CLASS 12 PHYSICS · CHAPTER 7 · READ
Alternating Currentप्रत्यावर्ती धाराAlternating Current
AC through R → rms values → phasors → AC through L (current lags) and C (current leads) → series LCR: impedance, phase, resonance → power and power factor → transformers.R में ac → rms मान → फ़ेज़र → L में ac (धारा पीछे) और C में ac (धारा आगे) → श्रेणी LCR: प्रतिबाधा, कला, अनुनाद → शक्ति और शक्ति गुणांक → ट्रांसफ़ॉर्मर।R mein ac → rms values → phasors → L mein ac (current peeche) aur C mein ac (current aage) → series LCR: impedance, phase, resonance → power aur power factor → transformers.
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Board Notes: Pass to Topperबोर्ड नोट्स: पास से टॉपर तकBoard Notes: Pass se Topper tak🔥 Top topic came in 7 of 7 years · see what to learn first🔥 सबसे ऊपर वाला टॉपिक 7 में से 7 वर्ष आया · देखें पहले क्या पढ़ें🔥 Top topic 7 mein se 7 saal aaya · dekhiye pehle kya padhna haiTry it yourself — live simulationsख़ुद करके देखें — लाइव सिमुलेशनKhud karke dekhiye — live simulations
Diagrams in this chapterइस अध्याय के चित्रIs chapter ke diagrams
▶ Animate: Phase in R, L and C▶ एनिमेट: R, L और C में कला▶ Animate: R, L aur C mein phaseAnimated
▶ Animate: Transformer▶ एनिमेट: ट्रांसफ़ॉर्मर▶ Animate: TransformerAnimated
AC voltage applied to a resistorप्रतिरोधक पर लगाई गई ac वोल्टताAC voltage applied to a resistorFig 7.1 / 7.2
Root mean square (rms) currentवर्ग माध्य मूल (rms) धाराRoot mean square (rms) currentFig 7.3
Phasors for an ac circuit with a resistorप्रतिरोधक वाले ac परिपथ के फ़ेज़रPhasors for an ac circuit with a resistorFig 7.4
AC voltage across a pure inductor (current lags)शुद्ध प्रेरक पर ac वोल्टता (धारा पीछे)AC voltage across a pure inductor (current lags)Fig 7.5 / 7.6
Power supplied to a pure inductor averages to zeroशुद्ध प्रेरक को दी गई शक्ति का औसत शून्यPower supplied to a pure inductor averages to zeroSec 7.4
AC voltage across a capacitor (current leads)संधारित्र पर ac वोल्टता (धारा आगे)AC voltage across a capacitor (current leads)Fig 7.7 / 7.8
Bulb and coil: effect of an iron rodबल्ब और कुंडली: लोहे की छड़ का प्रभावBulb and coil: effect of an iron rodFig 7.9
Series LCR circuit with an ac sourceac स्रोत के साथ श्रेणी LCR परिपथSeries LCR circuit with an ac sourceFig 7.10
Phasor diagram of a series LCR circuitश्रेणी LCR परिपथ का फ़ेज़र आरेखPhasor diagram of a series LCR circuitFig 7.11
Impedance diagram (right triangle)प्रतिबाधा आरेख (समकोण त्रिभुज)Impedance diagram (right triangle)Fig 7.12
Series LCR (XC > XL): V, I phasors and graphsश्रेणी LCR (XC > XL): V, I फ़ेज़र और ग्राफ़Series LCR (XC > XL): V, I phasors and graphsFig 7.13
Resonance curves of a series LCR circuitश्रेणी LCR परिपथ के अनुनाद वक्रResonance curves of a series LCR circuitFig 7.14
Power factor improvement with a capacitorसंधारित्र से शक्ति गुणांक में सुधारPower factor improvement with a capacitorFig 7.15
Transformer: coils one over the otherट्रांसफ़ॉर्मर: एक के ऊपर दूसरी कुंडलीTransformer: coils one over the otherFig 7.16(a)
Transformer: coils on separate limbsट्रांसफ़ॉर्मर: अलग-अलग भुजाओं पर कुंडलियाँTransformer: coils on separate limbsFig 7.16(b)
Power transmission with step-up and step-down transformersउच्चायी और अपचायी ट्रांसफ़ॉर्मर से शक्ति संचरणPower transmission with step-up and step-down transformersSec 7.8In this chapter (NCERT sections)इस अध्याय में (NCERT खंड)Is chapter mein (NCERT sections)
- 7.1IntroductionभूमिकाIntroduction
- 7.2AC Voltage Applied to a Resistorप्रतिरोधक पर लगाई गई ac वोल्टताResistor par AC Voltage
- 7.3Representation of AC Current and Voltage by Rotating Vectors — Phasorsघूर्णी सदिशों द्वारा ac धारा और वोल्टता का निरूपण — फ़ेज़रRotating Vectors se AC — Phasors
- 7.4AC Voltage Applied to an Inductorप्रेरक पर लगाई गई ac वोल्टताInductor par AC Voltage
- 7.5AC Voltage Applied to a Capacitorसंधारित्र पर लगाई गई ac वोल्टताCapacitor par AC Voltage
- 7.6AC Voltage Applied to a Series LCR Circuitश्रेणी LCR परिपथ पर लगाई गई ac वोल्टताSeries LCR Circuit par AC Voltage
- 7.6.1Phasor-diagram solutionफ़ेज़र-आरेख द्वारा हलPhasor-diagram se solution
- 7.6.2ResonanceअनुनादResonance
- 7.7Power in AC Circuit: The Power Factorac परिपथ में शक्ति: शक्ति गुणांकAC Circuit mein Power: Power Factor
- 7.8Transformersट्रांसफ़ॉर्मरTransformers
7.1IntroductionभूमिकाIntroduction
So far our currents never changed direction — that is direct current (dc). The mains supply in our homes is different: its voltage varies with time like a sine function. Such a voltage is an alternating voltage (ac voltage), and the current it drives is an alternating current (ac). Most devices we use run on ac, mainly because power companies transmit and distribute electrical energy as ac.अब तक हमारी धाराएँ कभी दिशा नहीं बदलती थीं — वह दिष्ट धारा (dc) है। हमारे घरों की मेन्स आपूर्ति अलग है: इसकी वोल्टता समय के साथ ज्या (sine) फलन की तरह बदलती है। ऐसी वोल्टता प्रत्यावर्ती वोल्टता (ac वोल्टता) है, और इससे चलने वाली धारा प्रत्यावर्ती धारा (ac) है। हमारे अधिकांश उपकरण ac पर चलते हैं, मुख्यतः क्योंकि बिजली कंपनियाँ विद्युत ऊर्जा का संचरण और वितरण ac के रूप में करती हैं।Ab tak hamare currents kabhi direction nahi badalte the — wo direct current (dc) hai. Ghar ki mains supply alag hai: iska voltage time ke saath sine function ki tarah badalta hai. Aise voltage ko alternating voltage (ac voltage) kehte hain, aur isse chalne wala current alternating current (ac) hai. Hamare zyadatar devices ac par chalte hain, mainly kyunki power companies bijli ko ac ke roop mein transmit aur distribute karti hain.Mains electricity keeps reversing direction many times a second. That is called alternating current (ac).
Why prefer ac? The main reason: ac voltages can be easily and efficiently changed from one value to another with transformers, and so electrical energy can be transmitted economically over long distances. AC circuits also have special properties used every day — for example, tuning a radio to your favourite station uses resonance, which you will meet in this chapter.ac को क्यों चुनें? मुख्य कारण: ac वोल्टता को ट्रांसफ़ॉर्मर से आसानी से और दक्षता से एक मान से दूसरे में बदला जा सकता है, इसलिए विद्युत ऊर्जा को लंबी दूरी तक मितव्ययिता से भेजा जा सकता है। ac परिपथों के कुछ विशेष गुण रोज़ काम आते हैं — जैसे रेडियो को पसंदीदा स्टेशन पर ट्यून करना अनुनाद का उपयोग है, जिसे आप इसी अध्याय में पढ़ेंगे।ac kyun? Main reason: ac voltage ko transformer se aasaani aur efficiency se ek value se doosri mein badla ja sakta hai, isliye bijli ko lambi doori tak sasti tarah se bheja ja sakta hai. ac circuits ki kuch khaas properties roz kaam aati hain — jaise radio ko favourite station par tune karna resonance ka use hai, jo aap isi chapter mein padhenge.Why do we prefer ac (alternating current)? The main reason: ac voltages can be changed easily and with little loss from one value to another using transformers. Because of this, electrical energy can be sent cheaply over long distances. AC circuits also have special properties we use every day. For example, when you tune a radio to your favourite station, you use resonance, which you will study in this chapter.
In India the mains supply is about 220–240 V at 50 Hz: the voltage completes 50 full cycles every second, so the current reverses direction 100 times a second. A bulb does not visibly flicker because its filament cannot cool down that fast.भारत में मेन्स आपूर्ति लगभग 220–240 V, 50 Hz है: वोल्टता हर सेकंड 50 पूरे चक्र पूरे करती है, यानी धारा सेकंड में 100 बार दिशा बदलती है। बल्ब टिमटिमाता नहीं दिखता क्योंकि उसका तंतु इतनी जल्दी ठंडा नहीं हो पाता।India mein mains supply lagbhag 220–240 V, 50 Hz hai: voltage har second 50 poore cycles poore karta hai, yaani current second mein 100 baar direction badalta hai. Bulb flicker karta nahi dikhta kyunki uska filament itni jaldi thanda nahi ho paata.In India the mains supply is about 220–240 V at 50 Hz. This means the voltage completes 50 full cycles every second, so the current changes direction 100 times a second. A bulb does not seem to flicker because its filament cannot cool down that fast.
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7.2AC Voltage Applied to a Resistorप्रतिरोधक पर लगाई गई ac वोल्टताResistor par AC Voltage
Connect a resistor R to an ac source. The source gives a sinusoidal potential difference v = vm sin ωt, where vm is the amplitude (peak value) and ω is the angular frequency (ω = 2πν). Applying Kirchhoff’s loop rule, vm sin ωt = iR at every instant.एक प्रतिरोधक R को ac स्रोत से जोड़िए। स्रोत ज्यावक्रीय विभवांतर v = vm sin ωt देता है, जहाँ vm आयाम (शिखर मान) है और ω कोणीय आवृत्ति (ω = 2πν) है। किरखोफ़ का पाश नियम लगाने पर हर क्षण vm sin ωt = iR।Ek resistor R ko ac source se jodiye. Source sinusoidal potential difference v = vm sin ωt deta hai, jahan vm amplitude (peak value) hai aur ω angular frequency (ω = 2πν) hai. Kirchhoff ka loop rule lagane par har instant vm sin ωt = iR.Connect a resistor R to an ac source. The source gives a potential difference that changes like a sine wave: v = vm sin ωt. Here vm is the amplitude (the peak or highest value) and ω is the angular frequency (ω = 2πν). Using Kirchhoff’s loop rule, vm sin ωt = iR at every instant.
