Alternating Current: Class 12 Physics Chapter 7 Hinglish mein

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

Published 12 Oct 2026 · Updated 12 Oct 2026

120 min
1 simulation
16 practice questions
34 MCQs

Is chapter mein aap kya seekhenge

Alternating voltage ko uske amplitude, angular frequency aur period se describe karo, aur samjho ki power supply ke liye ac kyun use hota hai
Resistor mein average power derive karo, aur current aur voltage ki rms values define karke use karo
Sinusoidal voltages aur currents ko rotating phasors se represent karo, aur phasor diagrams se phase relations padho
Pure inductor aur pure capacitor ke liye current, reactance aur phase derive karo, aur dikhao ki dono mein se koi bhi average power nahi leta
Phasor diagram se series LCR circuit ka impedance aur phase angle nikaalo
Series LCR circuit mein resonance samjho, resonant frequency calculate karo aur resonance ki sharpness ko R se relate karo
Average power aur power factor calculate karo, aur wattless current aur power-factor correction samjho
Samjho ki transformer kaise kaam karta hai, turns ratio apply karo, aur uske losses aur power transmission mein uska role describe karo

NCERT Physics Class 12, Chapter 7, pp. 177–200 par based original companion text. Yeh NCERT se juda nahi hai aur NCERT ne ise endorse nahi kiya hai.

7.1 Introduction

Tumhare phone ki battery charge ko uske circuits mein sirf ek hi direction mein, lagbhag constant voltage par, ghumati hai. Deewar ka socket bilkul alag tarah behave karta hai. Uske terminals ke beech ka potential difference maximum tak badhta hai, zero se guzar kar girta hai, negative ho jaata hai, aur phir wapas aata hai, India mein har second pachaas baar, aur jab bhi voltage reverse hota hai, uska chalaya current bhi reverse ho jaata hai. Battery jaisa steady, ek-taraf current direct current (dc) hai. Iske ulta socket ek alternating voltage (प्रत्यावर्ती वोल्टता) deta hai, jo time ke saath sinusoidally badhta aur ghatta hai, aur jo current yeh aage-peeche dhakelta hai woh alternating current, yaani ac (प्रत्यावर्ती धारा) hai.Yahan “ac” sirf yeh batata hai ki koi quantity time ke saath kaise badalti hai. Isliye “ac voltage” ek alternating potential difference hai aur “ac current” ek alternating current, bhale hi letters ka strict matlab “alternating current” hai.

3/4Har half cycle mein voltage ka sign badalta hai, aur uska chalaya current reverse ho jaata hai.

Fig. 7.1Battery steady voltage deti hai; mains voltage alternate karta hai.

Time ke saath voltage ke do graphs, ek ke upar doosra. Step 1: upar wala graph battery ka 12 volt output ek flat horizontal line ki tarah dikhata hai. Step 2: neeche wale graph mein mains voltage judta hai, ek smooth sine wave jo plus 325 volts aur minus 325 volts ke beech jhoolti hai. Step 3: wave ke axis se upar wale hisse blue shade kiye gaye hain, ek arrow aur one way shabdon ke saath, aur neeche wale hisse orange shade kiye gaye hain, ulte arrow ke saath; yeh dikhata hai ki current har half cycle mein direction badalta hai. Step 4: pehli poori wave ke upar ek bracket one cycle mark karta hai, jo 20 milliseconds chalti hai.

Power companies dc ki jagah ac kyun supply karti hain? Zyadatar transformer (Section 7.8) ki wajah se. Alternating voltage ke saath, bina kisi moving part wala device bahut kam waste ke saath voltage badha ya ghata sakta hai. Power stations iska use karke energy ko saikdon kilometres tak bahut high voltage aur chhote current par bhejte hain, jisse lines mein heating kam rehti hai, aur phir gharon mein pahunchne se pehle voltage wapas kam kar dete hain. Ac woh kaam bhi kar sakta hai jo dc nahi kar sakta: radio apne aerial tak pahunchne wale darjanon stations mein se ek ko ek aise circuit se chunta hai jo sirf ek frequency par zor se respond karta hai aur baaki par mushkil se (Section 7.6.2).

Hum sinusoidal voltage ko $v = v_{m} \sin ω t$ likhte hain. Yahan $v$ instantaneous value hai, $v_{m}$ peak value yaani amplitude hai, aur $ω$ radians per second mein angular frequency hai. Ek cycle ek period $T = 2 π / ω$ tak chalti hai, aur frequency $ν = 1 / T = ω / 2 π$ hai. Indian mains ke liye $ν = 50 H z$ , isliye $T = 20 m s$ aur $ω = 100 π \approx 314 r a d / s$ .

Aisa voltage aata kahan se hai? Chapter 6 mein tum ac generator se mile the: magnetic field mein steadily ghoomti coil mein induced emf $ε = N B A ω \sin ω t$ hota hai. Neeche wala simulation tumhe khud ek coil ghumane deta hai.

SimulationGhoomti coil ac banati hai

Ek rectangular coil do magnet poles ke beech ghoomti hai aur ek meter mein current chalati hai jiski needle left aur right jhoolti hai. Ek graph fixed chaar-second window mein time ke saath emf aur coil se guzarne wala flux trace karta hai. Readouts rpm mein speed, frequency, peak emf aur is pal ka emf dikhate hain; pause karne par coil angle dikhta hai aur coil ko kisi bhi angle par drag kiya ja sakta hai. Sliders speed, turns ki sankhya aur field badalte hain.

Fig. 7.2Steady rate se ghoomti coil ka emf ek sine wave hai: har half turn par yeh reverse hota hai.

Try karo

Play dabane se pehle coil ko tab tak drag karo jab tak EMF now reading sabse badi na ho jaaye. Coil angle note karo, phir coil ko aadha turn aur aage drag karo. Emf ko kya hua?
Ab Play dabao aur Speed ko 60 rpm par set karo. Frequency aur Peak EMF readings, aur graph window mein cycles ki sankhya, 30 rpm wali values se kaise compare karti hain?
Ek poore turn mein emf trace ko dekho. Turn ke kitne hisse mein yeh positive rehta hai? Poore turn par iska average kya hona chahiye?

Timeline

Ac ne “war of the currents” kaise jeeti

1882
Ek shehar ke liye dc power
New York mein Edison ka Pearl Street station paas ki buildings ko dc supply karta hai; yeh sirf thodi doori tak ke customers tak pahunch sakta hai.
1888
Induction motor
Tesla alternating current par chalne wali motors patent karte hain; Westinghouse patents khareed lete hain.
1893
Ac se roshan ek mela
Westinghouse ki company Chicago mein World's Columbian Exposition ko alternating current se roshan karti hai.
1896
Niagara se power
Niagara Falls ke generators transformers ki madad se kai kilometres door Buffalo shehar ko ac bhejte hain.

Quick check

3 mein se 1

Kuch countries apna mains

60 H z

par chalati hain. Us supply ka period kitna hai, milliseconds mein?

Answer dekho

Answer 16.667 ms

T = 1 / ν = 1 / 60 s \approx 0.0167 s = 16.7 m s

, jo

50 H z

supply ke

20 m s

se thoda chhota hai.

7.2 Resistor par lagaya gaya AC voltage

Ek ac source ke across resistor $R$ jodo. Current kitna bada hai, aur voltage ke mukable uski timing kaisi hai?

DerivationResistor se current

Maan lo source ka voltage hai

v = v_{m} \sin ω t

(7.1)

Kirchhoff's loop rule har pal lagta hai, bilkul dc ki tarah: source voltage resistor ke across potential drop $i R$ ke barabar hota hai. Isliye $v_{m} \sin ω t = i R$ , aur

i = i_{m} \sin ω t

(7.2)

jahan current amplitude hai

i_{m} = \frac{v_{m}}{R}

(7.3)

Equation (7.3) bas peak values ke liye likha $V = I R$ hai, isliye resistor ac par bhi Ohm's law follow karta hai jaise dc par. (7.2) mein bhi kuch current ko delay nahi karta. Jab bhi $v$ zero se guzarta hai, $i$ bhi guzarta hai, aur dono saath-saath apne crest par pahunchte hain: current aur voltage in phase hain.

3/3Peaks, troughs aur zero crossings line up hote hain: current aur voltage in phase hain.

Fig. 7.3Resistor mein current voltage ke saath-saath badhta aur ghatta hai.NCERT Fig. 7.2 · p. 178

Left mein ek circuit: ek ac source, jo chhoti wave wale circle ki tarah bana hai, 5 ohm resistor se juda hai, current i ke liye ek arrow aur 10 volts peak label ke saath. Right mein time ke saath ek graph. Step 1: source voltage v, ek sine wave jo plus aur minus 10 volts tak jaati hai. Step 2: current i ek dashed sine wave ki tarah judta hai jiska peak 2 amperes hai; yeh bilkul usi pal zero cross karta hai aur peaks par pahunchta hai jab voltage. Step 3: peaks aur troughs se guzarti dotted vertical lines aur zero crossings par dots dikhate hain ki dono curves saath-saath peak karte hain aur zero cross karte hain.

Jo ammeter current ko ek cycle par simply average karta, woh zero padhta, kyunki current aadhe time ek taraf behta hai aur baaki time doosri taraf. Phir bhi mains par electric iron zaroor garam hoti hai. Iska hal yeh hai ki heating $i^{2}$ ke saath chalti hai, jo positive hai, charge kisi bhi taraf chale.

DerivationResistor mein average power

Kisi bhi pal electrical energy jis rate se heat mein badalti hai, woh hai

p = i^{2} R = i_{m}^{2} R \sin^{2} ω t

(7.4)

Ek period $T$ par quantity $F (t)$ ka average aise define hota hai

⟨ F ⟩ = \frac{1}{T} \int_{0}^{T} F (t) d t

Average power ko $\overset{ˉ}{p}$ likhkar,

\overset{ˉ}{p} = ⟨ i^{2} R ⟩ = ⟨ i_{m}^{2} R \sin^{2} ω t ⟩

(7.5a)

Na $i_{m}$ time ke saath badalta hai na $R$ , isliye dono average ke bahar aa jaate hain:

\overset{ˉ}{p} = i_{m}^{2} R ⟨ \sin^{2} ω t ⟩

(7.5b)

$\sin^{2} ω t$ ka average nikaalne ke liye $\sin^{2} ω t = \frac{1}{2} (1 - \cos 2 ω t)$ use karo. Term $\cos 2 ω t$ time $T = 2 π / ω$ mein bilkul do poori cycles chalta hai, isliye uske positive aur negative hisse cancel ho jaate hain:

⟨ \cos 2 ω t ⟩ = \frac{1}{T} {[\frac{\sin 2 ω t}{2 ω}]}_{0}^{T} = \frac{\sin 4 π}{2 ω T} = 0

Isliye $⟨ \sin^{2} ω t ⟩ = \frac{1}{2}$ , aur

\overset{ˉ}{p} = \frac{1}{2} i_{m}^{2} R

(7.5c)

Average power peak power $i_{m}^{2} R$ ka bilkul aadha hai.

