko (11.11) mein rakho aur har term ko divide karo isse, $I :$

\begin{aligned} I R_{s} & = I R_{1} + I R_{2} \\ + I R_{3} \end{aligned}

R_{s} = R_{1} + R_{2} + R_{3}

(11.14)

Series mein resistors ka equivalent resistance unke resistances ka sum hota hai, isliye yeh hamesha unmein sabse bade se bhi bada hota hai.

Example 11.7NCERT Example 11.7 · p. 184

Dimming resistor ke saath ek lamp

Ek $15 Ω$ lamp aur ek $5 Ω$ resistor $9 V$ battery se series mein jode gaye hain. Nikaalo (a) total resistance, (b) current, aur (c) lamp ke across aur resistor ke across p.d.

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

SimulationSeries circuit explorer

4.5 volt ka cell series mein jude 5, 10 aur 15 ohms ke teen resistors se current chalata hai, saath mein ek ammeter aur ek voltmeter hai. Readouts equivalent resistance, current aur voltmeter reading dikhate hain. Learner chunta hai ki voltmeter kahan juda ho, ek resistor chunkar slider se uski value badalta hai, cell voltage badalta hai aur switch kholta ya band karta hai; chalte markers current dikhate hain.

Fig. 11.8Activities 11.4 aur 11.5 wali chain. Voltmeter ko ek part se doosre par le jao aur resistances badlo.

Try karo

Voltmeter ko baari-baari se har resistor ke across lagao. Kya teeno readings milkar cell ka p.d. banati hain?
R2 ko bahut bada kar do. Current ka kya hota hai, aur ab sabse zyada voltage kaun sa resistor leta hai?
Switch kholo. Loop mein har jagah ek saath kya hota hai?

Practice · 2 questions

Practice 11.4 (NCERT in-text questions, p. 185): answers ke saath

DiagramEasyNCERT in-text Questions p. 185, Q1
Ek battery (chaar $1.5 V$ cells), ek switch, aur resistors $2 Ω,$ $4 Ω$ aur $6 Ω,$ sab series mein jude hue: inka circuit diagram banao.
Solution dikhao
AnswerEk closed loop: chaar cell symbols + se − jude hue, phir switch, phir 2 Ω, 4 Ω aur 6 Ω label wale teen zigzags.
Ek rectangular loop banao. Ek side par chaar cell symbols ek line mein lagao, har lambi plate (+) agle cell ki chhoti plate (−) se judi ho; battery par likho $6 V .$ Baaki sides par ek switch aur teen zigzag resistors lagao, label $2 Ω,$ $4 Ω$ aur $6 Ω,$ ek ke baad ek. Koi branch nahi hai, isliye junction dots ki zaroorat nahi.
2
NumericalMediumNCERT in-text Questions p. 185, Q2
Q1 ke circuit mein do meters jodo: ek jo chain ka current padhe, aur ek jo $6 Ω$ resistor ke dono ends ke beech p.d. padhe.
1. Dono meters kahan lagenge?
  Solution dikhao
  AnswerAmmeter loop mein kahin bhi, series mein; voltmeter 6 Ω resistor ke dono ends ke across, parallel mein.
  Ammeter loop mein kahin bhi series mein lagaya jaata hai (current har jagah same hai). Voltmeter parallel mein joda jaata hai, ek lead $6 Ω$ resistor ke har end par.
2. Ammeter kya padhta hai?
  A
  Answer dekho
  Answer 0.5 A
  $R_{s} = 2 + 4 + 6 = 12 Ω$ aur $V = 4 \times 1.5 = 6 V,$ isliye $I = 6 / 12 = 0.5 A .$
3. Voltmeter kya padhta hai?
  V
  Answer dekho
  Answer 3 V
  $V = I R = 0.5 \times 6 = 3 V .$

Quick check

2 mein se 1

3 Ω,

4 Ω

aur

5 Ω

ke resistors

6 V

battery ke across series mein jode gaye hain. Kitna current behta hai?

Answer dekho

Answer 0.5 A

R_{s} = 3 + 4 + 5 = 12 Ω,

isliye

I = 6 / 12 = 0.5 A .

11.6.2 Parallel mein Resistors

Ab resistors ko aas-paas jodo, har ek ko same do points X aur Y ke beech. Har resistor ko apni branch milti hai. X par aane wala charge branches mein bant jaata hai aur Y par phir mil jaata hai.

Activity 11.6NCERT Activity 11.6 · p. 185

Current ka bantwaara

Tumhe chahiye

do $1.5 V$ cells ( $3 V$ )
$10 Ω,$ $15 Ω$ aur $30 Ω$ ke resistors
ek ammeter (0–1 A), ek voltmeter (0–5 V) aur ek switch

Pehle guess karo

Branch currents cells se aane wale current se kaise compare honge?

Branch currents milkar main current banate hainHar branch mein poora current behta hai

30 Ω

wali branch mein sabse zyada current behta haiSaare branch currents barabar hain

Teeno resistors ko do terminals X aur Y ke beech aas-paas jodo. X aur Y ko switch aur ammeter se hokar cells se jodo.
Switch band karo. Main current $I$ padho, aur voltmeter ko X aur Y ke across lagakar padho p.d. $V .$
Voltmeter ko baari-baari se har resistor ke across lagao. Kya uski reading badalti hai?
Ammeter ko baari-baari se har branch mein lagao aur padho $I_{1},$ $I_{2}$ aur $I_{3} .$

Table 11.8Tumhari readings

Quantity	Reading
Main current I (A)
X aur Y ke across p.d. (V)
10 Ω mein current (A)
15 Ω mein current (A)
30 Ω mein current (A)

Tumhari entries apne aap save hoti hain.

Abhi nahi kar sakte? Simulation try karo · Fig. 11.12

2/3 $0.3 + 0.2 + 0.1 = 0.6 A :$ sabse chhote resistor mein sabse zyada current behta hai.

Fig. 11.9Parallel mein: ek voltage, currents jud jaate hain, aur equivalent resistance har branch se chhota hota hai.NCERT Fig. 11.10 · p. 185

10, 15 aur 30 ohms ke teen zigzag resistors left aur right wire ke beech ek ladder ke teen dandon par lage hain, aur neeche wali wire mein 3 volt ka cell hai. Pehle step mein har dande par 3 volts likha hai, yaani har resistor ke across same p.d. Doosre step mein dandon par red arrows 0.3, 0.2 aur 0.1 amperes ke currents dikhate hain, aur left wire par ek arrow 0.6 amperes ka main current dikhata hai. Teesre step mein teeno dandon ki jagah ek single 5 ohm resistor aa jaata hai jo utna hi 0.6 amperes leta hai.