Tap to zoom · 7 labels to exploreज़ूम करने के लिए टैप करें · 7 लेबलTap kijiye → zoom · 7 labelsVoltage and current reach zero, maximum and minimum at the same time — they are in phase. Over a full cycle the current is positive for half the time and negative for the other half, so the average current is zero. But that does not mean no power is used: Joule heating is i²R, and i² is never negative.वोल्टता और धारा एक ही समय शून्य, अधिकतम और न्यूनतम होती हैं — वे समान कला में हैं। पूरे चक्र में धारा आधे समय धनात्मक और आधे समय ऋणात्मक रहती है, इसलिए औसत धारा शून्य है। पर इसका अर्थ यह नहीं कि कोई शक्ति व्यय नहीं होती: जूल तापन i²R है, और i² कभी ऋणात्मक नहीं होता।Voltage aur current ek hi time zero, maximum aur minimum hote hain — ye in phase hain. Poore cycle mein current aadha time positive aur aadha time negative rehta hai, isliye average current zero hai. Par iska matlab ye nahi ki power use nahi hoti: Joule heating i²R hai, aur i² kabhi negative nahi hota.v and i rise and fall together. The current’s average is zero, but the heating is not, because heating depends on i².
Average over one cycle: since sin²ωt = ½(1 − cos 2ωt) and the average of cos 2ωt over a cycle is 0, the average of sin²ωt is ½. So the average power is p̄ = ½ im²R.एक चक्र का औसत: क्योंकि sin²ωt = ½(1 − cos 2ωt) और एक चक्र में cos 2ωt का औसत 0 है, इसलिए sin²ωt का औसत ½ है। अतः औसत शक्ति p̄ = ½ im²R।Ek cycle ka average: kyunki sin²ωt = ½(1 − cos 2ωt) aur ek cycle mein cos 2ωt ka average 0 hai, isliye sin²ωt ka average ½ hai. Isliye average power p̄ = ½ im²R.Now take the average over one full cycle. We know sin²ωt = ½(1 − cos 2ωt), and the average of cos 2ωt over a cycle is 0. So the average of sin²ωt is ½. This gives the average power p̄ = ½ im²R.
To write ac power in the same form as dc power (P = I²R), we define the root mean square (rms) or effective current: I = √(mean of i²) = im/√2 = 0.707 im. Similarly the rms voltage V = vm/√2 = 0.707 vm. The rms current is the dc current that would produce the same average heating in the same resistor.ac शक्ति को dc शक्ति (P = I²R) के रूप में लिखने के लिए हम वर्ग माध्य मूल (rms) या प्रभावी धारा परिभाषित करते हैं: I = √(i² का माध्य) = im/√2 = 0.707 im। इसी प्रकार rms वोल्टता V = vm/√2 = 0.707 vm। rms धारा वह dc धारा है जो उसी प्रतिरोधक में समान औसत तापन करती।ac power ko dc power (P = I²R) ke form mein likhne ke liye hum root mean square (rms) ya effective current define karte hain: I = √(i² ka mean) = im/√2 = 0.707 im. Isi tarah rms voltage V = vm/√2 = 0.707 vm. rms current wo dc current hai jo usi resistor mein same average heating karta.rms value = the steady dc value that heats a resistor just as much. For a sine wave it is peak ÷ √2.
Tap to zoom · 6 labels to exploreज़ूम करने के लिए टैप करें · 6 लेबलTap kijiye → zoom · 6 labelsIt is customary to quote rms values for ac. The household line voltage of 220 V is an rms value; its peak is vm = √2 × 220 V ≈ 311 V. AC ammeters and voltmeters also read rms values.ac के लिए rms मान बताने की परिपाटी है। 220 V की घरेलू लाइन वोल्टता rms मान है; इसका शिखर vm = √2 × 220 V ≈ 311 V है। ac ऐमीटर और वोल्टमीटर भी rms मान पढ़ते हैं।ac ke liye rms values batane ka rivaaj hai. 220 V ki ghar ki line voltage rms value hai; iska peak vm = √2 × 220 V ≈ 311 V hai. ac ammeters aur voltmeters bhi rms values padhte hain.For ac, we usually give rms values (root mean square, a kind of effective value). The household line voltage of 220 V is an rms value. Its peak value is vm = √2 × 220 V ≈ 311 V. AC ammeters and voltmeters also show rms values.
Why ⟨cos 2ωt⟩ = 0: over one period T = 2π/ω, ∫cos 2ωt dt = [sin 2ωt / 2ω] from 0 to T = (sin 4π − 0)/2ω = 0. The same trick gives ⟨sin 2ωt⟩ = 0, which you will need for L and C. Also note: the rms value is defined for any waveform, but the factor 1/√2 holds only for a sine wave.⟨cos 2ωt⟩ = 0 क्यों: एक आवर्तकाल T = 2π/ω में ∫cos 2ωt dt = [sin 2ωt / 2ω] (0 से T) = (sin 4π − 0)/2ω = 0। इसी तरीके से ⟨sin 2ωt⟩ = 0 मिलता है, जो L और C में काम आएगा। ध्यान दें: rms मान किसी भी तरंग-रूप के लिए परिभाषित है, पर गुणक 1/√2 केवल ज्या तरंग के लिए सही है।⟨cos 2ωt⟩ = 0 kyun: ek period T = 2π/ω mein ∫cos 2ωt dt = [sin 2ωt / 2ω] (0 se T) = (sin 4π − 0)/2ω = 0. Isi trick se ⟨sin 2ωt⟩ = 0 milta hai, jo L aur C mein kaam aayega. Dhyaan dijiye: rms value kisi bhi waveform ke liye define hai, par factor 1/√2 sirf sine wave ke liye sahi hai.Why ⟨cos 2ωt⟩ = 0: over one period T = 2π/ω, ∫cos 2ωt dt = [sin 2ωt / 2ω] from 0 to T = (sin 4π − 0)/2ω = 0. In simple words, the positive and negative parts cancel out. The same method gives ⟨sin 2ωt⟩ = 0, which you will need for L and C circuits. Also note: an rms value can be found for any wave shape, but the factor 1/√2 is true only for a sine wave.
Change the peak current and watch i, i² and the running average of i² build up over a cycle. The equivalent dc current that heats the resistor equally is always 0.707 × peak.शिखर धारा बदलिए और एक चक्र में i, i² और i² का चलता औसत बनते देखिए। समान तापन करने वाली तुल्य dc धारा हमेशा 0.707 × शिखर रहती है।Peak current badaliye aur ek cycle mein i, i² aur i² ka running average bante dekhiye. Same heating karne wala equivalent dc current hamesha 0.707 × peak rehta hai.
A light bulb is rated at 100 W for a 220 V supply. Find (a) the resistance of the bulb; (b) the peak voltage of the source; and (c) the rms current through the bulb.एक बल्ब 220 V आपूर्ति के लिए 100 W अंकित है। ज्ञात कीजिए (a) बल्ब का प्रतिरोध; (b) स्रोत की शिखर वोल्टता; और (c) बल्ब में rms धारा।Ek bulb 220 V supply ke liye 100 W rated hai. Nikaaliye (a) bulb ka resistance; (b) source ka peak voltage; aur (c) bulb mein rms current.A light bulb is marked 100 W for a 220 V supply. Find (a) the resistance of the bulb, (b) the peak voltage of the source, and (c) the rms current through the bulb.
Show solutionहल देखेंSolution dekhiye
(a) P = V²/R, so R = V²/P = (220)²/100 = 484 Ω. (b) vm = √2 V = 1.414 × 220 ≈ 311 V. (c) P = IV, so I = P/V = 100/220 ≈ 0.454 A (rms).(a) P = V²/R, अतः R = V²/P = (220)²/100 = 484 Ω। (b) vm = √2 V = 1.414 × 220 ≈ 311 V। (c) P = IV, अतः I = P/V = 100/220 ≈ 0.454 A (rms)।(a) P = V²/R, isliye R = V²/P = (220)²/100 = 484 Ω. (b) vm = √2 V = 1.414 × 220 ≈ 311 V. (c) P = IV, isliye I = P/V = 100/220 ≈ 0.454 A (rms).(a) Use P = V²/R, so R = V²/P = (220)²/100 = 484 Ω. (b) Peak voltage vm = √2 V = 1.414 × 220 ≈ 311 V. (c) Use P = IV, so I = P/V = 100/220 ≈ 0.454 A (rms).
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7.3Representation of AC Current and Voltage by Rotating Vectors — Phasorsघूर्णी सदिशों द्वारा ac धारा और वोल्टता का निरूपण — फ़ेज़रRotating Vectors se AC — Phasors
For a resistor v and i are in phase, but for an inductor, a capacitor or their combinations they are not. To show phase relationships we use phasors. A phasor is a vector that rotates about the origin with angular speed ω. Its length is the amplitude (vm or im), and its vertical component (projection on the vertical axis) gives the instantaneous value (vm sin ωt or im sin ωt).प्रतिरोधक में v और i समान कला में हैं, पर प्रेरक, संधारित्र या इनके संयोजन में नहीं। कला-संबंध दिखाने के लिए हम फ़ेज़र का उपयोग करते हैं। फ़ेज़र एक सदिश है जो मूल बिंदु के परितः कोणीय चाल ω से घूमता है। इसकी लंबाई आयाम (vm या im) है, और इसका ऊर्ध्वाधर घटक (ऊर्ध्वाधर अक्ष पर प्रक्षेप) तात्क्षणिक मान (vm sin ωt या im sin ωt) देता है।Resistor mein v aur i in phase hain, par inductor, capacitor ya inke combinations mein nahi. Phase relationship dikhane ke liye hum phasors use karte hain. Phasor ek vector hai jo origin ke around angular speed ω se ghoomta hai. Iski length amplitude (vm ya im) hai, aur iska vertical component (vertical axis par projection) instantaneous value (vm sin ωt ya im sin ωt) deta hai.Imagine an arrow spinning like a clock hand, but anticlockwise. Its shadow on the vertical axis goes up and down like a sine wave.