Alternating current ko us steady current se describe karna convenient hai jo same resistor ko same average rate se garam kare. Ise $I$ kaho. $I^{2} R$ ko $\frac{1}{2} i_{m}^{2} R$ ke barabar rakhne se $I^{2} = \frac{1}{2} i_{m}^{2} = ⟨ i^{2} ⟩$ milta hai. Yaani $I$ nikaalne ke liye current ko square karo, ek cycle par square ka mean lo, aur phir square root lo. Isi recipe se iska naam pada: root-mean-square value, yaani rms value, jise effective value bhi kehte hain (Fig. 7.4).

4/4Mean square ka square root $I = i_{m} / \sqrt{2} \approx 0.707 i_{m}$ hai.

Fig. 7.4Square karo, mean lo, root lo: rms current peak current ka

0.707

hai.NCERT Fig. 7.3 · p. 179

Do cycles par time ke saath ek graph. Step 1: current i ke liye ek dashed sine curve, jo plus i m aur minus i m ke beech jhoolta hai. Step 2: current fade hota hai aur i squared ka ek solid curve dikhta hai; yeh kabhi negative nahi hota, aur zero se i m squared tak current se do guna baar uthta hai. Step 3: ek dashed horizontal line i squared ka mean, i m squared ke aadhe par, mark karti hai, aur line ke upar wale curve ke hisse shade kiye gaye hain yeh dikhane ke liye ki woh neeche ke gaddhon ko bilkul bhar dete hain. Step 4: current curve wapas aata hai aur I barabar i m over root 2 par ek solid horizontal line hai, peak ka lagbhag 0.707, jis par root of the mean square label hai.

I = \sqrt{⟨ i^{2} ⟩} = \sqrt{\frac{1}{2} i_{m}^{2}} = \frac{i_{m}}{\sqrt{2}} \approx 0.707 i_{m}

(7.6)

$I$ ke terms mein average power jaana-pehchana dc form le leta hai:

P = \overset{ˉ}{p} = \frac{1}{2} i_{m}^{2} R = I^{2} R

(7.7)

Definition

rms (effective) value

Us steady current ki value jo diye gaye resistor ko alternating current ke same average rate se garam kare: $I = \sqrt{⟨ i^{2} ⟩}$ , jo sine wave ke liye $i_{m} / \sqrt{2}$ hai. Isi tarah voltage ke liye $V = v_{m} / \sqrt{2}$ .

Example $10 A$ peak value wale current ki rms value $7.07 A$ hai.

Voltages ko bhi bilkul currents ki tarah treat kiya jaata hai, jisse rms voltage milta hai:

V = \frac{v_{m}}{\sqrt{2}} \approx 0.707 v_{m}

(7.8)

$v_{m} = i_{m} R$ ke dono sides ko $\sqrt{2}$ se divide karne par Ohm's law rms values ke beech ek relation ban jaata hai:

V = I R

(7.9)

Jab tum socket par “ $230 V$ ” padhte ho, ya meter par koi bhi ac reading, to woh number rms value hai; mains voltage asal mein $\sqrt{2} \times 230 V \approx 325 V$ tak peak karta hai. Rms values batane ka fayda yeh hai ki inke saath ac par resistor jaane-pehchane dc formulas bilkul follow karta hai: $V = I R$ aur $P = I^{2} R = V^{2} / R = V I$ . Isi tarah appliance par likhi power uski average power hoti hai.

Example 7.1NCERT Example 7.1 · p. 180

Mains par ek room heater

Ek room heater par $1000 W$ , $230 V$ , $50 H z$ likha hai. Uske element ko pure resistor maano. (a) Garam element ka resistance kya hai? (b) Yeh kitna rms current leta hai, aur current apne peak par kitna bada hota hai? (c) Kisi bhi pal yeh energy sabse zyada kis rate se leta hai? (d) Agar yeh 30 din ke mahine mein har din 3 ghante chale, to ₹8 per unit (kWh) par iska kharcha kitna hoga?

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

Example

Equation se ac current padhna

$8.0 Ω$ resistor mein current $i = 5.0 \sin (100 π t)$ hai, jahan $i$ amperes mein aur $t$ seconds mein hai. Nikaalo (a) frequency, (b) rms current, (c) average power aur (d) sabse badi instantaneous power.

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

Try karke dekho

Rating plates padho

Tumhe chahiye

rating labels wale teen ya chaar appliances (ya unke dabbe)
ek calculator

Pehle guess karo

Saare appliances

230 V

ke liye rated hain. Kaun sa sabse zyada rms current lega?

Sab same current lete hain, kyunki voltage same haiJiski power rating sabse zyada haiJiski power rating sabse kam haiJiski cable sabse lambi hai

Teen ya chaar appliances, ya unke dabbe, ikattha karo jin par rating label ho, jaise electric iron, kettle, LED bulb aur phone charger. Labels sirf unplugged appliances par hi padho.
Har ek ka rated voltage, frequency aur power note karo.
Har ek ka rms current $I = P / V$ nikaalo. Heating appliances (iron, kettle, heater) ke liye garam resistance $R = V^{2} / P$ bhi nikaalo.

Tumhare appliances

Appliance	Rated power P (W)	Rated voltage V (V)	I = P/V (A)

Tumhari entries apne aap save hoti hain.

Currents power ratings ke proportional hain: irons, kettles aur heaters ke liye lagbhag $4$ se $8 A$ , aur LED bulbs aur chargers ke liye ampere ka sirf kuch sauvaan hissa. Formula $R = V^{2} / P$ sirf un appliances ke liye sense banata hai jo simply heating elements hain, jo pure resistors ki tarah behave karte hain. Chargers, fans aur LED lamps mein doosre components hote hain, isliye unka current voltage ke saath in phase nahi hota (iska matlab tum Section 7.7 mein dekhoge).

Quick check

4 mein se 1

Lamp ke across juda ac voltmeter

120 V

padhta hai. Lamp ke across peak voltage kitna hai, volts mein?

Answer dekho

Answer 169.7 V

Meters rms values padhte hain, isliye

v_{m} = \sqrt{2} \times 120 V \approx 170 V

7.3 Rotating vectors se ac ko represent karna: phasors

Resistor current aur voltage ko saath-saath rakhta hai, lekin agle do elements aisa nahi karte. Inductor mein current voltage ke baad peak par pahunchta hai; capacitor mein pehle. Out of step sine waves ko point-by-point bana kar jodna slow hai aur galtiyan hone ka darr rehta hai. Ek simple geometric picture yeh kaam kahin zyada aasani se karti hai.

Socho ek arrow hai jiski length $v_{m}$ hai, jo origin par pivot hai aur steady angular speed $ω$ se anticlockwise ghoom raha hai. Time $t$ par yeh horizontal axis ke saath angle $ω t$ banata hai, isliye vertical axis par iska projection $v_{m} \sin ω t$ hai, jo bilkul instantaneous voltage hai. Jaise-jaise arrow ghoomta hai, uska projection sine curve banata jaata hai (Fig. 7.5). Is tarah use hone wale rotating arrow ko phasor (फेज़र) kehte hain. Hum phasors ko bold mein likhte hain: voltage ke liye $𝐕$ , current ke liye $𝐈$ .

1/5 $ω t = 30^{\circ}$ par projection $v_{m} \sin 30^{\circ} = v_{m} / 2$ hai.

Fig. 7.5Rotating phasor ka vertical projection instantaneous value

v_{m} \sin ω t

hai.NCERT Fig. 7.4 · p. 181

Left mein horizontal aur vertical axes wala ek circle, aur centre par pivot hua v m length ka blue arrow, ek curved arrow ke saath jo dikhata hai ki yeh angular speed omega se anticlockwise ghoomta hai. Right mein angle omega t ke saath v ka graph, 0 se 360 degrees tak, guide ke liye ek halke sine curve ke saath. Step 1: arrow 30 degrees par hai; vertical axis par ek purple bar uska vertical projection dikhata hai, v m ka aadha, aur ek dashed line yeh height graph par ek dot tak le jaati hai. Step 2: 90 degrees par arrow seedha upar point karta hai aur dot curve ke top par hai. Step 3: 180 degrees par arrow left point karta hai, projection zero hai aur dot axis par hai. Step 4: 270 degrees par arrow neeche point karta hai aur dot curve ke bottom par hai. Step 5: ek poore turn ke baad arrow wapas horizontal axis par hai aur bana hua curve ek poori sine wave hai.

Definition

phasor

Ek rotating arrow jo sinusoidally badalti quantity ko represent karta hai: iski length amplitude hai, yeh angular frequency $ω$ se anticlockwise ghoomta hai, aur vertical axis par iska projection instantaneous value deta hai.

ExampleVoltage $v = 325 \sin 314 t$ volts ko $325 V$ length ka phasor represent karta hai jo $314 r a d / s$ se ghoomta hai.

Current $i = i_{m} \sin (ω t + ϕ)$ ko ek doosra phasor represent karta hai, jiski length $i_{m}$ hai, jo same $ω$ se ghoomta hai lekin voltage phasor se angle $ϕ$ aage rehta hai. Kyunki dono phasors saath ghoomte hain, unke beech ka angle kabhi nahi badalta. Isliye hum aam taur par ek pal ka phasor diagram banate hain aur rotation ko samjha hua maan lete hain. Isse do cheezein seedhe padhi ja sakti hain. Lengths amplitudes deti hain, aur do phasors ke beech ka angle unka phase difference deta hai. Jo phasor doosre se anticlockwise ghooma hua hai, woh usse aage hai, yaani use lead karta hai; jo clockwise ghooma hai, woh peeche lag karta hai.

Resistor ke liye $𝐕$ aur $𝐈$ har pal same direction mein point karte hain, isliye unke beech phase difference zero hai. Baaki do cases, jahan phasors ek doosre se right angle par hote hain, humein agle do sections mein chahiye honge.

7.4 Inductor par lagaya gaya AC voltage

Ab resistor ki jagah ek coil lagao. Hum ek ideal coil lete hain, pure inductance $L$ aur bilkul resistance nahi (moti copper wire ki coil iske kareeb hoti hai), jo source $v = v_{m} \sin ω t$ ke across judi hai. Kaun sa current behta hai?