DerivationParallel mein equivalent resistance

Activity ne dikhaya ki main current branch currents ka sum hai:

I = I_{1} + I_{2} + I_{3}

(11.15)

Har branch ka p.d. same hai, $V .$ Ohm's law us single resistor $R_{p}$ par lagao jo utna hi main current lega:

I = \frac{V}{R_{p}}

(11.16)

aur har branch par alag se:

\begin{array}{c} I_{1} = \frac{V}{R_{1}} \\ I_{2} = \frac{V}{R_{2}} \\ I_{3} = \frac{V}{R_{3}} \end{array}

(11.17)

Inhe (11.15) mein rakho:

\begin{aligned} \frac{V}{R_{p}} & = \frac{V}{R_{1}} + \frac{V}{R_{2}} \\ + \frac{V}{R_{3}} \end{aligned}

Har term ko divide karo isse, $V :$

\begin{aligned} \frac{1}{R_{p}} & = \frac{1}{R_{1}} + \frac{1}{R_{2}} \\ + \frac{1}{R_{3}} \end{aligned}

(11.18)

Shabdon mein: resistances ke reciprocals jodo, aur total ko ulta karo (reciprocal lo) to milta hai $R_{p} .$ Har extra branch charge ko ek aur raasta deti hai, isliye $R_{p}$ hamesha group ke sabse chhote resistor se bhi chhota hota hai. Sirf do resistors ke liye yeh ban jaata hai $R_{p} = R_{1} R_{2} / (R_{1} + R_{2}),$ yaani 'product over sum'.

Example 11.8NCERT Example 11.8 · p. 186

Ek battery par teen branches

$4 Ω,$ $6 Ω$ aur $12 Ω$ ke resistors $6 V$ battery ke across parallel mein jode gaye hain. (a) Har branch mein kitna current behta hai? (b) Battery kul kitna current deti hai? (c) Kaun sa single resistor teeno ki jagah le sakta hai?

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

Fig. 11.10Group 1 (6 Ω ∥ 3 Ω) group 2 (12 Ω ∥ 4 Ω) ke saath series mein.NCERT Fig. 11.12 · p. 187

Circuit diagram. Left mein group 1 hai, jismein 6 ohm aur 3 ohm ke resistors do junctions ke beech aas-paas lage hain. Right mein group 2 hai, jismein 12 ohm aur 4 ohm ke resistors aas-paas lage hain. Ek wire dono groups ko jodti hai, aur neeche wali wire mein 10 volt ki battery loop poora karti hai.

Example 11.9NCERT Example 11.9 · p. 187

Series mein do parallel groups

Fig. 11.10 mein parallel mein jude $6 Ω$ aur $3 Ω$ ke resistors (group 1) parallel mein jude $12 Ω$ aur $4 Ω$ ke resistors (group 2) ke saath series mein jude hain. $10 V$ ki battery circuit chalati hai. Nikaalo (a) total resistance aur (b) battery se liya gaya current.

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

Tumhare ghar mein kuch bhi series mein wired kyun nahi hai? Socho ki school corridor ke paanch lamps ek series chain mein jude hain. Unhe aapas mein baantna padta $220 V,$ isliye kisi ko bhi woh voltage nahi milta jiske liye woh bana hai. Agar ek lamp fuse ho jaata, to akela path toot jaata aur poora corridor andhera ho jaata (Fig. 11.11). Aur lamp kabhi heater ke saath ek chain mein nahi lag sakta, kyunki dono ko bahut alag currents chahiye jabki series chain dono se same current guzarne par majboor karti hai.

2/3Series row andheri ho jaati hai; parallel row jalti rehti hai.

Fig. 11.11Ek kharaab lamp poori series chain ko andhera kar deta hai, lekin parallel row ko nahi.NCERT p. 187

Paanch lamps ki do rows. Upar wali row series mein wired hai, paancho ek wire par supply ke saath ek single loop mein. Neeche wali row parallel mein wired hai, har lamp do supply wires ke beech apne dande par. Pehle step mein saare das lamps jalte hain. Doosre step mein har row ka beech wala lamp kharaab hai: series row ka har lamp andhera ho jaata hai, jabki parallel row ke baaki chaar lamps jalte rehte hain. Teesre step mein arrows dikhate hain ki current apne-apne path se har chalte parallel lamp tak ab bhi pahunch raha hai, jabki series loop ek point par toota hua hai.

Parallel mein har appliance seedha supply ke across hota hai, isliye har ek ko poora voltage milta hai aur woh sirf utna current leta hai jitna use chahiye. Har ek ka apna switch hota hai, aur ek ke kharaab hone par baaki chalte rehte hain. Parallel mein zyada appliances jodne se total resistance ghatta hai aur supply se liya gaya total current badhta hai. Isiliye ghar ki wiring mein current ki limits aur fuses zaroori hain.

SimulationParallel circuit explorer

3 volt ka cell parallel mein jude 10, 15 aur 30 ohms ke teen resistors se current chalata hai, main line mein ek ammeter hai aur ek voltmeter hai jise cell ke across ya resistors ke group ke across lagaya ja sakta hai. Readouts equivalent resistance, main current, voltmeter reading aur chune gaye resistor ka current dikhate hain; chalte markers current ko branches mein bantte dikhate hain. Sliders har resistance aur cell voltage badalte hain, aur ek switch circuit kholta hai.

Fig. 11.12Activity 11.6 wala circuit. Kisi resistor ka current padhne ke liye use tap karo, resistances badlo aur dekho current kaise bantta hai.

Try karo

R1, R2 aur R3 ko baari-baari se tap karo aur har ek ka current note karo. Kya teeno milkar ammeter reading ke barabar hain?
R3 ko bahut bada kar do, maan lo 1000 Ω. Kya main current zyada badalta hai? Equivalent resistance ko sirf R1 aur R2 se compare karo.
R1 ko chhota karo aur main current dekho. Ab kaun si branch sabse zyada current le jaati hai?