Tap to zoom · 10 labels to exploreज़ूम करने के लिए टैप करें · 10 लेबलTap kijiye → zoom · 10 labelsAs the phasors rotate anticlockwise at ω, their projections trace out the v and i graphs. For a resistor, the phasors V and I always point in the same direction — the phase angle between voltage and current is zero.जैसे-जैसे फ़ेज़र ω से वामावर्त घूमते हैं, उनके प्रक्षेप v और i के ग्राफ़ बनाते हैं। प्रतिरोधक के लिए फ़ेज़र V और I हमेशा एक ही दिशा में रहते हैं — वोल्टता और धारा के बीच कला कोण शून्य है।Jaise-jaise phasors ω se anticlockwise ghoomte hain, unke projections v aur i ke graphs banate hain. Resistor ke liye phasors V aur I hamesha ek hi direction mein rehte hain — voltage aur current ke beech phase angle zero hai.Phasors are arrows that turn anticlockwise at speed ω. Their shadows (projections) on the vertical axis draw the V and i graphs. For a resistor, the phasors V and I always point in the same direction. So the phase angle (time gap) between voltage and current is zero: they rise and fall together. 👉 I
Phasors are not real vectors. Voltage and current are scalars. It just happens that the amplitudes and phases of sine-varying quantities combine exactly like the projections of rotating vectors — so we borrow the familiar rule of vector addition as a shortcut.फ़ेज़र वास्तविक सदिश नहीं हैं। वोल्टता और धारा अदिश हैं। बस ज्यावक्रीय राशियों के आयाम और कलाएँ ठीक घूमते सदिशों के प्रक्षेपों की तरह जुड़ती हैं — इसलिए हम सदिश योग का परिचित नियम एक आसान तरीके के रूप में उधार लेते हैं।Phasors asli vectors nahi hain. Voltage aur current scalars hain. Bas sine-varying quantities ke amplitudes aur phases bilkul ghoomte vectors ke projections ki tarah judte hain — isliye hum vector addition ka jaana-pehchaana rule ek shortcut ki tarah use karte hain.Phasors are not real vectors. Voltage and current are scalars (they have no direction in space). But the amplitudes and phases of sine-changing quantities happen to add exactly like the projections of turning vectors. So we borrow the easy rule of vector addition as a shortcut.
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7.4AC Voltage Applied to an Inductorप्रेरक पर लगाई गई ac वोल्टताInductor par AC Voltage
Now connect an ac source to an inductor L. Real coils have some resistance, but we take a pure inductor (negligible resistance). The inductor produces a self-induced (back) emf −L di/dt. Kirchhoff’s loop rule gives v − L di/dt = 0, soअब एक ac स्रोत को प्रेरक L से जोड़िए। वास्तविक कुंडलियों में कुछ प्रतिरोध होता है, पर हम शुद्ध प्रेरक (नगण्य प्रतिरोध) लेते हैं। प्रेरक स्वप्रेरित (विरोधी) emf −L di/dt उत्पन्न करता है। किरखोफ़ का पाश नियम देता है v − L di/dt = 0, अतःAb ek ac source ko inductor L se jodiye. Asli coils mein thoda resistance hota hai, par hum pure inductor (negligible resistance) lete hain. Inductor self-induced (back) emf −L di/dt banata hai. Kirchhoff ka loop rule deta hai v − L di/dt = 0, isliyeNow connect an ac source to an inductor L (a coil). Real coils have some resistance, but here we take a pure inductor (resistance is so small we ignore it). The inductor makes a self-induced (back) emf −L di/dt. Kirchhoff’s loop rule gives v − L di/dt = 0, so
Integrating: i = −(vm/ωL) cos ωt + constant. The source swings symmetrically about zero, so the current has no steady part and the constant is zero. Using −cos ωt = sin(ωt − π/2):समाकलन करने पर: i = −(vm/ωL) cos ωt + नियतांक। स्रोत शून्य के परितः सममित रूप से दोलन करता है, इसलिए धारा में कोई स्थिर भाग नहीं, और नियतांक शून्य है। −cos ωt = sin(ωt − π/2) का उपयोग करके:Integrate karne par: i = −(vm/ωL) cos ωt + constant. Source zero ke around symmetric swing karta hai, isliye current mein koi steady part nahi, aur constant zero hai. −cos ωt = sin(ωt − π/2) use karke:Integrate (add up) to find i: i = −(vm/ωL) cos ωt + constant. The source voltage swings equally above and below zero, so the current has no steady part. So the constant is zero. Then use −cos ωt = sin(ωt − π/2) to write it as a sine wave:
The inductive reactance XL = ωL limits the current the way a resistance does. It is directly proportional to L and to the frequency: a coil lets low-frequency ac through easily but strongly opposes high-frequency ac (for dc, ω = 0 and XL = 0).प्रेरकीय प्रतिघात XL = ωL धारा को वैसे ही सीमित करता है जैसे प्रतिरोध। यह L और आवृत्ति दोनों के अनुक्रमानुपाती है: कुंडली कम आवृत्ति की ac को आसानी से जाने देती है पर अधिक आवृत्ति की ac का प्रबल विरोध करती है (dc के लिए ω = 0 और XL = 0)।Inductive reactance XL = ωL current ko waise hi limit karta hai jaise resistance. Ye L aur frequency dono ke proportional hai: coil low-frequency ac ko aasaani se jaane deti hai par high-frequency ac ka zor se virodh karti hai (dc ke liye ω = 0 aur XL = 0).The inductive reactance XL = ωL limits the current, just like a resistance does. It is directly proportional to L and to the frequency. So a coil lets low-frequency ac pass easily but strongly blocks high-frequency ac. For dc, ω = 0, so XL = 0.
Tap to zoom · 12 labels to exploreज़ूम करने के लिए टैप करें · 12 लेबलTap kijiye → zoom · 12 labelsCompare v = vm sin ωt with i = im sin(ωt − π/2): in a pure inductor the current LAGS the voltage by π/2 (a quarter cycle). The current reaches its maximum T/4 later than the voltage. In the phasor diagram the current phasor I is π/2 behind (clockwise from) the voltage phasor V.v = vm sin ωt की तुलना i = im sin(ωt − π/2) से कीजिए: शुद्ध प्रेरक में धारा वोल्टता से π/2 पीछे (LAG) रहती है (चौथाई चक्र)। धारा अपना अधिकतम मान वोल्टता से T/4 बाद पाती है। फ़ेज़र आरेख में धारा फ़ेज़र I वोल्टता फ़ेज़र V से π/2 पीछे (दक्षिणावर्त) है।v = vm sin ωt ko i = im sin(ωt − π/2) se compare kijiye: pure inductor mein current voltage se π/2 peeche (LAG) rehta hai (quarter cycle). Current apna maximum voltage se T/4 baad paata hai. Phasor diagram mein current phasor I voltage phasor V se π/2 peeche (clockwise) hai.In a coil, the current is always a quarter cycle late. The coil resists change, so the current takes time to build up.
Memory trick — “ELI the ICE man”: in an inductor (L), the emf E comes before the current I → E leads I (current lags). In a capacitor (C), I comes before E → current leads.याद रखने की तरकीब — “ELI the ICE man”: प्रेरक (L) में emf E, धारा I से पहले → E आगे (धारा पीछे)। संधारित्र (C) में I, E से पहले → धारा आगे।Yaad rakhne ki trick — “ELI the ICE man”: inductor (L) mein emf E, current I se pehle → E aage (current peeche). Capacitor (C) mein I, E se pehle → current aage.Memory trick — “ELI the ICE man”: in an inductor (L), the emf E comes before the current I, so E leads I (the current lags behind). In a capacitor (C), I comes before E, so the current leads.
Does an inductor use up power? The instantaneous power is pL = iv = im sin(ωt − π/2) × vm sin ωt = −(im vm/2) sin 2ωt. The average of sin 2ωt over a cycle is zero, so the average power supplied to a pure inductor over a complete cycle is zero. In one quarter cycle energy is stored in its magnetic field; in the next it is given back.क्या प्रेरक शक्ति व्यय करता है? तात्क्षणिक शक्ति pL = iv = im sin(ωt − π/2) × vm sin ωt = −(im vm/2) sin 2ωt। एक चक्र में sin 2ωt का औसत शून्य है, इसलिए पूरे चक्र में शुद्ध प्रेरक को दी गई औसत शक्ति शून्य है। एक चौथाई चक्र में ऊर्जा इसके चुंबकीय क्षेत्र में संचित होती है; अगले में वापस लौट जाती है।Kya inductor power kharch karta hai? Instantaneous power pL = iv = im sin(ωt − π/2) × vm sin ωt = −(im vm/2) sin 2ωt. Ek cycle mein sin 2ωt ka average zero hai, isliye poore cycle mein pure inductor ko di gayi average power zero hai. Ek quarter cycle mein energy iske magnetic field mein store hoti hai; agle mein wapas laut jaati hai.Does an inductor use up power? The power at any instant is pL = iv = im sin(ωt − π/2) × vm sin ωt = −(im vm/2) sin 2ωt. The average of sin 2ωt over a full cycle is zero. So the average power given to a pure inductor over a complete cycle is zero. In one quarter of the cycle, energy is stored in its magnetic field. In the next quarter, it is given back to the source.
Tap to zoom · 6 labels to exploreज़ूम करने के लिए टैप करें · 6 लेबलTap kijiye → zoom · 6 labelsA pure inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz.25.0 mH का एक शुद्ध प्रेरक 220 V के स्रोत से जुड़ा है। यदि स्रोत की आवृत्ति 50 Hz हो, तो परिपथ का प्रेरकीय प्रतिघात और rms धारा ज्ञात कीजिए।25.0 mH ka ek pure inductor 220 V ke source se juda hai. Agar source ki frequency 50 Hz ho, to circuit ka inductive reactance aur rms current nikaaliye.A pure inductor of 25.0 mH is connected to a 220 V source. The frequency of the source is 50 Hz. Find the inductive reactance and the rms current in the circuit.
Show solutionहल देखेंSolution dekhiye
XL = 2πνL = 2 × 3.14 × 50 × 25 × 10⁻³ Ω = 7.85 Ω. rms current I = V/XL = 220/7.85 ≈ 28 A.XL = 2πνL = 2 × 3.14 × 50 × 25 × 10⁻³ Ω = 7.85 Ω। rms धारा I = V/XL = 220/7.85 ≈ 28 A।XL = 2πνL = 2 × 3.14 × 50 × 25 × 10⁻³ Ω = 7.85 Ω. rms current I = V/XL = 220/7.85 ≈ 28 A.Reactance: XL = 2πνL = 2 × 3.14 × 50 × 25 × 10⁻³ Ω = 7.85 Ω. rms current: I = V/XL = 220/7.85 ≈ 28 A.
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7.5AC Voltage Applied to a Capacitorसंधारित्र पर लगाई गई ac वोल्टताCapacitor par AC Voltage
With dc, a capacitor lets current flow only for the short time it takes to charge; then the current stops. With ac, the capacitor is charged and discharged alternately as the voltage reverses every half cycle, so current keeps flowing in the circuit — the capacitor limits the current but does not block it.dc के साथ संधारित्र केवल आवेशित होने के थोड़े समय तक धारा बहने देता है; फिर धारा रुक जाती है। ac के साथ वोल्टता हर अर्ध-चक्र में उलटती है, इसलिए संधारित्र बारी-बारी से आवेशित और अनावेशित होता रहता है और परिपथ में धारा बहती रहती है — संधारित्र धारा को सीमित करता है, रोकता नहीं।dc ke saath capacitor sirf charge hone ke thode time tak current behne deta hai; phir current ruk jaata hai. ac ke saath voltage har half cycle mein palatta hai, isliye capacitor baari-baari se charge aur discharge hota rehta hai aur circuit mein current behta rehta hai — capacitor current ko limit karta hai, rokta nahi.With dc, current flows through a capacitor circuit only for the short time the capacitor takes to charge. Then the current stops. With ac, the voltage changes direction every half cycle, so the capacitor is charged and discharged again and again. So current keeps flowing in the circuit. The capacitor limits the current but does not stop it.