DerivationPure inductor se current

Coil mein changing current ek back emf $- L d i / d t$ induce karta hai (Chapter 6). Circuit mein resistance na hone par loop rule deta hai

v - L \frac{d i}{d t} = 0

(7.10)

Back emf ka minus sign Lenz's law hai: coil apne current mein kisi bhi change ko oppose karti hai. Rearrange karne par,

\frac{d i}{d t} = \frac{v}{L} = \frac{v_{m}}{L} \sin ω t

(7.11)

Source voltage current ka rate of change fix karta hai, current khud nahi. $i$ nikaalne ke liye time ke respect mein integrate karo:

i = \int \frac{v_{m}}{L} \sin ω t d t = - \frac{v_{m}}{ω L} \cos ω t + constant

Constant ka kya? Woh oscillation ke upar savaar ek steady, dc jaisa current hota, aur is circuit mein kuch bhi use chalta nahi rakh sakta: har cycle mein source ek taraf utni hi der aur utne hi zor se dhakelta hai jitna doosri taraf. Isliye hum constant ko zero rakhte hain. $- \cos ω t = \sin (ω t - π / 2)$ likhkar,

i = i_{m} \sin (ω t - \frac{π}{2}), i_{m} = \frac{v_{m}}{ω L}

(7.12)

Ise resistor ke $i_{m} = v_{m} / R$ ke saath rakho: yahan $ω L$ resistance ka kaam karta hai. Hum ise inductive reactance (प्रेरकीय प्रतिघात) kehte hain,

X_{L} = ω L = 2 π ν L

(7.13)

taaki current amplitude ho

i_{m} = \frac{v_{m}}{X_{L}}

(7.14)

Reactance ki unit resistance wali hai, ohm: $1 H \times 1 s^{- 1} = 1 V s A^{- 1} \times 1 s^{- 1} = 1 Ω$ . Badi coil ya zyada frequency ka matlab bada reactance aur chhota current, jabki resistor har frequency ko barabar oppose karta hai.

(7.12) ko (7.1) se compare karo. Current har peak par voltage ke ek quarter period, $T / 4$ , baad pahunchta hai: pure inductor mein current voltage se $π / 2$ peeche lag karta hai. Fig. 7.6 sirf (7.11) use karke dikhata hai ki aisa kyun hai. Jahan $v$ sabse bada hai, wahan $d i / d t$ sabse bada hai, isliye current sabse tezi se chadh raha hai. Jahan $v$ zero se guzarta hai, wahan $d i / d t = 0$ , isliye current us pal flat hai, crest ya trough par. Phasor diagram mein $𝐈$ , $𝐕$ se $90^{\circ}$ peeche chalta hai.

4/4Current voltage ke $T / 4$ baad peak karta hai; phasor diagram mein $𝐈$ , $𝐕$ se $90^{\circ}$ peeche chalta hai.

Fig. 7.6Pure inductor mein voltage current ka slope set karta hai, isliye current quarter cycle baad peak karta hai.NCERT Fig. 7.6 · p. 183

Ek cycle par omega t ke saath v aur i ka graph. Step 1: voltage v ek blue sine wave ki tarah. Step 2: chhote orange line segments woh slope dikhate hain jo current ka hona chahiye: jahan voltage peak par hai wahan sabse steep aur upar jaata hua, jahan voltage zero cross karta hai wahan flat, aur jahan voltage sabse negative hai wahan sabse steep aur neeche jaata hua. Dotted lines voltage peak aur zero crossing ko in slopes se jodti hain. Step 3: dashed orange current curve in segments se hokar banaya gaya hai; yeh right shift hui sine wave hai. Step 4: T over 4 likha ek bracket voltage peak ko baad wale current peak se jodta hai, aur right mein ek chhota phasor diagram current phasor I ko voltage phasor V se right angle peeche dikhata hai, I lags V shabdon ke saath.

Definition

inductive reactance

Alternating current ke against inductor ka opposition, $X_{L} = ω L$ , ohms mein naapa jaata hai; yeh frequency ke proportion mein badhta hai.

Example $50 H z$ par $0.10 H$ coil ka $X_{L} = 2 π \times 50 \times 0.10 = 31.4 Ω$ hai.

Resistor milne wali energy ko heat mein badal deta hai. Ideal inductor us energy ka kya karta hai?

DerivationPure inductor ko di gayi power

Har pal inductor ko di gayi power hai

p_{L} = i v = i_{m} \sin (ω t - \frac{π}{2}) v_{m} \sin ω t = - i_{m} v_{m} \cos ω t \sin ω t

p_{L} = - \frac{i_{m} v_{m}}{2} \sin 2 ω t

Ek poori cycle par $\sin 2 ω t$ jitni der positive hota hai utni hi der negative, isliye uska average zero hai:

P_{L} = ⟨ p_{L} ⟩ = - \frac{i_{m} v_{m}}{2} ⟨ \sin 2 ω t ⟩ = 0

Ideal inductor koi average power nahi leta.

Energy ke saath kya hota hai, yeh Fig. 7.7 mein dikhaya gaya hai. Jis quarter cycle mein current ka size badhta hai, source work karta hai aur coil apne magnetic field mein energy $\frac{1}{2} L i^{2}$ store karti hai. Agle quarter mein current ghatta hai aur coil wahi energy source ko wapas de deti hai. Energy har cycle mein do baar aage-peeche jaati hai, lekin koi energy lost nahi hoti.

3/3Magnetic field mein store hui energy (+) quarter cycle baad source ko wapas (−) di jaati hai.

Fig. 7.7Ideal inductor mein jaane wali power ek quarter cycle positive aur agle mein negative hoti hai: average zero hai.

Ek cycle par omega t ke saath ek graph. Step 1: voltage v ek solid blue sine wave ki tarah aur current i ek dashed orange wave ki tarah, jo usse quarter cycle peeche hai. Step 2: voltage aur current fade hote hain aur ek moti green curve power p barabar v times i dikhati hai; yeh do guni frequency par oscillate karti hai, zero ke around symmetric. Step 3: power curve ke axis se upar wale loops green shade kiye gaye hain, plus signs ke saath aur stored label ke saath; neeche wale loops orange shade kiye gaye hain, minus signs ke saath aur returned label ke saath. Ek note kehta hai ki average zero hai.

Example 7.2NCERT Example 7.2 · p. 183

Kya yeh coil mains par safe hai?

Spare-parts box mein ek coil par likha hai “ $120 m H$ , maximum current $5.5 A$ ”; iska resistance negligible hai. Kya ise akele $230 V$ , $50 H z$ mains ke across jodna safe hoga? Kya yeh $230 V$ , $60 H z$ supply par safe hoga?

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

7.5 Capacitor par lagaya gaya AC voltage

Capacitor ki plates ke beech insulator hota hai, isliye pehli nazar mein lagta hai ki usse koi current guzarna hi nahi chahiye. Mains supply kuch aur dikhati hai: $230 V$ socket ke across juda capacitor jab tak juda rehta hai, usmein steady alternating current behta rehta hai. Trick yeh hai ki charge ko kabhi gap cross karne ki zaroorat nahi padti. Ek plate par aata charge doosri plate se utna hi charge dhakel deta hai, isliye jab tak capacitor charge ho raha hai, dono taraf ki wires mein same current behta hai. Battery ke saath yeh sirf ek baar hota hai, ek chhota surge jo plates bhar jaane par khatam ho jaata hai. Alternating voltage baar-baar reverse hota rehta hai, isliye plates bharti hain, khaali hoti hain, ulti taraf se bharti hain aur phir khaali hoti hain, second mein pachaas baar, aur wires mein current kabhi nahi rukta.

DerivationCapacitor circuit mein current

Maan lo time $t$ par capacitor par charge $q$ hai. Tab uski plates ke across voltage hai

v = \frac{q}{C}

(7.15)

Loop rule ise source voltage ke barabar rakhta hai, isliye $q = C v_{m} \sin ω t$ . Current is charge ka rate of change hai:

i = \frac{d q}{d t} = \frac{d}{d t} (C v_{m} \sin ω t) = ω C v_{m} \cos ω t

$\cos ω t = \sin (ω t + π / 2)$ use karke,

i = i_{m} \sin (ω t + \frac{π}{2}), i_{m} = ω C v_{m} = \frac{v_{m}}{1 / ω C}

(7.16)

Capacitor ke liye, resistor mein jahan $R$ tha wahan aane wali quantity $1 / ω C$ hai, yaani capacitive reactance (धारितीय प्रतिघात),

X_{C} = \frac{1}{ω C} = \frac{1}{2 π ν C}

(7.17)

aur current amplitude hai

i_{m} = \frac{v_{m}}{X_{C}}

(7.18)

$X_{C}$ bhi ohms mein naapa jaata hai, lekin yeh $X_{L}$ ka ulta behave karta hai: frequency ya capacitance mein se kisi ko bhi badhane par yeh chhota ho jaata hai, isliye $ν$ ya $C$ ko double karne se $X_{C}$ aadha ho jaata hai. High frequency par bada capacitor ac ko aasani se guzarne deta hai; zero frequency par, yaani dc par, $X_{C}$ infinite hota hai aur koi steady current nahi behta.

Is baar current har peak par voltage se ek quarter period pehle pahunchta hai: pure capacitor mein current voltage se $π / 2$ aage chalta hai. Fig. 7.8 ek cycle mein plates ko follow karta hai. Current tab sabse bada hota hai jab plates uncharged hoti hain aur voltage sabse tezi se badal raha hota hai; jis pal plates par sabse zyada charge hota hai, current zero hota hai. Phasor diagram mein $𝐈$ , $𝐕$ se $90^{\circ}$ aage hai.

5/5Current voltage se $T / 4$ pehle peak karta hai: $𝐈$ , $𝐕$ se $90^{\circ}$ aage hai.

Fig. 7.8Capacitor mein current tab sabse bada hota hai jab plates khaali hoti hain: yeh voltage se quarter cycle aage chalta hai.NCERT Fig. 7.8 · p. 185

Left mein ek parallel-plate capacitor jiske wires do terminals tak jaate hain; right mein ek cycle par omega t ke saath voltage v (solid blue) aur current i (dashed orange) ka graph, aur dots current pal ko mark karte hain. Step 1, 0 degrees par: plates par koi charge nahi, voltage zero hai, aur ek bada arrow upar wali plate par current aata dikhata hai; current apne peak par hai. Step 2, 90 degrees par: upar wali plate plus signs se aur neeche wali minus signs se bhari hai, voltage peak par hai aur current zero hai. Step 3, 180 degrees par: plates phir uncharged hain aur arrow doosri taraf point karta hai; current apne negative peak par hai. Step 4, 270 degrees par: charges ulat gaye hain, upar minus, voltage sabse negative hai aur current zero hai. Step 5: capacitor ki jagah ek phasor diagram aata hai jismein current phasor I voltage phasor V se right angle aage hai, aur graph par T over 4 likha bracket dikhata hai ki current peak voltage peak se pehle aata hai.