Practice · 5 questions

Practice 11.5 (NCERT in-text questions, p. 188): answers ke saath

1
Short answerMediumNCERT in-text Questions p. 188, Q1
Exact calculation ke bina, inhe parallel mein jodne par equivalent resistance ka andaaza lagao.
1. $5 Ω$ aur $10^{5} Ω$
  Solution dikhao
  Answer5 Ω se zara sa kam (lagbhag 4.9998 Ω).
  $R_{p}$ ko kam hona hi chahiye isse: $5 Ω,$ lekin bahut badi $10^{5} Ω$ wali branch lagbhag koi current nahi le jaati, isliye $R_{p}$ $5 Ω$ se bas baal barabar kam hai (lagbhag $4.9998 Ω$ ).
2. $5 Ω,$ $500 Ω$ aur $10^{5} Ω$
  Solution dikhao
  Answer5 Ω se thoda kam (lagbhag 4.95 Ω).
  Yeh bhi kam hai is value se: $5 Ω;$ $500 Ω$ wali branch $5 Ω$ wali ke current ka lagbhag 1 % leti hai, isliye $R_{p}$ lagbhag 1 % kam hai: $\approx 4.95 Ω .$
NumericalMediumNCERT in-text Questions p. 188, Q2
Ek tube light jiska resistance hai $880 Ω,$ $440 Ω$ ka fan aur $220 Ω$ ka TV $220 V$ supply par parallel mein chalte hain (har ek ko simple resistance maano). Usi supply par ek geyser akele utna current leta hai jitna teeno ka total hai. Woh current aur geyser ka resistance nikaalo.
Solution dikhao
Answer1.75 A; lagbhag 126 Ω.
Har ek ke across $220 V$ hai: $I = 0.25 + 0.5 + 1.0 = 1.75 A .$
Geyser ka resistance $R = V / I = 220 / 1.75 \approx 126 Ω$ hai (teeno ka equivalent resistance).
Short answerEasyNCERT in-text Questions p. 188, Q3
Ghar ke appliances ko series ke bajaay parallel mein jodne ke teen fayde batao.
Solution dikhao
AnswerHar ek ko poora mains voltage milta hai; har ek ko alag se on ya off kiya ja sakta hai; ek fault baaki ko nahi rokta (aur har ek sirf utna current leta hai jitna use chahiye).
(i) Har appliance ke across poora supply voltage hota hai, isliye har ek waise hi kaam karta hai jaise design kiya gaya. (ii) Har ek ka apna switch ho sakta hai aur use alag se use kiya ja sakta hai. (iii) Agar ek kharaab ho jaaye, to baaki chalte rehte hain. Saath hi, har ek utna hi current leta hai jitna use chahiye; series mein sabko zabardasti same current le jaana padta.
4
DiagramMediumNCERT in-text Questions p. 188, Q4
Tumhare paas yeh resistors hain: $3 Ω,$ $6 Ω$ aur $9 Ω .$ Dikhao ki teeno ko kaise jodein taaki total ho:
1. $11 Ω$
  Solution dikhao
  Answer9 Ω, (3 Ω ∥ 6 Ω) ke saath series mein.
  $3 Ω ∥ 6 Ω = 18 / 9 = 2 Ω;$ phir $9 + 2 = 11 Ω .$
2. $2.5 Ω$
  Solution dikhao
  Answer(6 Ω + 9 Ω), 3 Ω ke saath parallel mein.
  $6 + 9 = 15 Ω$ series mein; phir $15 ∥ 3 = 45 / 18 = 2.5 Ω .$
5
NumericalMediumNCERT in-text Questions p. 188, Q5
Chaar coils ke resistances hain $5 Ω,$ $10 Ω,$ $15 Ω$ aur $30 Ω .$ Chaaron ko use karke:
1. Sabse bada possible equivalent resistance kya hai?
  Ω
  Answer dekho
  Answer 60 Ω
  Sab series mein: $5 + 10 + 15 + 30 = 60 Ω .$
2. Sabse chhota kya hai?
  Ω
  Answer dekho
  Answer 2.5 Ω
  Sab parallel mein: $1 / R_{p} = (6 + 3 + 2 + 1) / 30 = 12 / 30,$ isliye $R_{p} = 2.5 Ω .$

11.7 Electric Current ka Heating Effect

Ek ghante use ke baad phone charger chhookar dekho: woh garam hai. Mixer-grinder apni zyadatar energy blades ko ghumane mein lagata hai, phir bhi uski body bhi garam ho jaati hai. Geyser element sirf ek resistor hai, isliye use milne wala har joule heat ban jaata hai. Jab bhi resistance wale conductor se current behta hai, source ki kuch ya saari energy heat mein badal jaati hai. Ise electric current ka heating effect kehte hain.

DerivationCurrent kitni heat banata hai?

Maan lo ek steady current $I$ time $t$ tak ek resistor $R$ se behta hai jiske across p.d. $V$ hai. Is time mein charge $Q = I t$ guzarta hai. Charge $Q$ ko p.d. $V$ se le jaane mein laga work hai $V Q,$ isliye source is rate se energy deta hai

P = \frac{V Q}{t} = V I

(11.19)

Time $t$ mein source jo energy deta hai woh hai $V I t .$ Purely resistive circuit mein yeh saari resistor mein heat $H$ ban jaati hai:

H = V I t

(11.20)

Ise current ke terms mein likhne ke liye Ohm's law, $V = I R,$ use karo:

H = I^{2} R t

(11.21)

LawJoule's law of heating

Resistor mein bani heat hai $H = I^{2} R t :$ kisi diye gaye resistor ke liye yeh current ke square ke saath badhti hai; diye gaye current ke liye yeh resistance ke proportional hai; aur yeh current behne ke time ke proportional hai.

Joule's law mein square par dhyaan do. Same resistor se current double karne par chaar guna heat banti hai, jabki resistance ya time double karne se sirf double hoti hai (Fig. 11.13). Known supply voltage wale appliance ke liye pehle current nikaalo is se, $I = V / R,$ phir use karo $H = I^{2} R t;$ ya seedha $H = V^{2} t / R$ use karo.

4/4Double current: $(2 I)^{2} = 4 I^{2},$ isliye 4H.

Fig. 11.13

H = I^{2} R t :

t

R

double karne se heat double hoti hai,

I

double karne se chaar guna.NCERT p. 189

Ek baseline par heat blocks ke chaar columns khade hain. Pehle column mein, current I, resistance R aur time t ke liye, H label wala ek block hai. Doosre step mein double time ka column aata hai jismein do blocks hain, 2H. Teesre step mein double resistance ke column mein bhi do blocks hain, 2H. Chauthe step mein double current ke column mein chaar blocks hain, 4H, jo sabse ooncha hai.

Example 11.10NCERT Example 11.10 · p. 189

Hair dryer ki hot aur warm settings

$220 V$ supply par ek hair dryer hot setting par $1100 W$ aur warm setting par $550 W$ use karta hai. Har setting mein heating element ka current aur resistance nikaalo.

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

Example 11.11NCERT Example 11.11 · p. 189

Heat se voltage nikaalna

Ek $10 Ω$ resistor $360 J$ heat banata hai, aur ismein time lagta hai $4 s .$ Uske across p.d. kya hai?

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

Quick check

2 mein se 1

Heater element ka current double kar diya jaata hai jabki uska resistance wahi rehta hai. Har second banne wali heat ho jaati hai:

doublechaar gunaaadhiutni hi

Answer dekho

Answer (B) chaar guna

H = I^{2} R t,

isliye

I

double karne se heat itne guna ho jaati hai:

2^{2} = 4.

'Double' wala jawab square bhool jaata hai.

11.7.1 Electric Current ke Heating Effect ke Practical Applications

Current se bani heat kabhi-kabhi pareshaani hoti hai. Yeh power lines aur chargers mein energy waste karti hai, aur extra garmi components ka behaviour badal sakti hai ya unhe kharaab bhi kar sakti hai. Lekin hum ise roz jaan-boojhkar use bhi karte hain: irons, toasters, ovens, kettles, geysers aur room heaters mein high-resistance element electrical energy ko heat mein badalta hai.

Purane style ka filament bulb asal mein ek heater hai jo saath mein chamak bhi jaata hai: iski use ki gayi energy ka 90 % se zyada roshni nahi, heat bankar nikalta hai. White chamakne ke liye iske filament ko paar karna padta hai $2500^{\circ} C,$ isliye yeh tungsten se banta hai, jo sirf lagbhag is temperature par pighalta hai: $3400^{\circ} C .$ Do design tricks madad karti hain. Patle supports heat ko base mein leak hone se rokte hain, aur glass mein hawa ki jagah unreactive nitrogen aur argon bhari jaati hai, taaki white-hot metal jal na sake.