If q is the charge on the capacitor, the voltage across it is v = q/C, and by the loop rule vm sin ωt = q/C. The current is i = dq/dt = d(vm C sin ωt)/dt = ωC vm cos ωt. Using cos ωt = sin(ωt + π/2):यदि संधारित्र पर आवेश q हो, तो इसके सिरों पर वोल्टता v = q/C, और पाश नियम से vm sin ωt = q/C। धारा i = dq/dt = d(vm C sin ωt)/dt = ωC vm cos ωt। cos ωt = sin(ωt + π/2) का उपयोग करके:Agar capacitor par charge q ho, to iske across voltage v = q/C, aur loop rule se vm sin ωt = q/C. Current i = dq/dt = d(vm C sin ωt)/dt = ωC vm cos ωt. cos ωt = sin(ωt + π/2) use karke:Let q be the charge on the capacitor. The voltage across it is v = q/C. By the loop rule, vm sin ωt = q/C. The current is the rate of flow of charge: i = dq/dt = d(vm C sin ωt)/dt = ωC vm cos ωt. Then use cos ωt = sin(ωt + π/2) to write it as a sine wave:
The capacitive reactance XC = 1/ωC is inversely proportional to the frequency and to the capacitance. High-frequency ac passes through a capacitor easily; for dc (ω → 0) XC → ∞, so no steady current flows.धारितीय प्रतिघात XC = 1/ωC आवृत्ति और धारिता के व्युत्क्रमानुपाती है। अधिक आवृत्ति की ac संधारित्र से आसानी से गुज़रती है; dc (ω → 0) के लिए XC → ∞, इसलिए कोई स्थायी धारा नहीं बहती।Capacitive reactance XC = 1/ωC frequency aur capacitance ke inversely proportional hai. High-frequency ac capacitor se aasaani se guzarti hai; dc (ω → 0) ke liye XC → ∞, isliye koi steady current nahi behta.The capacitive reactance XC = 1/ωC is inversely proportional to the frequency and to the capacitance. So high-frequency ac passes through a capacitor easily. For dc (ω → 0), XC → ∞ (becomes infinitely large), so no steady current flows.
Tap to zoom · 12 labels to exploreज़ूम करने के लिए टैप करें · 12 लेबलTap kijiye → zoom · 12 labelsCompare with v = vm sin ωt: in a capacitor the current LEADS the voltage by π/2. The current reaches its maximum T/4 earlier than the voltage; the current phasor I is π/2 ahead (anticlockwise) of V. As for an inductor, the instantaneous power is pC = (im vm/2) sin 2ωt, whose average is zero: a pure capacitor also takes zero average power.v = vm sin ωt से तुलना कीजिए: संधारित्र में धारा वोल्टता से π/2 आगे (LEAD) रहती है। धारा अपना अधिकतम मान वोल्टता से T/4 पहले पाती है; धारा फ़ेज़र I, V से π/2 आगे (वामावर्त) है। प्रेरक की तरह तात्क्षणिक शक्ति pC = (im vm/2) sin 2ωt है, जिसका औसत शून्य है: शुद्ध संधारित्र भी शून्य औसत शक्ति लेता है।v = vm sin ωt se compare kijiye: capacitor mein current voltage se π/2 aage (LEAD) rehta hai. Current apna maximum voltage se T/4 pehle paata hai; current phasor I, V se π/2 aage (anticlockwise) hai. Inductor ki tarah instantaneous power pC = (im vm/2) sin 2ωt hai, jiska average zero hai: pure capacitor bhi zero average power leta hai.In a capacitor the current is a quarter cycle early: current must flow in first to build up the charge, and only then does the voltage rise.
| ElementअवयवElement | OppositionविरोधOpposition | Phase of i w.r.t. vv के सापेक्ष i की कलाv ke saapeksh i ka phase | With frequencyआवृत्ति के साथFrequency ke saath | Average powerऔसत शक्तिAverage power |
|---|---|---|---|---|
| Resistor Rप्रतिरोधक RResistor R | RRR | in phase (φ = 0)समान कला (φ = 0)in phase (φ = 0) | does not depend on νν पर निर्भर नहींν par depend nahi | ½ im²R = I²R½ im²R = I²R½ im²R = I²R |
| Inductor Lप्रेरक LInductor L | XL = ωLXL = ωLXL = ωL | current lags by π/2धारा π/2 पीछेcurrent π/2 peeche | ∝ ν∝ ν∝ ν | zeroशून्यzero |
| Capacitor Cसंधारित्र CCapacitor C | XC = 1/ωCXC = 1/ωCXC = 1/ωC | current leads by π/2धारा π/2 आगेcurrent π/2 aage | ∝ 1/ν∝ 1/ν∝ 1/ν | zeroशून्यzero |
Pick R, L or C. The red voltage phasor and blue current phasor rotate anticlockwise; their shadows draw the graphs live. Change the frequency and watch the reactance and current change — XL grows with frequency, XC falls.R, L या C चुनिए। लाल वोल्टता फ़ेज़र और नीला धारा फ़ेज़र वामावर्त घूमते हैं; उनकी छायाएँ लाइव ग्राफ़ बनाती हैं। आवृत्ति बदलिए और प्रतिघात व धारा बदलते देखिए — XL आवृत्ति के साथ बढ़ता है, XC घटता है।R, L ya C chuniye. Laal voltage phasor aur neela current phasor anticlockwise ghoomte hain; unki shadows live graph banati hain. Frequency badaliye aur reactance aur current badalte dekhiye — XL frequency ke saath badhta hai, XC ghatta hai.
A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced?एक लैंप संधारित्र के साथ श्रेणी में जुड़ा है। dc और ac संयोजनों के लिए अपने प्रेक्षणों का अनुमान लगाइए। प्रत्येक स्थिति में संधारित्र की धारिता घटाने पर क्या होगा?Ek lamp capacitor ke saath series mein juda hai. dc aur ac connections ke liye apne observations predict kijiye. Har case mein capacitor ki capacitance kam karne par kya hoga?A lamp is joined in series with a capacitor. What will you see when it is connected to dc, and when it is connected to ac? What happens in each case if the capacitance of the capacitor is made smaller?
Show solutionहल देखेंSolution dekhiye
DC: the capacitor charges and then no current flows, so the lamp does not glow. Reducing C changes nothing. AC: the capacitor offers reactance 1/ωC and current flows, so the lamp glows. Reducing C increases the reactance, so the lamp glows less brightly than before.dc: संधारित्र आवेशित होता है और फिर कोई धारा नहीं बहती, इसलिए लैंप नहीं जलता। C घटाने से कुछ नहीं बदलता। ac: संधारित्र 1/ωC प्रतिघात देता है और धारा बहती है, इसलिए लैंप जलता है। C घटाने से प्रतिघात बढ़ता है, इसलिए लैंप पहले से कम चमकता है।dc: capacitor charge hota hai aur phir koi current nahi behta, isliye lamp nahi jalta. C kam karne se kuch nahi badalta. ac: capacitor 1/ωC reactance deta hai aur current behta hai, isliye lamp jalta hai. C kam karne se reactance badhta hai, isliye lamp pehle se kam chamakta hai.DC: the capacitor charges and then no current flows, so the lamp does not glow. Making C smaller changes nothing. AC: the capacitor has a reactance 1/ωC, and current flows, so the lamp glows. Making C smaller increases the reactance, so the lamp glows less brightly than before.
A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is doubled, what happens to the capacitive reactance and the current?15.0 μF का संधारित्र 220 V, 50 Hz स्रोत से जुड़ा है। धारितीय प्रतिघात और परिपथ में धारा (rms और शिखर) ज्ञात कीजिए। यदि आवृत्ति दुगुनी कर दी जाए, तो धारितीय प्रतिघात और धारा का क्या होगा?15.0 μF ka capacitor 220 V, 50 Hz source se juda hai. Capacitive reactance aur circuit mein current (rms aur peak) nikaaliye. Agar frequency double kar di jaaye, to capacitive reactance aur current ka kya hoga?A 15.0 μF capacitor is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) in the circuit. If the frequency is made double, what happens to the capacitive reactance and to the current?
Show solutionहल देखेंSolution dekhiye
XC = 1/(2πνC) = 1/(2π × 50 × 15.0 × 10⁻⁶) ≈ 212 Ω. rms current I = V/XC = 220/212 ≈ 1.04 A. Peak current im = √2 I = 1.41 × 1.04 ≈ 1.47 A — it oscillates between +1.47 A and −1.47 A and is ahead of the voltage by π/2. If the frequency is doubled, XC is halved (≈ 106 Ω) and the current doubles (≈ 2.07 A rms).XC = 1/(2πνC) = 1/(2π × 50 × 15.0 × 10⁻⁶) ≈ 212 Ω। rms धारा I = V/XC = 220/212 ≈ 1.04 A। शिखर धारा im = √2 I = 1.41 × 1.04 ≈ 1.47 A — यह +1.47 A और −1.47 A के बीच दोलन करती है और वोल्टता से π/2 आगे है। आवृत्ति दुगुनी करने पर XC आधा (≈ 106 Ω) हो जाता है और धारा दुगुनी (≈ 2.07 A rms)।XC = 1/(2πνC) = 1/(2π × 50 × 15.0 × 10⁻⁶) ≈ 212 Ω. rms current I = V/XC = 220/212 ≈ 1.04 A. Peak current im = √2 I = 1.41 × 1.04 ≈ 1.47 A — ye +1.47 A aur −1.47 A ke beech oscillate karta hai aur voltage se π/2 aage hai. Frequency double karne par XC aadha (≈ 106 Ω) ho jaata hai aur current double (≈ 2.07 A rms).Reactance: XC = 1/(2πνC) = 1/(2π × 50 × 15.0 × 10⁻⁶) ≈ 212 Ω. rms current: I = V/XC = 220/212 ≈ 1.04 A. Peak current: im = √2 I = 1.41 × 1.04 ≈ 1.47 A. So the current swings between +1.47 A and −1.47 A, and it is ahead of the voltage by π/2. If the frequency is doubled, XC becomes half (≈ 106 Ω) and the current becomes double (≈ 2.07 A rms).