Definition

capacitive reactance

Alternating current ke against capacitor ka opposition, $X_{C} = 1 / ω C$ , ohms mein naapa jaata hai; frequency badhne par yeh ghatta hai.

Example $50 H z$ par $10 μ F$ capacitor ka $X_{C} = 1 / (2 π \times 50 \times 10^{- 5}) = 318 Ω$ hai.

DerivationPure capacitor ko di gayi power

Instantaneous power hai

p_{C} = i v = i_{m} \cos ω t v_{m} \sin ω t = \frac{i_{m} v_{m}}{2} \sin 2 ω t

(7.19)

Ek baar phir ek cycle par $\sin 2 ω t$ ka average zero hai, isliye $P_{C} = 0$ . Capacitor quarter cycle tak apne electric field mein energy $q^{2} / 2 C$ store karta hai aur agle quarter mein use wapas kar deta hai. Power curve ki shape Fig. 7.7 mein inductor wali jaisi hi hai, bas store aur return wale quarters aapas mein badle hue hain.

Compare

Ac par resistor, inductor aur capacitor

Resistor

Opposition $R$ , har frequency par same
Current voltage ke saath in phase
Average power $I^{2} R$ , heat mein badalti hai

Inductor

Reactance $X_{L} = ω L$ , frequency ke saath badhta hai
Current voltage se $π / 2$ lag karta hai
Average power zero hai

Capacitor

Reactance $X_{C} = 1 / ω C$ , frequency ke saath ghatta hai
Current voltage se $π / 2$ lead karta hai
Average power zero hai

Fig. 7.9 teeno ko $R = 100 Ω$ , $L = 0.25 H$ aur $C = 20 μ F$ ke liye compare karta hai.Yaad rakhne ki trick: “ELI the ICE man” mein inductor (L) mein voltage (E) current (I) se pehle aata hai, aur capacitor (C) mein current voltage se pehle aata hai. Low frequencies par capacitor ka reactance bahut bada hota hai aur inductor ka bahut chhota; high frequencies par ulta. Lagbhag $71 H z$ par dono reactances barabar hote hain, ek sanyog jo Section 7.6.2 mein bahut important nikalta hai.

4/4 $ν \approx 71 H z$ par dono reactances barabar hain, har ek lagbhag $112 Ω$ .

Fig. 7.9

X_{L}

frequency ke saath badhta hai,

X_{C}

ghatta hai, aur

R

same rehta hai; ek frequency par

X_{L} = X_{C}

Frequency (0 se 200 hertz) ke saath ohms mein opposition (0 se 300) ka graph. Step 1: 100 ohms par ek dotted horizontal line, R label ke saath, dikhati hai ki resistance frequency par depend nahi karta. Step 2: origin se guzarti ek seedhi orange line dikhati hai ki inductive reactance X L barabar omega L frequency ke proportion mein badhta hai. Step 3: ek dashed blue curve capacitive reactance X C barabar 1 over omega C dikhata hai, jo low frequency par bahut bada hai aur frequency badhne par lagatar ghatta hai. Step 4: ek dot woh jagah mark karta hai jahan line aur curve cross karte hain, lagbhag 71 hertz aur 112 ohms par, X L barabar X C label ke saath.

Example 7.3NCERT Example 7.3 · p. 185

Resistor ya capacitor? Do lamp boxes

Do identical filament lamps mein se har ek ek sealed box mein chhupe component ke saath series mein juda hai. Box P mein resistor aur box Q mein capacitor chhupa hai, aise chune gaye ki $230 V$ , $50 H z$ mains par dono lamps barabar chamakte hain. Predict karo kya hota hai (a) jab dono boxes ko $230 V$ dc supply par le jaaya jaata hai; (b) jab, wapas mains par, har box ke chhupe part ki jagah uski do identical copies parallel mein jodi jaati hain; (c) use ke dauraan kaun sa box garam hota hai.

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

Example 7.4NCERT Example 7.4 · p. 185

Appliance ke andar mains ke across laga capacitor

Kai appliances mein ek chhota “X-capacitor” mains leads ke across hamesha ke liye juda hota hai, taaki high-frequency interference short-circuit ho jaaye. Aam taur par yeh $0.47 μ F$ ka hota hai. $230 V$ , $50 H z$ mains par nikaalo (a) iska reactance, (b) isse guzarne wale rms aur peak currents aur (c) yeh kitni average power leta hai. (d) Wahi appliance ek aise country mein becha jaata hai jahan $230 V$ , $60 H z$ supply hai. Current kaise badalta hai?

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

Example 7.5NCERT Example 7.5 · p. 186

Atka hua solenoid kyun jal jaata hai

Kai machines valve ya lock ko ac solenoid se kholti hain: ek coil jo switch on hone par soft-iron plunger ko apne andar kheench leti hai. Aisi ek coil ka resistance $40 Ω$ hai aur yeh $230 V$ , $50 H z$ mains par chalti hai. Plunger andar hone par iska inductance $1.2 H$ hai, lekin plunger bahar hone par sirf $0.20 H$ . (a) Plunger inductance ko itna kyun badalta hai? (b) Plunger andar aur bahar, dono haalat mein rms current aur wire mein heating power nikaalo. (c) Maker warning deta hai ki agar plunger atak jaaye aur andar na ja sake, to coil kuch minutes mein jal jaati hai. Samjhao kyun.

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

(c) Normally plunger sirf us second ke hisse tak bahar rehta hai jitna use andar jhatke se aane mein lagta hai, isliye yeh chhota surge koi nuksaan nahi karta. Agar yeh atak jaaye, to coil design se lagbhag $381 / 14.7 \approx 26$ guna rate par garam hoti rehti hai, aur uski patli wire aur insulation overheat ho jaate hain. Dono cases mein coil ka resistance same hai: normally inductance, aur isliye reactance, hi current ko chhota rakhta hai.

Quick check

4 mein se 1

50 H z

par

10 μ F

capacitor ka reactance nikaalo, ohms mein.

Answer dekho

Answer 318.3 Ω

X_{C} = 1 / (2 π ν C) = 1 / (2 π \times 50 \times 10 \times 10^{- 6}) \approx 318 Ω

7.6 Series LCR circuit par lagaya gaya AC voltage

Zyadatar asli circuits mein ek se zyada tarah ke elements hote hain. Ek resistor $R$ , ek inductor $L$ aur ek capacitor $C$ ko source $v = v_{m} \sin ω t$ ke saath series mein jodo. Time $t$ par capacitor par charge $q$ hai aur current $i = d q / d t$ hai. Loop mein ghoomte hue, source voltage teeno potential drops ke sum ke barabar hona chahiye:

L \frac{d i}{d t} + i R + \frac{q}{C} = v

(7.20)

Humein current aur source voltage ke relative uska phase chahiye. Equation calculus se solve ho sakti hai, lekin phasor diagram sirf ek right-angled triangle se answer tak pahunch jaata hai.

7.6.1 Phasor-diagram solution

Series loop mein kahin bhi charge jama nahi ho sakta, isliye $R$ , $L$ aur $C$ mein current time ka ek hi function hai. Isliye phasor diagram ke liye current natural reference hai.

DerivationSeries LCR circuit ka impedance aur phase angle

Current ko aise likho

i = i_{m} \sin (ω t + ϕ)

(7.21)

jahan $ϕ$ , jo abhi pata nahi, source voltage ke relative current ka phase hai. Sections 7.2, 7.4 aur 7.5 ke results batate hain ki current phasor $𝐈$ ke against har element ka voltage phasor kaise draw karein:

resistor: $𝐕_{R}$ , $𝐈$ ki direction mein point karta hai;
inductor: $𝐕_{L}$ , $𝐈$ se $90^{\circ}$ anticlockwise ghooma hai;
capacitor: $𝐕_{C}$ , $𝐈$ se $90^{\circ}$ clockwise ghooma hai.

Inki lengths har element ke current ke opposition se milti hain:

v_{R m} = i_{m} R, v_{L m} = i_{m} X_{L}, v_{C m} = i_{m} X_{C}

(7.22)

Loop rule (7.20) kehta hai ki har pal teeno voltages jud kar source voltage dete hain,

v_{R} + v_{L} + v_{C} = v

(7.23)

Har instantaneous voltage apne phasor ka vertical projection hai, aur sum ka projection projections ka sum hota hai. Isliye phasors bhi isi tarah judne chahiye:

𝐕_{R} + 𝐕_{L} + 𝐕_{C} = 𝐕

(7.24)

Fig. 7.10 yeh addition tip to tail karta hai. Kyunki $𝐕_{L}$ aur $𝐕_{C}$ antiparallel hain, unka sum bas unka difference $∣ v_{C m} - v_{L m} ∣$ hai, jo $𝐈$ ke perpendicular draw hota hai. Ise $𝐕_{R}$ ki tip par rakhne se ek right-angled triangle band hota hai jiski sabse lambi side $𝐕$ hai, isliye

v_{m}^{2} = v_{R m}^{2} + (v_{C m} - v_{L m})^{2} = i_{m}^{2} [R^{2} + (X_{C} - X_{L})^{2}]

isliye current amplitude hai

i_{m} = \frac{v_{m}}{\sqrt{R^{2} + (X_{C} - X_{L})^{2}}}

(7.25a)

Iski shape Ohm's law jaisi hai,

i_{m} = \frac{v_{m}}{Z}

(7.25b)

jahan resistance ki jagah impedance (प्रतिबाधा) hai

Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}

(7.26)

Origin par is triangle ka angle, $𝐕$ aur current direction ke beech, phase angle hai. Opposite side over adjacent side se milta hai

\tan ϕ = \frac{v_{C m} - v_{L m}}{v_{R m}} = \frac{X_{C} - X_{L}}{R}

(7.27)

6/7 $𝐕$ triangle ko band karta hai: $v_{m} = \sqrt{240^{2} + 100^{2}} = 260 V$ ; current $ϕ = {22.6}^{\circ}$ se lag karta hai.

Fig. 7.10Series LCR circuit ke voltage phasors jodne se ek right-angled triangle milta hai;

i_{m}

se divide karne par impedance triangle milta hai.NCERT Fig. 7.11 · p. 187

Left mein ek series circuit: neeche wali wire par ek ac source, ek 120 ohm resistor, 90 ohm reactance wala inductor aur 40 ohm reactance wala capacitor, jismein 2.0 amperes peak ka current behta hai. Right mein ek phasor diagram. Step 1: current phasor I horizontal axis ke saath reference ki tarah draw kiya gaya hai. Step 2: resistor ka voltage phasor V R, 240 volts, isi direction mein joda gaya hai. Step 3: inductor ka voltage phasor V L, 180 volts, seedha upar point karta hai, I se 90 degrees aage. Step 4: capacitor ka voltage phasor V C, 80 volts, seedha neeche point karta hai, I se 90 degrees peeche. Step 5: teeno voltage phasors fade hote hain aur V L aur V C ka difference, 100 volts upar ki taraf, V R ki tip se draw kiya jaata hai. Step 6: source phasor V, 260 volts, origin se us point tak draw kiya jaata hai, I se angle phi upar. Step 7: har side ko current se divide kiya jaata hai, jisse impedance triangle milta hai: R barabar 120 ohm, X L minus X C barabar 50 ohm aur hypotenuse Z barabar 130 ohm.