Joule heating hume safe bhi rakhti hai. Maan lo bahut saare heaters ek saath on kar diye gaye, ya kharaab cord ki do wires aapas mein chhoo gayi. Current normal se bahut upar chala jaata hai, aur ghar ki wiring khud overheat hokar aag laga sakti hai. Fuse (फ़्यूज़) isse bachata hai. Yeh jaan-boojhkar banaya gaya kamzor link hai: kaafi kam melting point wali ek chhoti, patli wire, jo series mein lagi hoti hai taaki poora current usse guzre. Kyunki heating current ke square ke saath badhti hai, yaani $I^{2},$ isliye current rating paar karte hi fuse wire kuch hi pal mein pighal jaati hai, aur kuch aur khatarnaak had tak garam hone se pehle circuit toot jaata hai (Fig. 11.14). Wire aam taur par metal end caps wale porcelain ya glass cartridge mein hoti hai.

2/3Iron, heater aur kettle milkar lagbhag lete hain $20 A :$ fuse wire tezi se garam hoti hai.

Fig. 11.14Bahut bada current nuksaan kare usse pehle fuse pighalkar circuit khol deta hai.NCERT p. 190

Live wire upar aur neutral wire neeche chalti hai. 10 A marked fuse cartridge live wire mein teen appliances se pehle laga hai, yaani iron, heater aur kettle, jo wires ke beech parallel mein jude hain. Pehle step mein sirf iron on hai aur current 4.5 amperes hai, jo safe hai. Doosre step mein heater aur kettle bhi on ho jaate hain, current 20 amperes tak pahunchta hai aur fuse wire garam hokar chamakti hai. Teesre step mein fuse wire pighal chuki hai aur ek gap reh gaya hai, isliye current zero hai aur teeno appliances band hain.

Gharelu fuses tay currents ke liye bante hain, jaise 2 A, 5 A, 10 A ya 15 A. Appliance normally jitna current leta hai, usse thodi zyada rating chuno. Jaise, $220 V$ par $2 k W$ room heater leta hai $I = P / V = 2000 / 220 \approx 9.1 A,$ isliye use $10 A$ fuse chahiye. $5 A$ fuse heater on karte hi pighal jaata, jabki $15 A$ wala kam protection deta. Kai modern ghar iski jagah miniature circuit breaker (MCB) use karte hain: ek switch jo overload par trip ho jaata hai aur bas reset kiya ja sakta hai.

Practice · 3 questions

Practice 11.6 (NCERT in-text questions, p. 190): answers ke saath

Short answerMediumNCERT in-text Questions p. 190, Q1
Electric kettle ki flexible cable aur uske heating element mein same current behta hai. Element bahut garam kyun ho jaata hai jabki cable lagbhag thandi rehti hai?
Solution dikhao
AnswerHeat H = I²Rt; same current ke saath high-resistance element low-resistance copper cable se kahin zyada heat banata hai.
Dono mein same time tak same current behta hai, isliye $H = I^{2} R t$ se har ek mein bani heat uske resistance ke proportional hai. Element nichrome ka hai jiska resistance bada hai; cable moti copper ki hai jiska resistance bahut chhota hai, isliye lagbhag saari heat element mein banti hai.
Quick answerMediumNCERT in-text Questions p. 190, Q2
2 ghante mein ek heater se $18 000 C$ charge guzarta hai. Heater ke across p.d. hai $24 V .$ Kitni heat banti hai?
J
Answer dekho
Answer 432000 J
$H = V Q = 24 \times 18 000 = 4.32 \times 10^{5} J .$ Charge pata ho to time ki zaroorat nahi.
Quick answerEasyNCERT in-text Questions p. 190, Q3
$40 Ω$ resistance wale kettle element mein current hai $4 A .$ 2 minutes mein woh kitni heat banata hai?
J
Answer dekho
Answer 76800 J
$H = I^{2} R t = 4^{2} \times 40 \times 120 = 76 800 J .$

11.8 Electric Power

Geyser bill ko ceiling fan se itni tezi se kyun badhata hai? Kyunki woh electrical energy ko kahin zyada tezi se heat mein badalta hai. Koi appliance jis rate se electrical energy ko heat, light, sound ya motion mein badalta hai, woh uski electric power (विद्युत शक्ति) hai. Tum ise (11.19) mein mil chuke ho, $P = V I;$ Ohm's law do aur forms deta hai:

Electric power

\begin{aligned} P & = V I \\ = I^{2} R \\ = \frac{V^{2}}{R} \end{aligned}

(11.22)

Jab current pata ho, jaise series circuits mein, to $P = I^{2} R$ use karo, aur jab voltage pata ho, jaise mains par appliances ke liye, to $P = V^{2} / R$ use karo. Power ki SI unit watt (W) hai: one watt woh device use karta hai jo $1 A$ le jaata hai jab p.d. ho $1 V .$

\begin{aligned} 1 W & = 1 V \times 1 A \\ = 1 V A \end{aligned}

(11.23)

Definition

Electric power

Jis rate se electrical energy use hoti hai, $P = V I = I^{2} R = V^{2} / R .$ Iski SI unit watt (W) hai.

Example $0.25 A$ lene wala $220 V$ ka fan use karta hai $55 W .$

Appliance ratings aksar kilowatts mein di jaati hain, kyunki $1 k W = 1000 W .$ Tumhara bijli ka meter joules nahi ginta: poore mahine ke use ke liye joule bahut chhota hai. Kyunki energy = power × time, woh kilowatts mein power ko hours mein time se multiply karke ginta hai. Yeh unit, kilowatt hour (kW h), hi tumhare bill mein unit kehlaati hai. (Iska chhota rishtedaar, watt hour, W h, woh energy hai jo $1 W$ ka device ek ghante mein use karta hai.)

Ek kilowatt matlab har second $1000 J$ aur ek ghanta hai $3600 s,$ isliye ek kilowatt hour $1000 \times 3600 = 3 600 000$ joules hai:

1 k W h = 3.6 \times 10^{6} J

Definition

Kilowatt hour (ek unit)

$1 k W$ power wala device ek ghante mein jitni energy use karta hai: $1 k W h = 3.6 \times 10^{6} J .$ Yeh electrical energy ki commercial unit hai.

Example10 ghante on chhoda gaya $100 W$ bulb 1 unit use karta hai.

4/4Har rectangle ka area same hai: energy ki one unit.

Fig. 11.15Power × time = energy, isliye teeno rectangles same one unit hain.NCERT p. 191

Ek graph jismein aar-paar hours mein time aur upar ki taraf kilowatts mein power hai, isliye rectangle ka area energy hai. Pehle step mein ek square 1 ghante use kiya gaya 1 kilowatt heater dikhata hai. Doosre step mein ek lambi patli strip 10 ghante use kiya gaya 100 watt bulb dikhati hai. Teesre step mein ek oonchi patli column 30 minutes use kiya gaya 2 kilowatt geyser dikhati hai. Chauthe step mein teeno saath dikhte hain, is note ke saath ki har area 1 kilowatt hour, yaani one unit ke barabar hai.