Tap to zoom · 5 labels to exploreज़ूम करने के लिए टैप करें · 5 लेबलTap kijiye → zoom · 5 labelsA light bulb and an open coil inductor are connected to an ac source through a key (Fig 7.9). The switch is closed and after some time an iron rod is inserted into the interior of the inductor. The glow of the bulb (a) increases; (b) decreases; (c) is unchanged, as the iron rod is inserted. Give your answer with reasons.एक बल्ब और खुली कुंडली वाला प्रेरक कुंजी द्वारा ac स्रोत से जुड़े हैं (चित्र 7.9)। स्विच बंद करने के कुछ समय बाद प्रेरक के अंदर लोहे की छड़ डाली जाती है। छड़ डालने पर बल्ब की चमक (a) बढ़ती है; (b) घटती है; (c) अपरिवर्तित रहती है। कारण सहित उत्तर दीजिए।Ek bulb aur open coil inductor key ke through ac source se jude hain (Fig 7.9). Switch band karne ke kuch der baad inductor ke andar iron rod daali jaati hai. Rod daalne par bulb ki chamak (a) badhti hai; (b) ghatti hai; (c) same rehti hai. Reason ke saath answer dijiye.A light bulb and an open coil (inductor) are connected to an ac source through a key (Fig 7.9). The switch is closed. After some time, an iron rod is pushed inside the coil. As the iron rod goes in, does the glow of the bulb (a) increase, (b) decrease, or (c) stay the same? Give reasons.
Show solutionहल देखेंSolution dekhiye
(b) decreases. The iron is magnetised by the field inside the coil, which increases the magnetic field and hence the inductance L. So the inductive reactance XL = ωL increases, a larger fraction of the applied ac voltage appears across the inductor, and less is left across the bulb — the glow decreases.(b) घटती है। कुंडली के अंदर का क्षेत्र लोहे को चुंबकित करता है, जिससे चुंबकीय क्षेत्र और इसलिए प्रेरकत्व L बढ़ता है। अतः प्रेरकीय प्रतिघात XL = ωL बढ़ता है, लगाई गई ac वोल्टता का बड़ा भाग प्रेरक पर आता है और बल्ब के लिए कम बचता है — चमक घटती है।(b) ghatti hai. Coil ke andar ka field iron ko magnetise karta hai, jisse magnetic field aur isliye inductance L badhta hai. Isliye inductive reactance XL = ωL badhta hai, lagaye gaye ac voltage ka bada hissa inductor par aata hai aur bulb ke liye kam bachta hai — chamak ghatti hai.(b) It decreases. The field inside the coil magnetises the iron. This makes the magnetic field stronger, so the inductance L increases. So the inductive reactance XL = ωL also increases. Now a bigger share of the ac voltage is used up across the inductor, and less is left for the bulb. So the bulb glows less.
Didn’t get this part?समझ नहीं आया?Samajh nahi aaya?
7.6AC Voltage Applied to a Series LCR Circuitश्रेणी LCR परिपथ पर लगाई गई ac वोल्टताSeries LCR Circuit par AC Voltage
Now put an inductor L, a resistor R and a capacitor C in series with an ac source v = vm sin ωt. If q is the charge on the capacitor and i the current at time t, Kirchhoff’s loop rule gives the equation below. We solve it with phasors.अब एक प्रेरक L, प्रतिरोधक R और संधारित्र C को ac स्रोत v = vm sin ωt के साथ श्रेणी में जोड़िए। यदि समय t पर संधारित्र पर आवेश q और धारा i हो, तो किरखोफ़ का पाश नियम नीचे का समीकरण देता है। इसे हम फ़ेज़र से हल करते हैं।Ab ek inductor L, resistor R aur capacitor C ko ac source v = vm sin ωt ke saath series mein jodiye. Agar time t par capacitor par charge q aur current i ho, to Kirchhoff ka loop rule neeche wali equation deta hai. Ise hum phasors se solve karte hain.Now join an inductor L, a resistor R and a capacitor C in series (one after another) with an ac source v = vm sin ωt. Let q be the charge on the capacitor and i the current at time t. Kirchhoff’s loop rule gives the equation below. We solve it using phasors (rotating arrows).
Tap to zoom · 8 labels to exploreज़ूम करने के लिए टैप करें · 8 लेबलTap kijiye → zoom · 8 labels7.6.1Phasor-diagram solutionफ़ेज़र-आरेख द्वारा हलPhasor-diagram se solution
In a series circuit the same current flows through R, L and C at every instant, with the same amplitude and phase. Let it be i = im sin(ωt + φ), where φ is the phase difference between the source voltage and the current. From the previous sections: VR is parallel to I, VL is π/2 ahead of I, and VC is π/2 behind I. Their lengths are vRm = im R, vLm = im XL, vCm = im XC.श्रेणी परिपथ में हर क्षण R, L और C में एक ही धारा समान आयाम और कला के साथ बहती है। मान लीजिए i = im sin(ωt + φ), जहाँ φ स्रोत वोल्टता और धारा के बीच कलांतर है। पिछले खंडों से: VR, I के समांतर है, VL, I से π/2 आगे है, और VC, I से π/2 पीछे है। इनकी लंबाइयाँ vRm = im R, vLm = im XL, vCm = im XC।Series circuit mein har instant R, L aur C mein ek hi current same amplitude aur phase ke saath behta hai. Maan lijiye i = im sin(ωt + φ), jahan φ source voltage aur current ke beech phase difference hai. Pichhle sections se: VR, I ke parallel hai, VL, I se π/2 aage hai, aur VC, I se π/2 peeche hai. Inki lengths vRm = im R, vLm = im XL, vCm = im XC.In a series circuit the same current flows through R, L and C at every instant, with the same amplitude and phase. Let it be i = im sin(ωt + φ), where φ is the phase difference between the source voltage and the current. From the earlier sections we know: VR is parallel to I (in step), VL is π/2 ahead of I, and VC is π/2 behind I. Their lengths are vRm = im R, vLm = im XL and vCm = im XC.
Tap to zoom · 8 labels to exploreज़ूम करने के लिए टैप करें · 8 लेबलTap kijiye → zoom · 8 labelsThe loop equation vL + vR + vC = v is the vertical component of the phasor relation VL + VR + VC = V. Since VC and VL lie along the same line in opposite directions, they combine into one phasor of length |vCm − vLm|. Then V is the hypotenuse of a right triangle with sides VR and (VC + VL):पाश समीकरण vL + vR + vC = v, फ़ेज़र संबंध VL + VR + VC = V का ऊर्ध्वाधर घटक है। क्योंकि VC और VL एक ही रेखा पर विपरीत दिशाओं में हैं, वे |vCm − vLm| लंबाई के एक फ़ेज़र में मिल जाते हैं। तब V, VR और (VC + VL) भुजाओं वाले समकोण त्रिभुज का कर्ण है:Loop equation vL + vR + vC = v, phasor relation VL + VR + VC = V ka vertical component hai. Kyunki VC aur VL ek hi line par opposite directions mein hain, wo |vCm − vLm| length ke ek phasor mein mil jaate hain. Tab V, VR aur (VC + VL) sides wale right triangle ka hypotenuse hai:The loop equation VL + vR + VC = V is just the vertical part of the phasor equation VL + VR + VC = V. VC and VL lie on the same line but point in opposite directions. So they join into one phasor of length |vCm − vLm|. Then V is the hypotenuse (longest side) of a right-angled triangle whose other two sides are VR and (VC + VL):
By analogy with resistance, Z is called the impedance. The phase angle φ is the angle between VR (i.e. I) and V:प्रतिरोध से समानता के आधार पर Z को प्रतिबाधा कहते हैं। कला कोण φ, VR (यानी I) और V के बीच का कोण है:Resistance se analogy mein Z ko impedance kehte hain. Phase angle φ, VR (yaani I) aur V ke beech ka angle hai:Just like resistance, Z is called the impedance (the total opposition to ac). The phase angle φ is the angle between VR (which is along I) and V:
Tap to zoom · 5 labels to exploreज़ूम करने के लिए टैप करें · 5 लेबलTap kijiye → zoom · 5 labelsIf XC > XL, φ is positive: the circuit is predominantly capacitive and the current leads the source voltage. If XC < XL, φ is negative: the circuit is predominantly inductive and the current lags the source voltage. The figure below shows the case XC > XL.यदि XC > XL, तो φ धनात्मक है: परिपथ मुख्यतः धारितीय है और धारा स्रोत वोल्टता से आगे रहती है। यदि XC < XL, तो φ ऋणात्मक है: परिपथ मुख्यतः प्रेरकीय है और धारा स्रोत वोल्टता से पीछे रहती है। नीचे का चित्र XC > XL की स्थिति दिखाता है।Agar XC > XL, to φ positive hai: circuit mainly capacitive hai aur current source voltage se aage rehta hai. Agar XC < XL, to φ negative hai: circuit mainly inductive hai aur current source voltage se peeche rehta hai. Neeche ka figure XC > XL wala case dikhata hai.Bigger capacitor effect → current runs ahead. Bigger inductor effect → current falls behind.
Tap to zoom · 8 labels to exploreज़ूम करने के लिए टैप करें · 8 लेबलTap kijiye → zoom · 8 labelsA limitation: the phasor method says nothing about the initial conditions. What we found is the steady-state solution. The general solution also has a transient part (which exists even when v = 0); after a sufficiently long time the transient dies out and the circuit follows the steady-state solution.एक सीमा: फ़ेज़र विधि प्रारंभिक अवस्था के बारे में कुछ नहीं बताती। हमें जो मिला वह स्थायी-अवस्था हल है। व्यापक हल में एक क्षणिक भाग भी होता है (जो v = 0 होने पर भी रहता है); पर्याप्त समय बाद क्षणिक भाग समाप्त हो जाता है और परिपथ स्थायी-अवस्था हल के अनुसार चलता है।Ek limitation: phasor method initial conditions ke baare mein kuch nahi batata. Humein jo mila wo steady-state solution hai. General solution mein ek transient part bhi hota hai (jo v = 0 hone par bhi rehta hai); kaafi time baad transient khatam ho jaata hai aur circuit steady-state solution ke hisaab se chalta hai.One limit of this method: phasors tell us nothing about the starting conditions. What we found is the steady-state solution (how the circuit behaves after it settles). The full solution also has a transient part (a short-lived part that exists even when v = 0). After enough time the transient dies away, and the circuit follows the steady-state solution.
Do not add voltages like numbers. In an RC circuit with VR = 151 V and VC = 160 V, the source is not 311 V: since VC is π/2 out of phase with VR, V = √(VR² + VC²) ≈ 220 V. Likewise, in series LCR, V = √(VR² + (VC − VL)²).वोल्टताओं को संख्याओं की तरह मत जोड़िए। VR = 151 V और VC = 160 V वाले RC परिपथ में स्रोत 311 V नहीं है: VC, VR से π/2 कलांतर पर है, इसलिए V = √(VR² + VC²) ≈ 220 V। इसी तरह श्रेणी LCR में V = √(VR² + (VC − VL)²)।Voltages ko numbers ki tarah mat jodiye. VR = 151 V aur VC = 160 V wale RC circuit mein source 311 V nahi hai: VC, VR se π/2 out of phase hai, isliye V = √(VR² + VC²) ≈ 220 V. Isi tarah series LCR mein V = √(VR² + (VC − VL)²).Do not add these voltages like ordinary numbers. In an RC circuit with VR = 151 V and VC = 160 V, the source voltage is not 311 V. VC is π/2 out of step with VR, so V = √(VR² + VC²) ≈ 220 V. In the same way, for series LCR, V = √(VR² + (VC − VL)²).