Voltage triangle ki har side ko $i_{m}$ se divide karo, to tumhe impedance triangle milta hai jo Fig. 7.10 ke aakhri step mein dikhaya gaya hai: ek right-angled triangle jiski sides $R$ aur $∣ X_{C} - X_{L} ∣$ hain, hypotenuse $Z$ hai, aur $R$ aur $Z$ ke beech angle $ϕ$ hai. Kisi bhi series-circuit problem ko shuru karne ka sabse tez tareeka iska sketch banana hai, aur isse ek aur useful result milta hai, $\cos ϕ = R / Z$ .

Definition

impedance

Ac circuit ka current ke against total opposition, $Z = v_{m} / i_{m} = V / I$ , ohms mein naapa jaata hai; series LCR circuit ke liye $Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}$ .

ExampleSeries mein $R = 120 Ω$ , $X_{L} = 90 Ω$ aur $X_{C} = 40 Ω$ se $Z = 130 Ω$ milta hai.

Fig. 7.10 mein $X_{L} = 90 Ω$ , $X_{C} = 40 Ω$ se bada hai, isliye $\tan ϕ = (40 - 90) / 120 = - 0.417$ aur $ϕ = - {22.6}^{\circ}$ : current source voltage se peeche chalta hai. Kisi bhi series circuit ke liye $X_{C} - X_{L}$ ka sign yeh sawaal tay karta hai. Capacitive se zyada inductive reactance $ϕ$ ko negative banata hai; aise circuit ko inductive kehte hain, aur uska current lag karta hai. Zyada capacitive reactance $ϕ$ ko positive banata hai; circuit capacitive hai, aur uska current lead karta hai.

Upar ki saari baatein maan kar chalti hain ki circuit itni der se chal raha hai ki ek regular rhythm mein aa gaya hai, aur current source frequency par khud ko repeat kar raha hai. Is settled behaviour ko steady state kehte hain. Switch on karne ke baad pehle kuch palon mein current is par bhi depend karta hai ki shuruaat kaise hui: capacitor par kuch charge tha ya nahi, ya source apni cycle mein kahan tha. Yeh extra, irregular hissa transient hai, aur resistance jaise-jaise uski energy ko heat mein badalta hai, yeh kuch cycles mein fade ho jaata hai.Oscilloscope par transient switch band karne ke theek baad trace ke ek chhote, tedhe-medhe hisse ki tarah dikhta hai. Phasors sirf steady state describe karte hain, isiliye humein kabhi batana nahi pada ki diagram kis pal draw kiya gaya.

InteractivePhasor lab

Voltage aur current ke graphs ke saath ek interactive phasor diagram. Tabs se resistor, inductor, capacitor ya series LCR circuit chuna jaata hai, jise 10 volt peak source chalata hai. Phasors dheere ghoomte hain, ya pause karke 30 degrees ke steps mein aage badhaye ja sakte hain. Sliders frequency ko 10 se 250 hertz tak, aur resistance, inductance aur capacitance set karte hain. Readouts reactances, impedance, peak current, current ka phase, power factor aur average power dete hain; LCR tab mein yeh har element ke across peak voltages aur resonant frequency bhi dete hain, aur ek status line batati hai ki circuit capacitive hai, inductive hai ya resonance par hai.

Fig. 7.11Ek element chuno, frequency aur component values badlo, aur phasors aur voltage aur current ke waveforms dekho.

Try karke dekho

Phasor lab explore karo

L only chuno. Kaun sa phasor aage hai, $𝐕$ ya $𝐈$ ? Frequency ko $50 H z$ se $100 H z$ tak badhao aur $X_{L}$ aur peak current dekho.
C only chuno aur yahi karo. Is baar current kis taraf badalta hai?
Series LCR chuno. Status line batati hai ki $X_{C}$ bada hai ya $X_{L}$ . Frequency ko tab tak slide karo jab tak current “in phase” na padhe. Us frequency ko resonance readout $f_{0}$ se compare karo, aur $V_{L}$ aur $V_{C}$ readings ko $10 V$ source se compare karo.

L only: $𝐈$ , $𝐕$ se $90^{\circ}$ peeche chalta hai. Frequency double karne se $X_{L}$ double hota hai ( $31.4 Ω$ se $62.8 Ω$ ) aur peak current aadha ho jaata hai ( $318 m A$ se $159 m A$ ). C only: $𝐈$ $90^{\circ}$ aage hai, aur frequency double karne se $X_{C}$ aadha hota hai ( $159 Ω$ se $79.6 Ω$ ) aur current double ho jaata hai (lagbhag $63 m A$ se $126 m A$ ). Series LCR: current lagbhag $79.6 H z$ par in phase aata hai, jo $L = 200 m H$ aur $C = 20 μ F$ ke liye $1 / (2 π \sqrt{L C})$ hai. Wahan $V_{L}$ aur $V_{C}$ barabar hain, har ek lagbhag $16.7 V$ , jo source voltage se zyada hai, aur woh ek doosre ko cancel karte hain.

Example 7.6NCERT Example 7.6 · p. 190

Do voltmeter readings jo jud kar bahut zyada ho jaati hain

$100 Ω$ resistance ka ek lamp (ise pure resistor maano) $0.50 H$ inductance aur negligible resistance wale choke ke saath series mein $230 V$ , $50 H z$ mains ke across juda hai. (a) Rms current nikaalo. (b) Ek student ac voltmeter se lamp ke across aur choke ke across rms voltage naapta hai, dono readings jodta hai aur $230 V$ se kahin zyada paata hai. Kya meter kharab hai?

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

7.6.2 Resonance

Kisi purani bus mein baitho to shayad tum dekho ki ek khidki ka sheesha engine ki ek khaas speed par zor se kharkharata hai aur baaki speeds par chup rehta hai. Sheeshe ki apni ek natural frequency hoti hai. Jab engine use usi frequency par hilata hai, to har chhota jhatka pichhle mein jud jaata hai aur vibration bada ho jaata hai. Jo system apni natural frequency par ya uske paas drive kiya jaaye, woh zor se respond karta hai; yeh resonance (अनुनाद) hai. Series LCR circuit ki bhi ek natural frequency hoti hai.

DerivationResonant frequency

(7.25a) ke impedance ko dono reactances poore likh kar likho:

i_{m} = \frac{v_{m}}{Z} = \frac{v_{m}}{\sqrt{R^{2} + {(\frac{1}{ω C} - ω L)}^{2}}}

Squared bracket kabhi negative nahi ho sakta, isliye $Z$ kabhi $R$ se chhota nahi ho sakta. Low frequencies par $1 / ω C$ bada term hota hai aur high frequencies par $ω L$ (Fig. 7.9), isliye bilkul ek angular frequency $ω_{0}$ inhe balance karti hai aur bracket ko khaali kar deti hai:

ω_{0} L = \frac{1}{ω_{0} C}

ω_{0} = \frac{1}{\sqrt{L C}}, ν_{0} = \frac{1}{2 π \sqrt{L C}}

(7.28)

$ω_{0}$ par source ko sirf resistance dikhta hai. Current amplitude apni ceiling $v_{m} / R$ par pahunchta hai, aur kyunki $\tan ϕ = 0$ , current bilkul source voltage ke saath-saath badhta aur ghatta hai. $ω_{0}$ circuit ki resonant frequency hai.

Resistance kya badalta hai? Fig. 7.12 iska jawab $L = 0.50 m H$ aur $C = 8.0 n F$ wale circuit ke liye deta hai, jise $10 V$ peak voltage ka source chalata hai. Kyunki (7.28) mein $R$ kahin nahi aata, $R = 40 Ω$ aur $R = 100 Ω$ ke curves ek hi jagah peak karte hain, $ω_{0} = 1 / \sqrt{L C} = 5.0 \times 10^{5} r a d / s$ . $R$ peak ki height $v_{m} / R$ ( $0.25 A$ banaam $0.10 A$ ) aur uski width control karta hai: chhota resistance zyada ooncha, zyada patla peak deta hai. $ω_{0}$ ke left mein capacitor ka reactance jeetta hai aur current lead karta hai; right mein inductor ka jeetta hai aur current lag karta hai.

3/4 $R = 100 Ω$ ke saath peak usi $ω_{0}$ par hai lekin neecha ( $0.10 A$ ) aur zyada chauda hai.

Fig. 7.12Current

ω_{0} = 1 / \sqrt{L C}

par peak karta hai; chhota resistance zyada ooncha, zyada sharp peak deta hai.NCERT Fig. 7.14 · p. 189

Angular frequency (0 se 10 times 10 to the 5 radians per second) ke saath amperes mein peak current (0 se 0.3) ka graph, us circuit ke liye jismein L barabar 0.50 millihenry, C barabar 8.0 nanofarad aur 10 volt source hai. Step 1: R barabar 40 ohm ka blue curve dheere badhta hai, 0.25 ampere par sharp peak karta hai aur phir girta hai. Step 2: ek dotted line omega 0 barabar 5 times 10 to the 5 radians per second par peak mark karti hai, i m barabar v m over R label ke saath. Step 3: R barabar 100 ohm ka dashed orange curve usi frequency par peak karta hai lekin sirf 0.10 ampere tak, aur zyada chauda hai. Step 4: peak ke left wala region shade kiya gaya hai aur us par capacitive, i leads v likha hai, aur right wale region par inductive, i lags v likha hai.

Peak ki sharpness circuit ke quality factor se naapi jaati hai, jo ek pure number hai:

Q = \frac{ω_{0} L}{R} = \frac{1}{ω_{0} C R}

Upar ke curves ke liye $ω_{0} L = 250 Ω$ , isliye $R = 40 Ω$ hone par $Q = 6.25$ aur $R = 100 Ω$ hone par $Q = 2.5$ . Resonance par inductor ke across voltage $i_{m} X_{L} = (v_{m} / R) ω_{0} L = Q v_{m}$ hai, aur capacitor ke across bhi utna hi. High- $Q$ circuit $L$ aur $C$ ke across source voltage se kai guna bade voltages bana leta hai, bhale hi loop mein dono ek doosre ko cancel kar dete hain.