Example 11.12NCERT Example 11.12 · p. 191

Mixer-grinder ki power

Ek mixer-grinder $220 V$ supply se $2.5 A$ leta hai. Uski power kya hai?

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

Example 11.13NCERT Example 11.13 · p. 191

Garmi mein cooling ka bill

$180 W$ rating wala air cooler poore June (30 din) roz 10 ghante chalta hai. Bijli ₹7.50 per unit hai. Mahine bhar cooler chalane ka kharcha kitna hai?

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

Practice · 2 questions

Practice 11.7 (NCERT in-text questions, p. 192): answers ke saath

Short answerEasyNCERT in-text Questions p. 192, Q1
Do appliances same $220 V$ supply mein plug kiye gaye hain. Kaun tay karta hai ki unmein se kaun energy zyada tezi se use karta hai?
Solution dikhao
AnswerHar ek jitna current leta hai (jo uske resistance se tay hota hai): P = VI, isliye jo zyada current leta hai woh energy zyada tezi se use karta hai.
Energy use ka rate power hai, $P = V I .$ Agar same ho $V,$ to jo appliance zyada current leta hai, yaani jiska resistance chhota hai ( $P = V^{2} / R$ ), woh energy zyada tezi se use karta hai.
2
NumericalEasyNCERT in-text Questions p. 192, Q2
Ek water pump $220 V$ line se $4 A$ leta hai aur 90 minutes chalta hai.
1. Pump ki power nikaalo.
  W
  Answer dekho
  Answer 880 W
  $P = V I = 220 \times 4 = 880 W .$
2. Yeh kW h mein kitni energy use karta hai?
  kW h
  Answer dekho
  Answer 1.32 kW h
  $E = 0.88 k W \times 1.5 h = 1.32 k W h,$ yaani $4.75 \times 10^{6} J .$

Quick check

3 mein se 1

2.5 units electrical energy mein kitne joules hote hain?

Answer dekho

Answer 9000000 J

2.5 \times 3.6 \times 10^{6} = 9.0 \times 10^{6} J .

Recap

Summary aur key points

Electric current charge ke flow ka rate hai, $I = Q / t,$ jo amperes mein naapa jaata hai; metals mein ise electrons le jaate hain, aur source ke bahar conventional current + se − ki taraf maana jaata hai (Section 11.1).
Cell ya battery ek potential difference banaye rakhti hai, $V = W / Q,$ jo volts mein naapa jaata hai: $1 V = 1 J / C$ (Section 11.2).
Ammeter series mein joda jaata hai; voltmeter do points ke across parallel mein joda jaata hai (Section 11.2).
Circuit diagrams standard symbols use karte hain; cell ki lambi plate uska positive terminal hai (Section 11.3).
Ohm's law: constant temperature par metal conductor ke liye $V \propto I,$ isliye $V = I R;$ resistance ohms mein naapa jaata hai (Section 11.4).
Rheostat circuit mein resistance badalkar current badalta hai (Section 11.4).
$R = ρ l / A :$ resistance length ke saath badhta hai, area ke saath ghatta hai aur material par uski resistivity ke through depend karta hai, Ω m mein (Section 11.5).
Series mein current same rehta hai aur p.d.s jud jaate hain: $R_{s} = R_{1} + R_{2} + R_{3}$ (Section 11.6.1).
Parallel mein p.d. same rehta hai aur currents jud jaate hain: $1 / R_{p} = 1 / R_{1} + 1 / R_{2} + 1 / R_{3},$ jo sabse chhote resistor se bhi kam hai (Section 11.6.2).
Ghar parallel mein wired hote hain taaki har appliance ko poora voltage mile aur woh apne aap chale (Section 11.6.2).
Joule's law of heating kehta hai: $H = V I t = I^{2} R t$ (Section 11.7).
Heaters, filament bulbs aur fuses heating effect use karte hain; series mein laga fuse current bahut bada hone par pighal jaata hai (Section 11.7.1).
Electric power $P = V I = I^{2} R = V^{2} / R,$ watts mein; one unit $1 k W h = 3.6 \times 10^{6} J$ hai (Section 11.8).

Practice · 23 questions

Practice questions answers ke saath (NCERT “Exercises”)