7.6.2ResonanceअनुनादResonance
Systems that tend to oscillate at a particular natural frequency respond with a large amplitude when driven at (or near) that frequency. A child on a swing is the familiar example: pull the rope in step with the swing’s own rhythm and it goes higher and higher. This is resonance.जो निकाय किसी विशेष स्वाभाविक आवृत्ति पर दोलन करना चाहते हैं, उन्हें उसी (या उसके निकट की) आवृत्ति से चलाने पर बड़ा आयाम मिलता है। झूले पर बच्चा परिचित उदाहरण है: झूले की अपनी लय के साथ रस्सी खींचिए तो झूला ऊँचा और ऊँचा जाता है। यह अनुनाद है।Jo systems kisi khaas natural frequency par oscillate karna chahte hain, unhe usi (ya uske paas ki) frequency se chalane par bada amplitude milta hai. Jhoole par baccha jaana-pehchaana example hai: jhoole ki apni rhythm ke saath rassi kheenchiye to jhoola ooncha aur ooncha jaata hai. Ye resonance hai.Some systems like to swing (oscillate) at their own natural frequency. When they are pushed at (or near) that frequency, they swing with a very large amplitude. A child on a swing is a common example: if you pull the rope in step with the swing’s own rhythm, it goes higher and higher. This is called resonance.
For the LCR circuit, im = vm/√(R² + (XC − XL)²) with XC = 1/ωC and XL = ωL. As ω is varied, at one frequency ω0 we get XC = XL, the impedance is minimum, Z = R, and the current amplitude is maximum, im = vm/R. This is the resonant frequency:LCR परिपथ में im = vm/√(R² + (XC − XL)²), जहाँ XC = 1/ωC और XL = ωL। ω बदलने पर एक आवृत्ति ω0 पर XC = XL होता है, प्रतिबाधा न्यूनतम, Z = R होती है, और धारा आयाम अधिकतम, im = vm/R होता है। यह अनुनादी आवृत्ति है:LCR circuit mein im = vm/√(R² + (XC − XL)²), jahan XC = 1/ωC aur XL = ωL. ω badalne par ek frequency ω0 par XC = XL hota hai, impedance minimum, Z = R hoti hai, aur current amplitude maximum, im = vm/R hota hai. Ye resonant frequency hai:For the LCR circuit, im = vm/√(R² + (XC − XL)²), with XC = 1/ωC and XL = ωL. As we change ω, at one special frequency ω0 we get XC = XL. Then the impedance is smallest, Z = R, and the current amplitude is largest, im = vm/R. This special frequency is the resonant frequency:
Tap to zoom · 4 labels to exploreज़ूम करने के लिए टैप करें · 4 लेबलTap kijiye → zoom · 4 labelsThe figure plots im against ω for L = 1.00 mH, C = 1.00 nF and vm = 100 V, for two resistances: (i) R = 100 Ω and (ii) R = 200 Ω. Here ω0 = 1/√(LC) = 1.00 × 10⁶ rad/s for both. Since im = vm/R at resonance, the peak for R = 100 Ω (1.0 A) is twice that for R = 200 Ω (0.5 A).चित्र में L = 1.00 mH, C = 1.00 nF और vm = 100 V के लिए दो प्रतिरोधों (i) R = 100 Ω और (ii) R = 200 Ω पर im बनाम ω दिखाया गया है। यहाँ दोनों के लिए ω0 = 1/√(LC) = 1.00 × 10⁶ rad/s। अनुनाद पर im = vm/R, इसलिए R = 100 Ω का शिखर (1.0 A), R = 200 Ω वाले (0.5 A) का दुगुना है।Figure mein L = 1.00 mH, C = 1.00 nF aur vm = 100 V ke liye do resistances (i) R = 100 Ω aur (ii) R = 200 Ω par im vs ω dikhaya gaya hai. Yahan dono ke liye ω0 = 1/√(LC) = 1.00 × 10⁶ rad/s. Resonance par im = vm/R, isliye R = 100 Ω ka peak (1.0 A), R = 200 Ω wale (0.5 A) ka double hai.The figure shows im against ω for L = 1.00 mH, C = 1.00 nF and vm = 100 V, for two resistances: (i) R = 100 Ω and (ii) R = 200 Ω. For both, ω0 = 1/√(LC) = 1.00 × 10⁶ rad/s. At resonance im = vm/R, so the peak for R = 100 Ω (1.0 A) is twice the peak for R = 200 Ω (0.5 A).
Tuning a radio: the antenna picks up signals from many stations, which act as sources at many frequencies in the tuning circuit. Turning the tuning knob changes the capacitance C until the circuit’s resonant frequency matches the frequency of the station you want; the current at that frequency becomes maximum and you hear that station.रेडियो ट्यून करना: ऐन्टेना कई स्टेशनों के सिग्नल पकड़ता है, जो ट्यूनिंग परिपथ में अनेक आवृत्तियों के स्रोत की तरह काम करते हैं। ट्यूनिंग नॉब घुमाने से धारिता C तब तक बदलती है जब तक परिपथ की अनुनादी आवृत्ति आपके चाहे स्टेशन की आवृत्ति के बराबर न हो जाए; उस आवृत्ति की धारा अधिकतम हो जाती है और आपको वही स्टेशन सुनाई देता है।Radio tune karna: antenna kai stations ke signals pakadta hai, jo tuning circuit mein kai frequencies ke source ki tarah kaam karte hain. Tuning knob ghumane se capacitance C tab tak badalti hai jab tak circuit ki resonant frequency aapke chahe station ki frequency ke barabar na ho jaaye; us frequency ka current maximum ho jaata hai aur aapko wahi station sunai deta hai.Tuning a radio: the antenna catches signals from many stations. In the tuning circuit, these act like sources of many different frequencies. When you turn the tuning knob, you change the capacitance C. When the circuit’s resonant frequency matches the frequency of the station you want, the current at that frequency becomes the largest, and you hear that station.
Resonance needs both L and C. Only then can the voltages across L and C (π out of phase with each other) cancel, leaving the whole source voltage across R. An RL or RC circuit cannot show resonance.अनुनाद के लिए L और C दोनों चाहिए। तभी L और C पर की वोल्टताएँ (एक-दूसरे से π कलांतर पर) निरस्त हो सकती हैं और पूरी स्रोत वोल्टता R पर आती है। RL या RC परिपथ में अनुनाद नहीं हो सकता।Resonance ke liye L aur C dono chahiye. Tabhi L aur C par ke voltages (ek-doosre se π out of phase) cancel ho sakte hain aur poora source voltage R par aata hai. RL ya RC circuit mein resonance nahi ho sakta.Resonance needs both L and C. Only then can the voltages across L and C (which are π out of step with each other) cancel each other, so that the whole source voltage falls across R. An RL circuit or an RC circuit cannot show resonance.
Sharpness (quality factor Q): NCERT’s table of quantities lists Q = ω0L/R = 1/(ω0CR) for a series RLC circuit. A smaller R gives a larger Q — a taller, narrower peak, so the circuit picks one frequency more selectively (good tuning). Also note: at resonance the voltage across L alone (= I XL) can be much larger than the source voltage, because it is cancelled by the equal and opposite voltage across C.तीक्ष्णता (गुणता गुणांक Q): NCERT की राशियों की तालिका श्रेणी RLC परिपथ के लिए Q = ω0L/R = 1/(ω0CR) देती है। कम R → बड़ा Q — ऊँचा, संकरा शिखर, अतः परिपथ एक आवृत्ति को अधिक चयनात्मक रूप से चुनता है (अच्छी ट्यूनिंग)। ध्यान दें: अनुनाद पर अकेले L पर वोल्टता (= I XL) स्रोत वोल्टता से बहुत अधिक हो सकती है, क्योंकि C पर बराबर और विपरीत वोल्टता उसे निरस्त कर देती है।Sharpness (quality factor Q): NCERT ki quantities wali table series RLC circuit ke liye Q = ω0L/R = 1/(ω0CR) deti hai. Kam R → bada Q — ooncha, patla peak, isliye circuit ek frequency ko zyada selectively chunta hai (achhi tuning). Dhyaan dijiye: resonance par akele L par voltage (= I XL) source voltage se bahut zyada ho sakta hai, kyunki C par barabar aur opposite voltage use cancel kar deta hai.Sharpness (quality factor Q): NCERT’s table of quantities gives Q = ω0L/R = 1/(ω0CR) for a series RLC circuit. A smaller R gives a bigger Q. Then the peak is taller and narrower, so the circuit picks one frequency more sharply (good tuning). Also note: at resonance, the voltage across L alone (= I XL) can be much bigger than the source voltage. This is possible because it is cancelled by an equal and opposite voltage across C.
Set R, L, C and the source frequency. See XL, XC, Z, the current, phase φ, power factor and power update live, the phasor triangle redraw, and a marker move along the resonance curve. Defaults are NCERT Example 7.8 — then tap “Tune to resonance” for Example 7.9.R, L, C और स्रोत की आवृत्ति चुनिए। XL, XC, Z, धारा, कला φ, शक्ति गुणांक और शक्ति लाइव बदलते देखिए, फ़ेज़र त्रिभुज फिर से बनते और अनुनाद वक्र पर बिंदु खिसकते देखिए। शुरुआती मान NCERT उदाहरण 7.8 के हैं — फिर उदाहरण 7.9 के लिए “अनुनाद पर ट्यून करें” दबाइए।R, L, C aur source frequency set kijiye. XL, XC, Z, current, phase φ, power factor aur power live badalte dekhiye, phasor triangle dobara bante aur resonance curve par marker khisakte dekhiye. Default values NCERT Example 7.8 ki hain — phir Example 7.9 ke liye “Tune to resonance” dabaiye.
A resistor of 200 Ω and a capacitor of 15.0 μF are connected in series to a 220 V, 50 Hz ac source. (a) Calculate the current in the circuit; (b) calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox.200 Ω का प्रतिरोधक और 15.0 μF का संधारित्र श्रेणी में 220 V, 50 Hz ac स्रोत से जुड़े हैं। (a) परिपथ में धारा ज्ञात कीजिए; (b) प्रतिरोधक और संधारित्र पर (rms) वोल्टता ज्ञात कीजिए। क्या इन वोल्टताओं का बीजगणितीय योग स्रोत वोल्टता से अधिक है? यदि हाँ, तो इस विरोधाभास को सुलझाइए।200 Ω ka resistor aur 15.0 μF ka capacitor series mein 220 V, 50 Hz ac source se jude hain. (a) Circuit mein current nikaaliye; (b) resistor aur capacitor par (rms) voltage nikaaliye. Kya in voltages ka algebraic sum source voltage se zyada hai? Agar haan, to is paradox ko suljhaiye.A 200 Ω resistor and a 15.0 μF capacitor are joined in series to a 220 V, 50 Hz ac source. (a) Find the current in the circuit. (b) Find the rms voltage across the resistor and across the capacitor. Is the simple sum of these voltages more than the source voltage? If yes, explain this puzzle (paradox).