Radio ke aerial tak ek saath darjanon stations pahunchte hain, har ek apni carrier frequency par. Set apne tuning circuit se ek ko chunta hai, ek series LCR circuit jismein har station ka signal ek alag chhote source ki tarah kaam karta hai. $L = 0.20 m H$ aur $R = 20 Ω$ lo, aur tuning capacitor ko $156 p F$ par ghumao: isse $ν_{0}$ $900 k H z$ par aa jaata hai. Ab $900 k H z$ wala station $1250 k H z$ wale utne hi strong station se lagbhag 37 guna, aur $610 k H z$ wale se lagbhag 45 guna zyada current chalata hai, isliye sirf wahi sunai deta hai. Knob ghumane se $C$ badalta hai aur $ν_{0}$ dial par khisakta hai; zyada $Q$ choice ko aur sharp banata hai.

InteractiveResonance lab

Do tabs. Resonance curve: 0.50 millihenry inductor aur 10 volt source wale series LCR circuit ke liye angular frequency ke saath peak current ka graph, drive frequency, resistance aur capacitance ke sliders ke saath. Ek dot drive frequency par current mark karta hai aur ek dotted line resonant frequency mark karti hai; ek button abhi ke curve ko comparison ke liye dashed outline ki tarah pin kar deta hai. Readouts reactances, impedance, current, phase, har element ke across peak voltages aur quality factor dete hain. Radio tuner: 500 se 1500 kilohertz tak ka dial jis par teen stations hain, ek tuning-capacitor slider aur ek resistance slider; bars dikhate hain ki har station circuit mein kitna current chalata hai aur readouts sabse loud station ka naam batate hain.

Fig. 7.13Drive frequency ko resonance peak ke aar-paar sweep karo, resistances compare karo, phir ek radio tune karo.

Try karke dekho

Peak dhoondho, phir radio tune karo

Resonance curve mein drive frequency ko tab tak drag karo jab tak curve par dot peak ke upar na baith jaaye. Drive $ω$ ko $ω_{0}$ readout se, aur $V_{L}$ aur $V_{C}$ readings ko $10 V$ source se compare karo.
Pin this curve dabao, phir $R$ ko $40 Ω$ se $100 Ω$ tak badhao. Peak ki height aur width kaise badalti hai? Kya $ω_{0}$ khisakta hai?
Ab $C$ ko chhota karo. Peak kis taraf khisakta hai?
Radio tuner kholo aur tuning capacitor ko tab tak ghumao jab tak Station B ( $900 k H z$ ) sabse loud na ho jaaye. Iske liye kitna capacitance chahiye? Phir $R$ ko $300 Ω$ tak badhao. Kya ab bhi tum sirf B ko sun sakte ho?

Dot curve ke top par tab pahunchta hai jab drive frequency $ω_{0} = 5.00 \times 10^{5} r a d / s$ ke barabar hoti hai. Wahan $R = 40 Ω$ ke saath $V_{L} = V_{C} = 62.5 V$ , jo source voltage ka chhe guna se zyada hai, aur current in phase hai. $R = 100 Ω$ ke saath peak girkar $0.10 A$ ho jaata hai aur phail jaata hai, lekin usi $ω_{0}$ par rehta hai. Chhota $C$ , $ω_{0} = 1 / \sqrt{L C}$ ko badhata hai, isliye peak right ki taraf khisakta hai. Station B ko $C = 1 / (4 π^{2} ν^{2} L)$ se lagbhag $156 p F$ chahiye. $R = 20 Ω$ ke saath iska current agle station ka lagbhag 37 guna hai; $R = 300 Ω$ ke saath sirf lagbhag 2.7 guna, isliye paas ke stations bhi sunai denge.

7.7 Ac circuit mein power: power factor

Ek kettle, ceiling-fan motor aur tube light ka choke, sab same $230 V$ supply se same current le sakte hain aur phir bhi bilkul alag-alag power le sakte hain. Resistor ke liye $P = V I$ . Lekin general case mein current voltage se phase mein shift hota hai, aur yeh shift average power badal deta hai.

DerivationAc circuit mein average power

Maan lo source voltage $v = v_{m} \sin ω t$ hai aur current $i = i_{m} \sin (ω t + ϕ)$ , jaise series LCR circuit mein. Har pal source jo power deta hai, woh hai

p = v i = v_{m} i_{m} \sin ω t \sin (ω t + ϕ)

Identity $2 \sin A \sin B = \cos (A - B) - \cos (A + B)$ use karke,

p = \frac{v_{m} i_{m}}{2} [\cos ϕ - \cos (2 ω t + ϕ)]

(7.29)

Pehla term constant hai. Doosra source frequency ke do gune par oscillate karta hai aur ek cycle par uska average zero hai. Isliye average power hai

P = \frac{v_{m} i_{m}}{2} \cos ϕ = \frac{v_{m}}{\sqrt{2}} \frac{i_{m}}{\sqrt{2}} \cos ϕ = V I \cos ϕ

(7.30a)

Kyunki $V = I Z$ , ise aise bhi likh sakte hain

P = I^{2} Z \cos ϕ

(7.30b)

Yaani same rms voltage aur same rms current wale do circuits bilkul alag powers le sakte hain: phase angle apne cosine ke through aata hai.

Factor $\cos ϕ$ ko power factor (शक्ति गुणांक) kehte hain. Supply ko product $V I$ dene laayak hona chahiye, lekin uska sirf $\cos ϕ$ hissa hi asal mein use hota hai. Fig. 7.14 dikhata hai ki badhta phase angle average power ko kaise kha jaata hai: har cycle ka zyada se zyada hissa energy ko source ko wapas dene mein jaata hai.

2/3Current $60^{\circ}$ se lag karta hai ( $ϕ = - 60^{\circ}$ ). Har cycle ke kuch hisse mein $p < 0$ hota hai aur energy source ko wapas jaati hai: $P = V I \cos 60^{\circ} = V I / 2$ .

Fig. 7.14Phase difference jitna bada, har cycle ka utna zyada hissa power negative rehti hai, aur average power

V I \cos ϕ

utni chhoti.

Ek cycle par omega t ke saath graph, jismein voltage v solid blue sine wave hai, current i dashed orange wave hai, aur instantaneous power p moti green curve hai, uske average par ek dashed black line ke saath. Step 1: phi barabar 0, cos phi barabar 1; current voltage ke saath-saath hai, p kabhi negative nahi hota aur uska average uske peak ka aadha hai. Step 2: current 60 degrees se lag karta hai, isliye phi barabar minus 60 degrees aur cos phi barabar 0.5; p ke chhote loops axis ke neeche jaate hain, orange shade mein, aur average line neeche hai, pichhli height ke aadhe par. Step 3: current 90 degrees se lag karta hai, phi barabar minus 90 degrees aur cos phi barabar 0; axis ke upar aur neeche ke loops barabar hain aur average line axis par hai.

Definition

power factor

Average power $P = V I \cos ϕ$ mein factor $\cos ϕ$ , jahan $ϕ$ voltage aur current ke beech phase difference hai; series LCR circuit ke liye $\cos ϕ = R / Z$ .

Example $230 V$ se power factor $0.8$ par $10 A$ lene wali motor $1840 W$ leti hai.

Table 7.1 un cases ko ikattha karta hai jo tumhe sabse zyada milenge.

Table 7.1Common circuits ke liye power factor aur average powerNCERT p. 191

Circuit	Phase angle φ	Power factor	Average power	Comment
Pure resistor	$0$	$1$	$V I = I^{2} R$	Poora VI use hota hai
Pure inductor ya capacitor	$\pm 90^{\circ}$	$0$	$0$	Current wattless hai
Series RL, RC ya LCR	$\tan^{- 1} \frac{X_{C} - X_{L}}{R}$	$R / Z$	$I^{2} R$	Sirf resistor energy dissipate karta hai
Resonance par series LCR	$0$	$1$	$I^{2} R$ , jahan $I = V / R$	Us circuit ke liye sabse zyada power

Do entries par comment banta hai. Jo current voltage se quarter cycle out of step hai, jaise ideal inductor ya capacitor mein, woh energy ko bahar le jaata hai aur wapas laata hai bina kuch deliver kiye; ise wattless current (वाटहीन धारा) kehte hain. Aur kisi bhi series circuit ke liye impedance triangle $\cos ϕ = R / Z$ deta hai, isliye (7.30b) ban jaata hai $P = I^{2} Z (R / Z) = I^{2} R$ : reactances chahe jaise bhi arrange hon, har watt aakhir mein resistance mein heat ban jaata hai.

$R$ , $L$ aur $C$ ke kisi bhi combination ke liye $∣ ϕ ∣$ kabhi $90^{\circ}$ se zyada nahi hota, isliye $\cos ϕ$ kabhi negative nahi hota: aise circuit ki li hui average power kabhi negative nahi hoti, bhale hi instantaneous power aksar hoti hai. $\cos ϕ$ ko tum product $V I$ ka woh hissa samajh sakte ho jo circuit sach mein heat ya work mein badalta hai: pure resistance ke liye poora, pure reactance ke liye kuch bhi nahi.

Example 7.7NCERT Example 7.7 · p. 191

Kharab power factor theek karna

Ek workshop mein water pump $230 V$ , $50 H z$ supply se chalta hai. Uski motor $0.60$ power factor par $6.9 k W$ leti hai, aur current voltage se lag karta hai. Local transformer se aane wali cable ka total resistance $0.20 Ω$ hai. (a) Current aur cable mein waste hone wali power nikaalo. (b) Samjhao ki motor ke saath parallel mein juda capacitor supply ka power factor 1 tak kaise badha sakta hai, aur zaroori capacitance nikaalo. (c) Tab current aur cable loss kitne hain?

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

2/4 $𝐕$ ke saath $I_{p} = 30 A$ power le jaata hai; $𝐕$ se right angle peeche $I_{q} = 40 A$ wattless hai.

Fig. 7.15

𝐕

90^{\circ}

lead karne wala capacitor current jodne se lagging current ka wattless hissa cancel ho jaata hai.NCERT Fig. 7.15 · p. 192

Ek phasor diagram jismein voltage phasor V horizontal ke saath right ki taraf reference ki tarah bana hai; phasors anticlockwise ghoomte hain, isliye lagging current V ke neeche point karta hai. Step 1: motor current I, 50 amperes, V se 53 degrees ke angle phi neeche right ki taraf point karta hai, aur text kehta hai motor current 50 A, lags V by 53 degrees, power factor 0.60. Step 2: I ko do hisson mein baanta gaya hai, V ke saath 30 amperes ka green hissa I p aur 40 amperes ka dashed purple hissa I q jo I p ki tip se I ki tip tak seedha neeche point karta hai, right-angle mark ke saath; text kehta hai power part 30 A, wattless part 40 A. Step 3: 40 amperes ka capacitor current I C origin se seedha upar point karta hua draw kiya gaya hai, V se 90 degrees aage. Step 4: I C khisak kar I ki tip par head to tail rakha jaata hai, upar point karta hua; yeh bilkul V par khatam hota hai, isliye line current, yaani motor current plus capacitor current, sirf I p hai: V ke saath in phase 30 amperes, power factor 1. Motor current halka dikhaya gaya hai.