Quick answerMediumNCERT Exercises Q1
Resistance $R$ wali ek uniform wire ko chaar barabar tukdon mein kaata jaata hai, aur chaaron tukdon ko parallel mein joda jaata hai. Agar combination ka resistance hai $R^{'},$ to ratio $R / R^{'}$ hai:
$1 / 16$ $4$ $1 / 4$ $16$
Answer dekho
Answer (D) $16$
Har tukde ka resistance hai $R / 4.$ Parallel mein chaar barabar resistors ek ka ek-chauthaai dete hain: $R^{'} = R / 16,$ isliye $R / R^{'} = 16.$
Quick answerEasyNCERT Exercises Q2
Kaun sa expression time $t$ mein resistor dwara use ki gayi electrical energy nahi deta?
$I R^{2} t$ $V I t$ $I^{2} R t$ $V^{2} t / R$
Answer dekho
Answer (A) $I R^{2} t$
$V I t,$ $I^{2} R t$ aur $V^{2} t / R$ Ohm's law se barabar hain. $I R^{2} t$ ki units galat hain: yeh energy nahi hai.
Quick answerMediumNCERT Exercises Q3
$800 W,$ $220 V$ rating wali iron tab use hoti hai jab supply girkar ho jaati hai $176 V .$ Agar uska resistance nahi badalta, to woh jo power use karti hai woh hai:
$640 W$ $800 W$ $400 W$ $512 W$
Answer dekho
Answer (D) $512 W$
$P \propto V^{2},$ aur $(176 / 220)^{2} = {0.8}^{2} = 0.64,$ isliye $P = 0.64 \times 800 = 512 W .$ Square karna bhoolne par 640 W aata hai.
Quick answerHardNCERT Exercises Q4
Teen identical resistors pehle series mein aur phir parallel mein jode jaate hain, har baar same p.d. ke across same time ke liye. Bani heat ka ratio, series : parallel, hai:
$1 : 9$ $9 : 1$ $1 : 3$ $3 : 1$
Answer dekho
Answer (A) $1 : 9$
$R_{s} = 3 R$ aur $R_{p} = R / 3.$ $V$ aur $t$ fixed hon to $H \propto 1 / R,$ isliye $H_{s} : H_{p} = (R / 3) : 3 R = 1 : 9.$
Short answerEasyNCERT Exercises Q5
Asha bulb ke across p.d. naapna chahti hai, lekin woh voltmeter ko uske saath series mein jod deti hai. Kya galat hota hai, aur voltmeter kaise jodna chahiye?
Solution dikhao
AnswerSeries mein voltmeter current ko rok deta hai, isliye bulb bujh jaata hai aur reading galat aati hai; ise parallel mein, bulb ke dono ends ke across jodna chahiye.
Voltmeter do points ke across jodne ke liye bana hai. Series mein lagane par woh akele path ka hissa ban jaata hai aur uska bahut bada resistance current ko lagbhag rok deta hai: bulb bujh jaata hai aur meter lagbhag poore battery p.d. ko padhta hai. Asha ko voltmeter ki ek-ek lead bulb ke har end se jodni chahiye, uske parallel mein.
6
NumericalMediumNCERT Exercises Q6
Ek aluminium wire ka diameter hai $0.40 m m .$ Uski resistivity lo $2.6 \times 10^{- 8} Ω m .$
1. Is wire ki kitni length ka resistance $5.0 Ω$ hoga?
  m
  Answer dekho
  Answer 24.2 m
  $A = π (2.0 \times 10^{- 4})^{2} = 1.26 \times 10^{- 7} m^{2},$ isliye $l = R A / ρ = 5.0 \times 1.26 \times 10^{- 7} / (2.6 \times 10^{- 8}) \approx 24 m .$
2. Agar diameter aadha kar diya jaaye to usi length ka resistance kya hoga?
  Ω
  Answer dekho
  Answer 20 Ω
  Diameter aadha karne se area ek-chauthaai ho jaata hai, isliye resistance chaar guna ho jaata hai: $4 \times 5.0 = 20 Ω .$
NumericalMediumNCERT Exercises Q7
Ek resistor ke liye yeh readings li gayi thi. V (upar ki taraf) ko I ke against plot karo aur origin se guzarti best seedhi line ke slope se resistance nikaalo.
Table 11.9
I (A) V (V)
0.2 1.1
0.4 2.3
0.6 3.4
0.8 4.7
1.0 5.8
Ω
Solution dikhao
Paancho points plot karo: woh lagbhag origin se guzarti ek seedhi line par aate hain, jaisa Ohm's law kehta hai; thoda sa bikhraav experimental error hai.
Best line ka slope $= Δ V / Δ I \approx 5.8 / 1.0 = 5.8 Ω,$ jo resistance hai.
Quick answerEasyNCERT Exercises Q8
$9 V$ battery ek anjaan resistor se $1.5 m A$ current chalati hai. Ohms mein uska resistance nikaalo.
Ω
Answer dekho
Answer 6000 Ω
$R = V / I = 9 / (1.5 \times 10^{- 3}) = 6.0 \times 10^{3} Ω,$ yaani $6 k Ω .$
9
NumericalEasyNCERT Exercises Q9
$6 V$ battery series mein judi hai in resistors ke saath: $0.5 Ω,$ $1.0 Ω,$ $1.5 Ω,$ $2.0 Ω$ aur $10 Ω .$
1. $10 Ω$ resistor se kitna current behta hai?
  A
  Answer dekho
  Answer 0.4 A
  $R_{s} = 0.5 + 1.0 + 1.5 + 2.0 + 10 = 15 Ω,$ isliye $I = 6 / 15 = 0.4 A$ har resistor se.
2. $10 Ω$ resistor ke across p.d. kya hai?
  V
  Answer dekho
  Answer 4 V
  $V = I R = 0.4 \times 10 = 4 V .$
Quick answerMediumNCERT Exercises Q10
$220 V$ line se $4 A$ lene ke liye kitne $330 Ω$ resistors parallel mein jodne honge?
resistors
Answer dekho
Answer 6 resistors
Chahiye: $R_{p} = 220 / 4 = 55 Ω .$ Parallel mein $n$ barabar resistors dete hain $330 / n,$ isliye $n = 330 / 55 = 6.$
11
DiagramMediumNCERT Exercises Q11
Tumhare paas $4 Ω$ ke teen resistors hain. Dikhao ki teeno ko kaise jodein taaki total ho:
1. $6 Ω$
  Solution dikhao
  AnswerEk 4 Ω baaki do ke parallel combination ke saath series mein.
  $4 ∥ 4 = 2 Ω;$ phir $4 + 2 = 6 Ω .$
2. $8 / 3 Ω$ (lagbhag $2.7 Ω$ )
  Solution dikhao
  AnswerDo series mein (8 Ω), aur teesra unke across parallel mein.
  $4 + 4 = 8 Ω;$ phir $8 ∥ 4 = 32 / 12 = 8 / 3 Ω .$
Quick answerMediumNCERT Exercises Q12
$12 W,$ $220 V$ rating wale LED bulbs ek $220 V$ line ke across parallel mein jode jaate hain jo safely zyada se zyada le ja sakti hai $3 A .$ Kitne bulbs jode ja sakte hain?
bulbs
Answer dekho
Answer 55 bulbs
Ek bulb leta hai $I = P / V = 12 / 220 A .$ Sankhya hai $3 \div (12 / 220) = 660 / 12 = 55.$
13
NumericalMediumNCERT Exercises Q13
Ek room heater mein do identical elements, P aur Q, hain, har ek $55 Ω$ ka. Ek selector switch $220 V$ mains ke across ek element akele, dono series mein, ya dono parallel mein laga sakta hai.
1. Sirf ek coil ke saath current?
  A
  Answer dekho
  Answer 4 A
  $I = 220 / 55 = 4 A .$
2. Dono coils series mein hon to current?
  A
  Answer dekho
  Answer 2 A
  $R = 110 Ω,$ isliye $I = 220 / 110 = 2 A .$
3. Dono coils parallel mein hon to current?
  A
  Answer dekho
  Answer 8 A
  $R = 27.5 Ω,$ isliye $I = 220 / 27.5 = 8 A .$
4. Kaun si setting sabse tez garam karti hai, aur uski power kya hai?
  Solution dikhao