Show solutionहल देखेंSolution dekhiye
(a) XC = 1/(2πνC) = 1/(2 × 3.14 × 50 × 15.0 × 10⁻⁶) ≈ 212.3 Ω. Z = √(R² + XC²) = √(200² + 212.3²) ≈ 291.7 Ω. I = V/Z = 220/291.7 ≈ 0.755 A (more precisely 0.754 A).(a) XC = 1/(2πνC) = 1/(2 × 3.14 × 50 × 15.0 × 10⁻⁶) ≈ 212.3 Ω। Z = √(R² + XC²) = √(200² + 212.3²) ≈ 291.7 Ω। I = V/Z = 220/291.7 ≈ 0.755 A (अधिक सटीक 0.754 A)।(a) XC = 1/(2πνC) = 1/(2 × 3.14 × 50 × 15.0 × 10⁻⁶) ≈ 212.3 Ω. Z = √(R² + XC²) = √(200² + 212.3²) ≈ 291.7 Ω. I = V/Z = 220/291.7 ≈ 0.755 A (zyada precise 0.754 A).(a) First the reactance: XC = 1/(2πνC) = 1/(2 × 3.14 × 50 × 15.0 × 10⁻⁶) ≈ 212.3 Ω. Then the impedance: Z = √(R² + XC²) = √(200² + 212.3²) ≈ 291.7 Ω. Current: I = V/Z = 220/291.7 ≈ 0.755 A (more exactly 0.754 A).
(b) The same current flows through both: VR = IR ≈ 0.755 × 200 ≈ 151 V; VC = I XC ≈ 0.755 × 212.3 ≈ 160 V. Their algebraic sum ≈ 311 V, which is more than 220 V! The paradox goes away because the two voltages are 90° out of phase — they cannot be added like ordinary numbers. Using Pythagoras: V = √(VR² + VC²) = √(151² + 160²) ≈ 220 V, exactly the source voltage.(b) दोनों में एक ही धारा: VR = IR ≈ 0.755 × 200 ≈ 151 V; VC = I XC ≈ 0.755 × 212.3 ≈ 160 V। इनका बीजगणितीय योग ≈ 311 V, जो 220 V से अधिक है! विरोधाभास इसलिए दूर होता है कि दोनों वोल्टताएँ 90° कलांतर पर हैं — इन्हें साधारण संख्याओं की तरह नहीं जोड़ा जा सकता। पाइथागोरस से: V = √(VR² + VC²) = √(151² + 160²) ≈ 220 V, ठीक स्रोत वोल्टता।(b) Dono mein ek hi current: VR = IR ≈ 0.755 × 200 ≈ 151 V; VC = I XC ≈ 0.755 × 212.3 ≈ 160 V. Inka algebraic sum ≈ 311 V, jo 220 V se zyada hai! Paradox isliye door hota hai ki dono voltages 90° out of phase hain — inhe normal numbers ki tarah nahi joda ja sakta. Pythagoras se: V = √(VR² + VC²) = √(151² + 160²) ≈ 220 V, bilkul source voltage.(b) The same current flows through both parts. VR = IR ≈ 0.755 × 200 ≈ 151 V. VC = I XC ≈ 0.755 × 212.3 ≈ 160 V. If we just add them, we get ≈ 311 V, which is more than 220 V! The puzzle is solved because the two voltages are 90° out of phase (out of step). So they cannot be added like ordinary numbers. Use Pythagoras instead: V = √(VR² + VC²) = √(151² + 160²) ≈ 220 V, which is exactly the source voltage.
Didn’t get this part?समझ नहीं आया?Samajh nahi aaya?
7.7Power in AC Circuit: The Power Factorac परिपथ में शक्ति: शक्ति गुणांकAC Circuit mein Power: Power Factor
A voltage v = vm sin ωt across a series RLC circuit drives i = im sin(ωt + φ), with im = vm/Z and φ = tan⁻¹[(XC − XL)/R]. The instantaneous power supplied by the source isश्रेणी RLC परिपथ पर वोल्टता v = vm sin ωt धारा i = im sin(ωt + φ) चलाती है, जहाँ im = vm/Z और φ = tan⁻¹[(XC − XL)/R]। स्रोत द्वारा दी गई तात्क्षणिक शक्तिSeries RLC circuit par voltage v = vm sin ωt current i = im sin(ωt + φ) chalata hai, jahan im = vm/Z aur φ = tan⁻¹[(XC − XL)/R]. Source ki di gayi instantaneous powerA voltage v = vm sin ωt across a series RLC circuit drives a current i = im sin(ωt + φ), where im = vm/Z and φ = tan⁻¹[(XC − XL)/R]. The power given by the source at any instant is
So the average power depends not only on V and I but also on cos φ, called the power factor. From the impedance triangle, cos φ = R/Z. The cases:अतः औसत शक्ति केवल V और I पर नहीं, बल्कि cos φ पर भी निर्भर है, जिसे शक्ति गुणांक कहते हैं। प्रतिबाधा त्रिभुज से cos φ = R/Z। स्थितियाँ:Isliye average power sirf V aur I par nahi, balki cos φ par bhi depend karti hai, jise power factor kehte hain. Impedance triangle se cos φ = R/Z. Cases:Power factor tells what fraction of V × I is really used up. It is 1 for a pure resistor and 0 for a pure coil or capacitor.
- (i) Resistive circuit (only R): φ = 0, cos φ = 1 — maximum power dissipation.(i) प्रतिरोधी परिपथ (केवल R): φ = 0, cos φ = 1 — अधिकतम शक्ति क्षय।(i) Resistive circuit (sirf R): φ = 0, cos φ = 1 — maximum power dissipation.(i) Resistive circuit (only R): φ = 0 and cos φ = 1. The power used up is the maximum.
- (ii) Purely inductive or capacitive circuit: φ = ±π/2, cos φ = 0 — no power is dissipated even though a current flows. Such a current is called a wattless current.(ii) शुद्ध प्रेरकीय या धारितीय परिपथ: φ = ±π/2, cos φ = 0 — धारा बहने पर भी कोई शक्ति क्षय नहीं। ऐसी धारा को वाटहीन धारा कहते हैं।(ii) Pure inductive ya capacitive circuit: φ = ±π/2, cos φ = 0 — current behne par bhi koi power dissipate nahi hoti. Aise current ko wattless current kehte hain.(ii) Purely inductive or purely capacitive circuit: φ = ±π/2 and cos φ = 0. No power is used up, even though a current flows. Such a current is called a wattless current.
- (iii) LCR series circuit: P = VI cos φ with φ = tan⁻¹[(XC − XL)/R]; φ may be non-zero in an RL, RC or RLC circuit. Even then, power is dissipated only in the resistor.(iii) LCR श्रेणी परिपथ: P = VI cos φ, जहाँ φ = tan⁻¹[(XC − XL)/R]; RL, RC या RLC परिपथ में φ शून्येतर हो सकता है। तब भी शक्ति केवल प्रतिरोधक में क्षय होती है।(iii) LCR series circuit: P = VI cos φ, jahan φ = tan⁻¹[(XC − XL)/R]; RL, RC ya RLC circuit mein φ non-zero ho sakta hai. Tab bhi power sirf resistor mein dissipate hoti hai.(iii) LCR series circuit: P = VI cos φ, with φ = tan⁻¹[(XC − XL)/R]. In an RL, RC or RLC circuit, φ may not be zero. Even so, power is used up only in the resistor.
- (iv) At resonance: XC − XL = 0, φ = 0, cos φ = 1 and P = I²Z = I²R — maximum power (through R).(iv) अनुनाद पर: XC − XL = 0, φ = 0, cos φ = 1 और P = I²Z = I²R — अधिकतम शक्ति (R में)।(iv) Resonance par: XC − XL = 0, φ = 0, cos φ = 1 aur P = I²Z = I²R — maximum power (R mein).(iv) At resonance: XC − XL = 0, so φ = 0 and cos φ = 1. Then P = I²Z = I²R. The power is the maximum (all used in R).
Power ratings of ac appliances are average power ratings, and the average power in an ac circuit is never negative. Only the resistive part of a circuit uses up energy; pure L and pure C just store energy and give it back.ac उपकरणों की शक्ति रेटिंग औसत शक्ति होती है, और ac परिपथ में औसत शक्ति कभी ऋणात्मक नहीं होती। केवल परिपथ का प्रतिरोधी भाग ऊर्जा व्यय करता है; शुद्ध L और शुद्ध C बस ऊर्जा संचित करके लौटा देते हैं।ac appliances ki power rating average power hoti hai, aur ac circuit mein average power kabhi negative nahi hoti. Sirf circuit ka resistive part energy kharch karta hai; pure L aur pure C bas energy store karke lauta dete hain.The power rating of an ac appliance is its average power. The average power in an ac circuit is never negative. Only the resistive part of a circuit uses up energy. A pure L and a pure C only store energy for a while and then give it back.
(a) For circuits used for transporting electric power, a low power factor implies large power loss in transmission. Explain. (b) Power factor can often be improved by the use of a capacitor of appropriate capacitance in the circuit. Explain.(a) विद्युत शक्ति के परिवहन वाले परिपथों में कम शक्ति गुणांक का अर्थ संचरण में अधिक शक्ति हानि है। समझाइए। (b) परिपथ में उचित धारिता का संधारित्र लगाकर प्रायः शक्ति गुणांक सुधारा जा सकता है। समझाइए।(a) Bijli transport karne wale circuits mein low power factor ka matlab transmission mein zyada power loss hai. Samjhaiye. (b) Circuit mein sahi capacitance ka capacitor lagakar aksar power factor sudhara ja sakta hai. Samjhaiye.(a) In lines that carry electric power, a low power factor means a big power loss during transmission. Explain why. (b) The power factor can often be improved by adding a capacitor of the right capacitance to the circuit. Explain how.
Show solutionहल देखेंSolution dekhiye
(a) P = IV cos φ. To supply a given power at a given voltage with a small cos φ, the current must be larger. A larger current means a larger I²R loss in the transmission lines.(a) P = IV cos φ। दी गई वोल्टता पर दी गई शक्ति देने के लिए कम cos φ हो तो धारा अधिक चाहिए। अधिक धारा का अर्थ संचरण लाइनों में अधिक I²R हानि।(a) P = IV cos φ. Di gayi voltage par di gayi power dene ke liye kam cos φ ho to current zyada chahiye. Zyada current ka matlab transmission lines mein zyada I²R loss.(a) Power P = IV cos φ. To deliver a fixed power at a fixed voltage with a small cos φ, the current has to be bigger. A bigger current means a bigger I²R loss (heating) in the transmission wires.