Example 7.8NCERT Example 7.8 · p. 192

Rheostat kitna garam hoga?

Ek laboratory mein $10 Ω$ rheostat, $95.5 m H$ inductance ki coil (negligible resistance) aur $531 μ F$ capacitor $230 V$ , $50 H z$ supply ke across series mein jude hain. Rheostat zyada se zyada $1.0 k W$ dissipate karne ke liye rated hai. (a) Circuit kitna rms current leta hai? (b) Kya rheostat overheat hoga? (c) Power factor kya hai, aur current supply voltage se lead karta hai ya lag, aur kitna?

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

Example 7.9NCERT Example 7.9 · p. 193

Supply ko circuit ke hisaab se tune karna

Example 7.8 wali laboratory supply mein ek frequency control bhi hai; iska voltage $230 V$ par rehta hai. (a) Kis frequency par rheostat sabse zyada power lega? (b) Woh power nikaalo. Kya rheostat ab bhi safe hai? (c) Tab capacitor ke across kitna rms voltage aata hai?

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

Example 7.10NCERT Example 7.10 · p. 193

Traffic signal ko kaise pata chalta hai ki car intezaar kar rahi hai

Kai traffic signals aur toll plazas par sadak ki surface ke theek neeche wire ka ek rectangular loop dabaya hota hai. Yeh loop ek tuned LC circuit ka inductor hai, aur detector circuit ki resonant frequency par nazar rakhta hai. Jab koi car loop ke upar rukti hai, to signal controller ko bataya jaata hai ki ek vehicle intezaar kar raha hai. Woh physics samjhao jisse daba hua loop car ko sense kar paata hai.

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

7.8 Transformers

Doorbell ko $12 V$ chahiye; national grid saikdon kilovolts le jaata hai. Dono ka kaam ek hi tarah ka device karta hai: transformer (ट्रांसफ़ॉर्मर), jo mutual induction (Chapter 6) se kaam karta hai aur ek alternating voltage ko bade ya chhote voltage mein badal sakta hai.

Transformer ke andar dekho to tumhe ek iron core aur insulated copper wire ki do alag windings milengi (Fig. 7.16). Windings electrically aapas mein judi nahi hoti. Unhe magnetic flux jodta hai, jise soft-iron core ek winding se doosri tak le jaata hai. Supply se judi winding ko primary ( $N_{p}$ turns) kehte hain; load se judi winding secondary ( $N_{s}$ turns) hai. Designers ya to dono windings ko band iron frame ki ulti sides par rakhte hain, jaise figure mein, ya ek ko seedhe doosri ke upar lapet dete hain taaki unke beech mushkil se koi flux nikal sake.

4/5Changing flux secondary coil ( $N_{s}$ turns) mein emf induce karta hai, jo load mein current chalata hai.

Fig. 7.16Primary ka banaya flux iron core secondary ke har turn se hokar le jaata hai.NCERT Fig. 7.16 · p. 194

Ek moti grey ring soft-iron core ko represent karti hai. Step 1: sirf core, soft-iron core label ke saath. Step 2: kai turns wali primary winding, ring ki left side par orange strokes ki tarah bani, ek ac source se judi hai, uske current ke liye ek arrow ke saath; is par primary, N p label hai. Step 3: arrowheads wala ek dashed blue circle ring ke andar ghoomta hai, jo core se guide hone wala magnetic flux Phi dikhata hai. Step 4: kam turns wali secondary winding ring ki right side par dikhti hai, ek jalte lamp se judi, secondary, N s label ke saath. Step 5: ring ke beech mein ratio V s over V p barabar N s over N p likha hai.

DerivationIdeal transformer

Ek ideal transformer se shuru karo: bina resistance wali windings, ek core jo koi energy waste nahi karta, aur flux ko guide karne mein itna achha core ki flux ki har line dono windings ke har turn se guzarti hai. Maan lo $Φ$ woh flux hai jo time $t$ par core mein hai, aur isliye har ek turn se bhi guzarta hai.

Faraday's law har turn par alag se lagta hai. Changing flux se guzarte ek loop ka emf $- d Φ / d t$ hai, aur kyunki core har turn ko same flux deta hai, core par har turn ka, chahe primary ho ya secondary, emf same hai. Engineers ise transformer ka volts per turn kehte hain. Series mein jude turns apne emfs jodte hain, isliye har winding ka emf uske turns ki sankhya times volts per turn hai:

ε_{s} = - N_{s} \frac{d Φ}{d t}

(7.31)

ε_{p} = - N_{p} \frac{d Φ}{d t}

(7.32)

Ab in emfs ko terminals ke voltages se jodo. Secondary side par agar load sirf chhota current le, to winding ke andar lagbhag koi voltage lost nahi hota, isliye terminal voltage $v_{s}$ bas emf hai. Primary side par supply bina resistance wali winding mein current chalati hai, isliye loop rule kuch bhi baaki nahi chhodta: har pal primary ka back emf bilkul applied voltage $v_{p}$ jitna hona chahiye. Core mein flux khud ko tab tak adjust karta hai jab tak yeh sach na ho jaaye, yaani flux kitni tezi se badlega yeh supply tay karti hai, load nahi. Har terminal voltage ko uski apni winding ke emf ke same sense mein lene par dono statements aise padhe jaate hain

v_{s} = - N_{s} \frac{d Φ}{d t}

(7.31a)

v_{p} = - N_{p} \frac{d Φ}{d t}

(7.32a)

Volts per turn dono sides par same hai, isliye ek equation ko doosri se divide karne par yeh hat jaata hai:

\frac{v_{s}}{v_{p}} = \frac{N_{s}}{N_{p}}

(7.33)

Shabdon mein, voltage turns ki sankhya ke proportion mein baant diya jaata hai. Currents nikaalne ke liye flux se energy par aao. Ideal transformer ke paas energy lose karne ki koi jagah nahi, isliye load secondary se jo bhi power leta hai, primary ko usi pal utni hi power supply se leni padti hai:

i_{p} v_{p} = i_{s} v_{s}

(7.34)

(7.33) aur (7.34) ko milakar,

\frac{i_{p}}{i_{s}} = \frac{v_{s}}{v_{p}} = \frac{N_{s}}{N_{p}}

(7.35)

Yeh saare voltages aur currents supply frequency par saath-saath oscillate karte hain, isliye yahi ratios unki peak values aur rms values ko bhi jodte hain:

V_{s} = \frac{N_{s}}{N_{p}} V_{p}, I_{s} = \frac{N_{p}}{N_{s}} I_{p}

(7.36)

(7.36) ko ek trade ki tarah padho: voltage jis factor se multiply hota hai, current usi factor se divide ho jaata hai. Secondary par zyada turns ( $N_{s} > N_{p}$ ) hone par transformer voltage badhata hai aur current ghatata hai; yeh step-up transformer hai. Secondary par kam turns hone par yeh step-down transformer hai, jo voltage ghatata hai aur current badhata hai. Aksar wahi transformer dono kaam kar sakta hai, is par depend karta hai ki kaun si winding supply se judi hai.

Example

Doorbell transformer

Ek bell transformer $230 V$ mains ko $12 V$ tak step down karta hai. Iski primary mein 2300 turns hain. (a) Secondary ko kitne turns chahiye? (b) Jab bell $1.5 A$ leti hai, to ideal transformer ki primary mein kitna current behta hai? (c) $90 %$ efficient transformer ke liye primary current kitna hoga?

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

Derivation ki har simplification ek aisi baat chhupati hai jismein asli transformer peeche reh jaata hai. Primary ka resistance supply voltage ka kuch hissa kha jaata hai, isliye emf ko balance karne ke liye kam bachta hai. Core se bahar bhatka flux primary se to guzarta hai lekin secondary ko miss kar deta hai. Aur bada load current secondary winding ke andar kuch voltage kho deta hai, isiliye load badhne par chhote adapter ka output thoda gir jaata hai. Phir bhi asli transformers ideal ke kaafi kareeb aate hain: bade transformers milne wali power ka $95 %$ se zyada aage bhejte hain, aur sabse bade $99 %$ se zyada. Jo energy woh khote hain, lagbhag saari heat ki tarah, woh chaar main tareekon se jaati hai (Table 7.2).

Table 7.2Asli transformer mein energy lossesNCERT p. 195

Loss	Kya hota hai	Ise kaise kam karte hain
Flux leakage	Kuch field lines core se bahar leak ho jaati hain aur secondary se kabhi nahi guzarti, isliye uska emf turns ratio ke predict kiye emf se thoda kam reh jaata hai.	Band core par ek coil ko seedhe doosri ke upar lapeto, taaki dono lagbhag saara flux share karein.
Windings ka resistance	Current copper wire ko rate $I^{2} R$ par garam karta hai.	Moti copper use karo, khaas kar us winding par jo zyada current le jaati hai (low-voltage side).
Eddy currents	Changing flux iron ke andar current ke loops ghumata hai, aur iron garam ho jaata hai.	Core ko patli, insulated sheets (laminations) se banao jo un loops ko tod deti hain.
Hysteresis	Core ka magnetisation har cycle mein do baar ulta hota hai, aur har baar thodi energy heat mein badal jaati hai.	Silicon steel jaisa soft magnetic material use karo, jiska magnetisation kam loss ke saath ulta hota hai.

Transformers hi long-distance transmission ko practical banate hain. Resistance $R_{ℓ}$ wali line jo current $I$ le jaati hai, power $I^{2} R_{ℓ}$ khoti hai. Di gayi power $P = V I$ ke liye voltage badhane se current usi proportion mein ghatta hai aur loss us factor ke square se kam ho jaata hai. Practice mein safar ke kai stages hote hain (Fig. 7.17). Power station ke paas ek step-up transformer long-distance lines ko saikdon kilovolts par feed karta hai. Shehar ke paas substations ise ek ya do steps mein local distribution ke liye lagbhag $11 k V$ tak kam karte hain, aur pole par ya street kiosk mein laga ek chhota transformer ise ghar mein use hone wale $230 V$ tak le aata hai.

4/4Pole par laga transformer aakhri step down $230 V$ tak karta hai.

Fig. 7.17Transmission ke liye voltage step up karne se current ghatta hai, aur uske saath lines mein heating loss bhi.NCERT p. 196

Left se right symbols ki ek chain, har stage ke upar voltage aur neeche current likha hai. Step 1: power station par generator G 11 kilovolts banata hai, aur ek mota arrow 909 amperes ka bada current dikhata hai. Step 2: step-up transformer do towers ke beech tani transmission line par voltage ko 220 kilovolts tak badhata hai, aur ek patla arrow dikhata hai ki current ab sirf 45 amperes hai. Step 3: substation transformer voltage ko 11 kilovolts tak kam karta hai. Step 4: pole transformer ise ghar ke liye phir 230 volts tak kam karta hai.