  AnswerParallel setting: 1760 W.
  Power $P = V I :$ ek coil $880 W,$ series $440 W,$ parallel $220 \times 8 = 1760 W .$ Parallel setting sabse tez garam karti hai.
14
NumericalMediumNCERT Exercises Q14
In do circuits mein se kismein $3 Ω$ resistor zyada power use karta hai, aur kitne guna zyada?
1. $3 Ω$ aur $6 Ω$ resistors series mein, saath mein $9 V$ battery.
  W
  Answer dekho
  Answer 3 W
  $I = 9 / 9 = 1 A,$ isliye $P = I^{2} R = 1 \times 3 = 3 W .$
2. $3 Ω$ aur $6 Ω$ resistors parallel mein, saath mein $6 V$ battery.
  W
  Answer dekho
  Answer 12 W
  $3 Ω$ resistor ke across poora hai $6 V :$ $P = V^{2} / R = 36 / 3 = 12 W,$ jo (a) se chaar guna hai.
Quick answerEasyNCERT Exercises Q15
$40 W$ ki tube light aur $70 W$ ka ceiling fan, dono ki rating hai $220 V,$ aur dono $220 V$ mains par parallel mein chalte hain. Woh kul kitna current lete hain?
A
Answer dekho
Answer 0.5 A
Total power $110 W,$ isliye $I = P / V = 110 / 220 = 0.5 A .$
Quick answerEasyNCERT Exercises Q16
Kaun zyada energy use karta hai: 6 minutes chalaya gaya $1500 W$ microwave oven, ya 3 ghante chalaya gaya $75 W$ ceiling fan?
Microwave ovenCeiling fanDono barabar use karte hainVoltage jaane bina nahi bata sakte
Answer dekho
Answer (B) Ceiling fan
Microwave: $1500 \times 0.1 h = 150 W h .$ Fan: $75 \times 3 = 225 W h .$ Kam power wala fan zyada use karta hai kyunki woh kahin zyada der chalta hai.
17
NumericalMediumNCERT Exercises Q17
$22 Ω$ resistance wala geyser element mains se 30 minutes tak $10 A$ leta hai.
1. Heat kis rate se banti hai?
  W
  Answer dekho
  Answer 2200 W
  $P = I^{2} R = 10^{2} \times 22 = 2200 W$ (yaani joules per second).
2. Kul kitni heat banti hai?
  J
  Answer dekho
  Answer 3960000 J
  $H = P t = 2200 \times 1800 = 3.96 \times 10^{6} J,$ yaani $1.1 k W h .$
18
Short answerMediumNCERT Exercises Q18 (a)–(d)
Thode mein samjhao:
1. bulb filaments tungsten ke kyun bante hain
  Solution dikhao
  AnswerYeh sirf lagbhag 3400 °C par pighalta hai, isliye bina pighle white-hot chamak sakta hai.
  Roshni dene ke liye filament ko $2500^{\circ} C$ se kaafi upar pahunchna padta hai. Tungsten ka melting point bahut high hai (lagbhag $3400^{\circ} C$ ) aur yeh garam hone par bhi mazboot rehta hai.
2. toasters aur irons ke elements pure metal ke bajaay alloy ke kyun bante hain
  Solution dikhao
  AnswerAlloys ki resistivity zyada hoti hai aur woh high temperature par aasaani se oxidise nahi hote.
  Nichrome jaise alloy ki resistivity pure metals se kahin zyada hoti hai, isliye chhota element bhi bada resistance aur khoob heat deta hai; yeh red-hot hone par aasaani se oxidise (burn) bhi nahi hota.
3. ghar ke appliances series mein kyun nahi jode jaate
  Solution dikhao
  AnswerSeries mein woh voltage baantte, same current le jaate aur ek ke kharaab hone par sab band ho jaate.
  Series mein mains voltage bant jaata hai, isliye kisi appliance ko poora 220 V nahi milta; sabko same current le jaana padta hai jabki unhe alag-alag currents chahiye; aur ek fault ya switch-off sabka circuit tod deta hai.
4. agar wire ka area of cross-section double kar diya jaaye to uska resistance kaise badalta hai
  Solution dikhao
  AnswerAadha ho jaata hai, kyunki R ∝ 1/A.
  $R = ρ l / A,$ isliye same length aur material ke saath $A$ double karne se aadha ho jaata hai $R .$
Short answerEasyNCERT Exercises Q18 (e)
Desh bhar mein aur gharon tak bijli le jaane wali wires ke liye copper aur aluminium kyun use hote hain?
Solution dikhao
AnswerInki resistivities bahut kam hain, isliye heat ke roop mein kam energy waste hoti hai; aluminium halka aur sasta bhi hai.
Copper aur aluminium ki resistivities sabse kam mein se hain, isliye lambi wires ka resistance chhota hota hai aur heat ( $I^{2} R$ ) mein kam energy waste hoti hai. Aluminium halka aur sasta bhi hai, jo overhead lines ke liye sahi hai; copper ghar ki wiring mein use hota hai.
20
Quick answerEasy
Har quantity ko uski SI unit se match karo.
Left mein koi item tap karo, phir right mein uska match.
Answer dekho
Answer
charge : coulomb
potential difference : volt
resistance : ohm
resistivity : ohm metre
power : watt
C, V, Ω, Ω m aur W. Resistivity mein extra metre hai kyunki $ρ = R A / l .$
Quick answerMedium
Assertion (A): Parallel mein jude do resistors ka equivalent resistance hamesha dono mein se har ek se chhota hota hai. Reason (R): Parallel mein resistances ke reciprocals jud jaate hain.
A aur R dono sahi hain, aur R, A ko samjhata haiA aur R dono sahi hain, lekin R, A ko nahi samjhataA sahi hai lekin R galat haiA galat hai lekin R sahi hai
Answer dekho
Answer (A) A aur R dono sahi hain, aur R, A ko samjhata hai
$1 / R_{p} = 1 / R_{1} + 1 / R_{2}$ dono mein se har ek se bada hai, $1 / R_{1}$ ya $1 / R_{2},$ isliye $R_{p}$ dono se chhota hai: R sahi hai aur A ko samjhata hai.
Quick answerHard
Assertion (A): $100 W$ bulb $40 W$ bulb se zyada chamakta hai jab dono series mein jode jaate hain across $220 V .$ Reason (R): Series mein zyada resistance wala bulb zyada power use karta hai.
A aur R dono sahi hain, aur R, A ko samjhata haiA aur R dono sahi hain, lekin R, A ko nahi samjhataA sahi hai lekin R galat haiA galat hai lekin R sahi hai
Answer dekho
Answer (D) A galat hai lekin R sahi hai
Series mein dono mein same current behta hai, isliye bade resistance ke liye $P = I^{2} R$ bada hota hai. $40 W$ bulb ka resistance hai $R = 1210 Ω,$ jo zyada hai $100 W$ bulb ke resistance se, yaani $484 Ω,$ isliye 40 W bulb zyada chamakta hai. A galat hai, R sahi hai.
23
NumericalMedium
Ek parivaar ka bijli meter 1 May ko 4512 units aur 1 June ko 4728 units padhta hai.
1. May mein kitne units use hue?
  kW h
  Answer dekho
  Answer 216 kW h
  $4728 - 4512 = 216$ units.
2. ₹6.50 per unit par bill kitna hai?
  ₹
  Answer dekho
  Answer 1404 ₹
  $216 \times 6.50 = 1404,$ isliye ₹1404.