Example

Grid high voltage par kyun chalta hai

Ek power station $2.0 Ω$ total resistance wali line se $10 M W$ bhejta hai. Power factor 1 maano. Line mein lost power compare karo agar energy (a) generator voltage $11 k V$ par, aur (b) step-up transformer ke baad $220 k V$ par bheji jaaye.

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

Chalo explore karein

Charger ke andar kya hai?

Tumhe chahiye

rating label wala ek unplugged mains adapter ya charger
ek calculator
ek kitchen scale (optional)

Pehle guess karo

Ek label par likha hai INPUT

230 V

~, OUTPUT

12 V

1.0 A

. Agar adapter mein ideal transformer hota, to full load par woh mains se kitna current leta?

Lagbhag

19 A

Lagbhag

4.3 m A

1.0 A

Lagbhag

52 m A

Ek unplugged mains adapter ya charger dhoondho aur uska label padho. Purane, bhaari adapters (toys, radios ya door bells ke) par aksar kuch aisa likha hota hai: INPUT 230 V~ 50 Hz, OUTPUT 12 V~ 1.0 A.
Andar ke ideal transformer ko kitna turns ratio $N_{s} / N_{p} = V_{s} / V_{p}$ chahiye, yeh nikaalo.
Output power $V_{s} I_{s}$ aur full load par ideal transformer jitna primary current leta, woh nikaalo.
Purane adapter ke weight ko milti-julti power wale modern phone charger ke weight se compare karo.

Tumhara adapter

Quantity	Value
Input voltage (V)
Output voltage (V)
Turns ratio Ns/Np
Output power (W)
Ideal primary current (A)

Tumhari entries apne aap save hoti hain.

$12 V$ output ke liye turns ratio lagbhag $1 : 19$ hai, aur full load par primary sirf kuch das milliamperes le jaati hai. Purane adapters bhaari hote hain kyunki unmein $50 H z$ par kaam karne wala iron-core transformer hota hai. Modern chargers halke hote hain kyunki woh pehle current ko second mein das-hazaaron baar on aur off karte hain aur phir ek chhota sa transformer use karte hain. Emf per turn flux ke rate of change ke proportional hai, isliye high frequency par kahin chhota core, kahin kam flux ke saath, wahi kaam kar deta hai.

Table 7.3Is chapter mein aayi quantitiesNCERT p. 198

Quantity	Symbol	SI unit	Dimensions	Remarks
rms voltage	$V$	volt (V)	$[M L^{2} T^{- 3} A^{- 1}]$	Amplitude $v_{m}$ wale sinusoidal voltage ke liye $V = v_{m} / \sqrt{2}$
rms current	$I$	ampere (A)	$[A]$	$I = i_{m} / \sqrt{2}$
inductive reactance	$X_{L}$	ohm (Ω)	$[M L^{2} T^{- 3} A^{- 2}]$	$X_{L} = ω L$
capacitive reactance	$X_{C}$	ohm (Ω)	$[M L^{2} T^{- 3} A^{- 2}]$	$X_{C} = 1 / ω C$
impedance	$Z$	ohm (Ω)	$[M L^{2} T^{- 3} A^{- 2}]$	elements par depend karta hai; series LCR ke liye $\sqrt{R^{2} + (X_{C} - X_{L})^{2}}$
resonant frequency	$ω_{0}$ (ya $ν_{0}$ )	rad/s (ya Hz)	$[T^{- 1}]$	Series LCR ke liye $ω_{0} = 1 / \sqrt{L C}$
quality factor	$Q$	koi nahi	dimensionless	Series LCR ke liye $Q = ω_{0} L / R = 1 / (ω_{0} C R)$
power factor	$\cos ϕ$	koi nahi	dimensionless	$ϕ$ voltage aur current ke beech phase difference hai

Recap

Summary aur key points

Alternating voltage $v = v_{m} \sin ω t$ har half cycle mein direction badalta hai; Indian mains mein $ν = 50 H z$ aur $V = 230 V$ rms hai (Section 7.1).
Resistor mein $i = (v_{m} / R) \sin ω t$ : current voltage ke saath in phase hai (Section 7.2).
Resistor mein average power $\frac{1}{2} i_{m}^{2} R$ hai. Rms values $I = i_{m} / \sqrt{2}$ aur $V = v_{m} / \sqrt{2}$ ke saath $P = I^{2} R = V^{2} / R = V I$ , bilkul dc ki tarah (Section 7.2).
Meter readings, supply voltages aur power ratings rms ya average values hoti hain: $230 V$ rms ka peak lagbhag $325 V$ hai.
Phasor ek rotating arrow hai jiska vertical projection instantaneous value deta hai; lengths amplitudes deti hain aur angles phase differences (Section 7.3).
Pure inductor mein $X_{L} = ω L$ aur current voltage se $π / 2$ peeche hai (Section 7.4).
Pure capacitor mein $X_{C} = 1 / ω C$ aur current voltage se $π / 2$ aage hai. Na pure inductor koi average power leta hai, na pure capacitor (Section 7.5).
Series LCR circuit: $Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}$ aur $\tan ϕ = (X_{C} - X_{L}) / R$ . Elements ke across voltages phasors ki tarah judte hain, numbers ki tarah nahi (Section 7.6.1).
Resonance $ω_{0} = 1 / \sqrt{L C}$ par hoti hai: $X_{L} = X_{C}$ , $Z = R$ aur current sabse bada aur voltage ke saath in phase hota hai. Chhota $R$ zyada sharp peak deta hai, jise $Q = ω_{0} L / R$ se naapte hain (Section 7.6.2).
Average power $P = V I \cos ϕ$ ; $\cos ϕ$ power factor hai. $\cos ϕ = 0$ wala current wattless hai (Section 7.7).
Low power factor ka matlab same power ke liye zyada current, aur zyada line losses; parallel mein laga capacitor ise correct kar sakta hai (Example 7.7).
Ideal transformer mein $V_{s} / V_{p} = N_{s} / N_{p}$ aur $I_{s} / I_{p} = N_{p} / N_{s}$ . High-voltage transmission $I^{2} R$ losses kam karta hai; asli transformers flux leakage, winding resistance, eddy currents aur hysteresis se thodi energy khote hain (Section 7.8).

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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 Exercises 7.1	Practice Q1
NCERT Exercises 7.2	Practice Q2
NCERT Exercises 7.3	Practice Q3
NCERT Exercises 7.4	Practice Q4
NCERT Exercises 7.5	Practice Q5
NCERT Exercises 7.6	Practice Q6
NCERT Exercises 7.7	Practice Q7
NCERT Exercises 7.8	Practice Q8

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7.1 Introduction3 questions
7.2 Resistor par lagaya gaya AC voltage4 questions
7.3 Rotating vectors se ac ko represent karna: phasors3 questions
7.4 Inductor par lagaya gaya AC voltage4 questions
7.5 Capacitor par lagaya gaya AC voltage4 questions
7.6.1 Phasor-diagram solution4 questions
7.6.2 Resonance4 questions
7.7 Ac circuit mein power: power factor4 questions
7.8 Transformers4 questions

Glossary · 12

Key terms (English + Hindi)

alternating current (ac)प्रत्यावर्ती धाराpratyavarti dhara: Aisa current (ya voltage) jo time ke saath sinusoidally badalta hai aur har half cycle mein direction badalta hai, jaise $i = i_{m} \sin ω t$ .Chapter mein pehli baar yahan aaya →

capacitive reactanceधारितीय प्रतिघातdharitiya pratighat: $X_{C} = 1 / ω C$ , ac ke against capacitor ka opposition, ohms mein; frequency badhne par ghatta hai.Chapter mein pehli baar yahan aaya →

impedanceप्रतिबाधाpratibadha: Ac circuit ke liye ratio $Z = V / I$ ; series LCR ke liye $Z = \sqrt{R^{2} + (X_{C} - X_{L})^{2}}$ .Chapter mein pehli baar yahan aaya →
inductive reactanceप्रेरकीय प्रतिघातprerakiya pratighat: $X_{L} = ω L$ , ac ke against inductor ka opposition, ohms mein; frequency ke saath badhta hai.Chapter mein pehli baar yahan aaya →

phasorफेज़रphezar: Sinusoidal quantity ko represent karne wala rotating arrow: iski length amplitude hai aur iska vertical projection instantaneous value.Chapter mein pehli baar yahan aaya →
power factorशक्ति गुणांकshakti gunank: $P = V I \cos ϕ$ mein $\cos ϕ$ ; series LCR circuit ke liye $R / Z$ ke barabar.Chapter mein pehli baar yahan aaya →

quality factor (Q): $Q = ω_{0} L / R$ , ek pure number jo batata hai ki resonance peak kitna sharp hai.Chapter mein pehli baar yahan aaya →

resonanceअनुनादanunaad: Apni natural frequency par drive kiye gaye system ka bada response; series LCR circuit mein $X_{L} = X_{C}$ hone par current ka peak.Chapter mein pehli baar yahan aaya →
resonant frequencyअनुनादी आवृत्तिanunadi aavritti: Woh frequency jis par series LCR circuit resonate karta hai: $ω_{0} = 1 / \sqrt{L C}$ .Chapter mein pehli baar yahan aaya →
rms valueवर्ग माध्य मूल मानvarg madhya mool maan: Ek cycle par square ke mean ka square root; sine wave ke liye peak value ko $\sqrt{2}$ se divide karke.

transformerट्रांसफ़ॉर्मरtransformer: Shared iron core par do coils wala device jo mutual induction se ac voltage badalta hai: $V_{s} / V_{p} = N_{s} / N_{p}$ .Chapter mein pehli baar yahan aaya →

wattless currentवाटहीन धाराwatt-heen dhara: Aisa current jo voltage se $90^{\circ}$ out of phase hai, isliye koi average power nahi le jaata.Chapter mein pehli baar yahan aaya →

Alternating Current: Class 12 Physics Chapter 7 Hinglish mein

Is chapter mein aap kya seekhenge

7.1 Introduction

7.2 Resistor par lagaya gaya AC voltage

7.3 Rotating vectors se ac ko represent karna: phasors

7.4 Inductor par lagaya gaya AC voltage

7.5 Capacitor par lagaya gaya AC voltage

7.6 Series LCR circuit par lagaya gaya AC voltage

7.6.1 Phasor-diagram solution

7.6.2 Resonance

7.7 Ac circuit mein power: power factor

7.8 Transformers

Summary aur key points

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Key terms (English + Hindi)

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