I (A)	V (V)
0.2	1.1
0.4	2.3
0.6	3.4
0.8	4.7
1.0	5.8

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 in-text Questions p. 172, Q1	Practice 11.1 (NCERT in-text questions, p. 172) · Q1
NCERT in-text Questions p. 172, Q2	Practice 11.1 (NCERT in-text questions, p. 172) · Q2
NCERT in-text Questions p. 172, Q3	Practice 11.1 (NCERT in-text questions, p. 172) · Q3
NCERT in-text Questions p. 174, Q1	Practice 11.2 (NCERT in-text questions, p. 174) · Q1
NCERT in-text Questions p. 174, Q2	Practice 11.2 (NCERT in-text questions, p. 174) · Q2
NCERT in-text Questions p. 174, Q3	Practice 11.2 (NCERT in-text questions, p. 174) · Q3
NCERT in-text Questions p. 181, Q1	Practice 11.3 (NCERT in-text questions, p. 181) · Q1
NCERT in-text Questions p. 181, Q2	Practice 11.3 (NCERT in-text questions, p. 181) · Q2
NCERT in-text Questions p. 181, Q3	Practice 11.3 (NCERT in-text questions, p. 181) · Q3
NCERT in-text Questions p. 181, Q4	Practice 11.3 (NCERT in-text questions, p. 181) · Q4
NCERT in-text Questions p. 181, Q5	Practice 11.3 (NCERT in-text questions, p. 181) · Q5
NCERT in-text Questions p. 185, Q1	Practice 11.4 (NCERT in-text questions, p. 185) · Q1
NCERT in-text Questions p. 185, Q2	Practice 11.4 (NCERT in-text questions, p. 185) · Q2
NCERT in-text Questions p. 188, Q1	Practice 11.5 (NCERT in-text questions, p. 188) · Q1
NCERT in-text Questions p. 188, Q2	Practice 11.5 (NCERT in-text questions, p. 188) · Q2
NCERT in-text Questions p. 188, Q3	Practice 11.5 (NCERT in-text questions, p. 188) · Q3
NCERT in-text Questions p. 188, Q4	Practice 11.5 (NCERT in-text questions, p. 188) · Q4
NCERT in-text Questions p. 188, Q5	Practice 11.5 (NCERT in-text questions, p. 188) · Q5
NCERT in-text Questions p. 190, Q1	Practice 11.6 (NCERT in-text questions, p. 190) · Q1
NCERT in-text Questions p. 190, Q2	Practice 11.6 (NCERT in-text questions, p. 190) · Q2
NCERT in-text Questions p. 190, Q3	Practice 11.6 (NCERT in-text questions, p. 190) · Q3
NCERT in-text Questions p. 192, Q1	Practice 11.7 (NCERT in-text questions, p. 192) · Q1
NCERT in-text Questions p. 192, Q2	Practice 11.7 (NCERT in-text questions, p. 192) · Q2
NCERT Exercises Q1	Exercises · Q1
NCERT Exercises Q2	Exercises · Q2
NCERT Exercises Q3	Exercises · Q3
NCERT Exercises Q4	Exercises · Q4
NCERT Exercises Q5	Exercises · Q5
NCERT Exercises Q6	Exercises · Q6
NCERT Exercises Q7	Exercises · Q7
NCERT Exercises Q8	Exercises · Q8
NCERT Exercises Q9	Exercises · Q9
NCERT Exercises Q10	Exercises · Q10
NCERT Exercises Q11	Exercises · Q11
NCERT Exercises Q12	Exercises · Q12
NCERT Exercises Q13	Exercises · Q13
NCERT Exercises Q14	Exercises · Q14
NCERT Exercises Q15	Exercises · Q15
NCERT Exercises Q16	Exercises · Q16
NCERT Exercises Q17	Exercises · Q17
NCERT Exercises Q18 (a)–(d)	Exercises · Q18
NCERT Exercises Q18 (e)	Exercises · Q19

Quick check: MCQs

11.1 Electric Current aur Circuit3 questions
11.2 Electric Potential aur Potential Difference3 questions
11.3 Circuit Diagram2 questions
11.4 Ohm's Law3 questions
11.5 Conductor ka Resistance kin Factors par Depend karta hai3 questions
11.6.1 Series mein Resistors2 questions
11.6.2 Parallel mein Resistors3 questions
11.7 Electric Current ka Heating Effect2 questions
11.7.1 Electric Current ke Heating Effect ke Practical Applications3 questions
11.8 Electric Power3 questions

Glossary · 15

Key terms (English + Hindi)

ammeterऐमीटरaimeetar: Current naapne wala meter. Ise series mein joda jaata hai, taaki naapa ja raha current isse hokar guzre.Chapter mein pehli baar yahan aaya →

electric circuitविद्युत परिपथvidyut paripath: Ek bina toota, closed conducting path jo charge ko source ke ek terminal se wapas doosre tak behne deta hai.Chapter mein pehli baar yahan aaya →
electric currentविद्युत धाराvidyut dhara: Conductor se electric charge ke flow ka rate, $I = Q / t,$ jo amperes (A) mein naapa jaata hai.Chapter mein pehli baar yahan aaya →
electric powerविद्युत शक्तिvidyut shakti: Jis rate se electrical energy use ya supply hoti hai, $P = V I,$ jo watts (W) mein naapa jaata hai.Chapter mein pehli baar yahan aaya →

fuseफ़्यूज़fyooz: Kam melting point wali chhoti wire jo circuit ke saath series mein judi hoti hai; current bahut bada hone par yeh pighalkar circuit tod deti hai.Chapter mein pehli baar yahan aaya →

Joule's law of heatingजूल का तापन नियमjool ka tapan niyam: Resistor mein bani heat hai $H = I^{2} R t .$ Chapter mein pehli baar yahan aaya →

kilowatt hourकिलोवाट घंटाkilowatt ghanta: Electrical energy ki commercial unit (one unit): 1 h ke liye use hua 1 kW, yaani $3.6 \times 10^{6} J .$ Chapter mein pehli baar yahan aaya →

Ohm's lawओम का नियमom ka niyam: Steady temperature par metal conductor ke liye current potential difference ke proportional hota hai: $V = I R .$ Chapter mein pehli baar yahan aaya →

parallel connectionपार्श्वक्रमparshvakram: Same do points ke beech jude parts, jisse har ek ke across same potential difference hota hai.Chapter mein pehli baar yahan aaya →
potential differenceविभवांतरvibhavantar: Do points ke beech ek coulomb charge le jaane mein kiya gaya work, $V = W / Q,$ jo volts (V) mein naapa jaata hai.Chapter mein pehli baar yahan aaya →

resistanceप्रतिरोधpratirodh: Conductor charge ke flow ko kitna rokta hai, $R = V / I,$ jo ohms (Ω) mein naapa jaata hai.Chapter mein pehli baar yahan aaya →
resistivityप्रतिरोधकताpratirodhakta: Material ki property, $ρ = R A / l,$ jo diye gaye size ki wire ka resistance tay karti hai; unit ohm metre (Ω m).Chapter mein pehli baar yahan aaya →
rheostatधारा नियंत्रकdhara niyantrak: Sliding contact wala variable resistor, jo source badle bina current badalne ke kaam aata hai.Chapter mein pehli baar yahan aaya →

series connectionश्रेणीक्रमshrenikram: Ek chain mein end to end jude parts, jisse har ek se same current behta hai.Chapter mein pehli baar yahan aaya →

voltmeterवोल्टमीटरvoltmeetar: Potential difference naapne wala meter. Iski do leads do points ke across parallel mein judi hoti hain.Chapter mein pehli baar yahan aaya →

11.1 Electric Current aur Circuit

11.2 Electric Potential aur Potential Difference

11.3 Circuit Diagram

11.4 Ohm's Law

11.5 Conductor ka Resistance kin Factors par Depend karta hai

11.6 Resistors ke System ka Resistance

11.6.1 Series mein Resistors

11.6.2 Parallel mein Resistors

11.7 Electric Current ka Heating Effect

11.7.1 Electric Current ke Heating Effect ke Practical Applications

11.8 Electric Power

Apna NCERT exercise question dhoondo

Quick check: MCQs

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