(b) Suppose the current I lags the voltage by φ (an inductive load). Resolve I into Ip = I cos φ along V (the power component, which does the work) and Iq = I sin φ perpendicular to V (the wattless component, no power). Connect a capacitor in parallel that draws a leading wattless current I′q equal to Iq. They cancel, the line current becomes just Ip, and the power factor approaches 1 while P = V Ip is unchanged.(b) मान लीजिए धारा I वोल्टता से φ पीछे है (प्रेरकीय भार)। I को V के अनुदिश Ip = I cos φ (शक्ति घटक, जो कार्य करता है) और V के लंबवत Iq = I sin φ (वाटहीन घटक, कोई शक्ति नहीं) में वियोजित कीजिए। पार्श्व में एक संधारित्र जोड़िए जो Iq के बराबर अग्रगामी वाटहीन धारा I′q ले। दोनों निरस्त हो जाती हैं, लाइन धारा केवल Ip रह जाती है, और P = V Ip वही रहते हुए शक्ति गुणांक 1 के निकट पहुँचता है।(b) Maan lijiye current I voltage se φ peeche hai (inductive load). I ko V ke along Ip = I cos φ (power component, jo kaam karta hai) aur V ke perpendicular Iq = I sin φ (wattless component, koi power nahi) mein todiye. Parallel mein ek capacitor jodiye jo Iq ke barabar leading wattless current I′q le. Dono cancel ho jaate hain, line current sirf Ip reh jaata hai, aur P = V Ip same rehte hue power factor 1 ke paas pahunchta hai.(b) Suppose the current I lags behind the voltage by φ (this happens with an inductive load, like a motor). Split I into two parts. Ip = I cos φ is along V. This is the power component, which does the useful work. Iq = I sin φ is at right angles to V. This is the wattless component, which gives no power. Now connect a capacitor in parallel. It draws a leading wattless current I′q equal to Iq. The two wattless currents cancel. The line current becomes just Ip, and the power factor comes close to 1, while the power P = V Ip stays the same.
Tap to zoom · 7 labels to exploreज़ूम करने के लिए टैप करें · 7 लेबलTap kijiye → zoom · 7 labelsA sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 μF. Find (a) the impedance of the circuit; (b) the phase difference between the voltage across the source and the current; (c) the power dissipated in the circuit; and (d) the power factor.283 V शिखर मान और 50 Hz आवृत्ति की ज्यावक्रीय वोल्टता एक श्रेणी LCR परिपथ पर लगाई गई है जिसमें R = 3 Ω, L = 25.48 mH और C = 796 μF। ज्ञात कीजिए (a) परिपथ की प्रतिबाधा; (b) स्रोत वोल्टता और धारा के बीच कलांतर; (c) परिपथ में क्षयित शक्ति; और (d) शक्ति गुणांक।283 V peak value aur 50 Hz frequency ka sinusoidal voltage ek series LCR circuit par lagaya gaya hai jismein R = 3 Ω, L = 25.48 mH aur C = 796 μF. Nikaaliye (a) circuit ki impedance; (b) source voltage aur current ke beech phase difference; (c) circuit mein dissipate hui power; aur (d) power factor.A sine-wave voltage with peak value 283 V and frequency 50 Hz is applied to a series LCR circuit with R = 3 Ω, L = 25.48 mH and C = 796 μF. Find (a) the impedance of the circuit, (b) the phase difference between the source voltage and the current, (c) the power used up in the circuit, and (d) the power factor.
Show solutionहल देखेंSolution dekhiye
(a) XL = 2πνL = 2 × 3.14 × 50 × 25.48 × 10⁻³ ≈ 8 Ω; XC = 1/(2πνC) = 1/(2 × 3.14 × 50 × 796 × 10⁻⁶) ≈ 4 Ω. Z = √(R² + (XL − XC)²) = √(3² + 4²) = 5 Ω.(a) XL = 2πνL = 2 × 3.14 × 50 × 25.48 × 10⁻³ ≈ 8 Ω; XC = 1/(2πνC) = 1/(2 × 3.14 × 50 × 796 × 10⁻⁶) ≈ 4 Ω। Z = √(R² + (XL − XC)²) = √(3² + 4²) = 5 Ω।(a) XL = 2πνL = 2 × 3.14 × 50 × 25.48 × 10⁻³ ≈ 8 Ω; XC = 1/(2πνC) = 1/(2 × 3.14 × 50 × 796 × 10⁻⁶) ≈ 4 Ω. Z = √(R² + (XL − XC)²) = √(3² + 4²) = 5 Ω.(a) First find both reactances. XL = 2πνL = 2 × 3.14 × 50 × 25.48 × 10⁻³ ≈ 8 Ω. XC = 1/(2πνC) = 1/(2 × 3.14 × 50 × 796 × 10⁻⁶) ≈ 4 Ω. Then the impedance: Z = √(R² + (XL − XC)²) = √(3² + 4²) = 5 Ω.
(b) φ = tan⁻¹[(XC − XL)/R] = tan⁻¹[(4 − 8)/3] = −53.1°. φ is negative, so the current lags the source voltage (the circuit is inductive).(b) φ = tan⁻¹[(XC − XL)/R] = tan⁻¹[(4 − 8)/3] = −53.1°। φ ऋणात्मक है, अतः धारा स्रोत वोल्टता से पीछे है (परिपथ प्रेरकीय है)।(b) φ = tan⁻¹[(XC − XL)/R] = tan⁻¹[(4 − 8)/3] = −53.1°. φ negative hai, isliye current source voltage se peeche hai (circuit inductive hai).(b) φ = tan⁻¹[(XC − XL)/R] = tan⁻¹[(4 − 8)/3] = −53.1°. φ is negative, so the current lags behind the source voltage. The circuit behaves as inductive.
(c) rms current I = im/√2 = (1/√2)(283/5) ≈ 40 A. P = I²R = (40)² × 3 = 4800 W. (d) Power factor = cos φ = cos(−53.1°) = 0.6 (check: R/Z = 3/5).(c) rms धारा I = im/√2 = (1/√2)(283/5) ≈ 40 A। P = I²R = (40)² × 3 = 4800 W। (d) शक्ति गुणांक = cos φ = cos(−53.1°) = 0.6 (जाँच: R/Z = 3/5)।(c) rms current I = im/√2 = (1/√2)(283/5) ≈ 40 A. P = I²R = (40)² × 3 = 4800 W. (d) Power factor = cos φ = cos(−53.1°) = 0.6 (check: R/Z = 3/5).(c) rms current: I = im/√2 = (1/√2)(283/5) ≈ 40 A. Power used: P = I²R = (40)² × 3 = 4800 W. (d) Power factor = cos φ = cos(−53.1°) = 0.6. (Check: R/Z = 3/5 = 0.6.)
Suppose the frequency of the source in the previous example can be varied. (a) What is the frequency of the source at which resonance occurs? (b) Calculate the impedance, the current, and the power dissipated at the resonant condition.मान लीजिए पिछले उदाहरण में स्रोत की आवृत्ति बदली जा सकती है। (a) किस आवृत्ति पर अनुनाद होगा? (b) अनुनाद की स्थिति में प्रतिबाधा, धारा और क्षयित शक्ति ज्ञात कीजिए।Maan lijiye pichhle example mein source ki frequency badli ja sakti hai. (a) Kis frequency par resonance hoga? (b) Resonance par impedance, current aur dissipate hui power nikaaliye.Now suppose we can change the frequency of the source in the last example. (a) At what frequency does resonance happen? (b) At resonance, find the impedance, the current and the power used up.
Show solutionहल देखेंSolution dekhiye
(a) ω0 = 1/√(LC) = 1/√(25.48 × 10⁻³ × 796 × 10⁻⁶) ≈ 222 rad/s, so ν0 = ω0/2π ≈ 222/(2 × 3.14) ≈ 35.4 Hz (35.3 Hz with π = 3.1416).(a) ω0 = 1/√(LC) = 1/√(25.48 × 10⁻³ × 796 × 10⁻⁶) ≈ 222 rad/s, अतः ν0 = ω0/2π ≈ 222/(2 × 3.14) ≈ 35.4 Hz (π = 3.1416 लेने पर 35.3 Hz)।(a) ω0 = 1/√(LC) = 1/√(25.48 × 10⁻³ × 796 × 10⁻⁶) ≈ 222 rad/s, isliye ν0 = ω0/2π ≈ 222/(2 × 3.14) ≈ 35.4 Hz (π = 3.1416 lene par 35.3 Hz).(a) ω0 = 1/√(LC) = 1/√(25.48 × 10⁻³ × 796 × 10⁻⁶) ≈ 222 rad/s. To get the frequency in hertz, divide by 2π: ν0 = ω0/2π ≈ 222/(2 × 3.14) ≈ 35.4 Hz (35.3 Hz if you use π = 3.1416).
(b) At resonance Z = R = 3 Ω. rms current I = V/Z = (283/√2)/3 ≈ 66.7 A. P = I²R = (66.7)² × 3 ≈ 13.35 kW — more than the 4.8 kW of Example 7.8, as expected at resonance.(b) अनुनाद पर Z = R = 3 Ω। rms धारा I = V/Z = (283/√2)/3 ≈ 66.7 A। P = I²R = (66.7)² × 3 ≈ 13.35 kW — उदाहरण 7.8 के 4.8 kW से अधिक, जैसा अनुनाद पर अपेक्षित है।(b) Resonance par Z = R = 3 Ω. rms current I = V/Z = (283/√2)/3 ≈ 66.7 A. P = I²R = (66.7)² × 3 ≈ 13.35 kW — Example 7.8 ke 4.8 kW se zyada, jaisa resonance par expected hai.(b) At resonance, Z = R = 3 Ω. rms current: I = V/Z = (283/√2)/3 ≈ 66.7 A. Power: P = I²R = (66.7)² × 3 ≈ 13.35 kW. This is more than the 4.8 kW of Example 7.8, just as we expect at resonance.
At an airport, a person is made to walk through the doorway of a metal detector for security reasons. If she/he is carrying anything made of metal, the detector emits a sound. On what principle does this detector work?हवाई अड्डे पर सुरक्षा के लिए व्यक्ति को धातु संसूचक के द्वार से गुज़ारा जाता है। यदि उसके पास धातु की कोई वस्तु हो, तो संसूचक ध्वनि करता है। यह संसूचक किस सिद्धांत पर कार्य करता है?Airport par security ke liye vyakti ko metal detector ke darwaze se guzaara jaata hai. Agar uske paas metal ki koi cheez ho, to detector awaaz karta hai. Ye detector kis principle par kaam karta hai?At an airport, a person is made to walk through the doorway of a metal detector for safety checks. If she or he carries anything made of metal, the detector makes a sound. What principle does this detector work on?
Show solutionहल देखेंSolution dekhiye
Resonance in ac circuits. The doorway contains a coil of many turns connected to a capacitor, tuned so the circuit is at resonance. Metal passing through the coil changes its impedance, which causes a significant change in the current. Electronic circuitry detects this change and sounds the alarm.ac परिपथों में अनुनाद। द्वार में अनेक फेरों की कुंडली होती है जो संधारित्र से जुड़ी होती है और परिपथ अनुनाद पर ट्यून रहता है। कुंडली से धातु गुज़रने पर इसकी प्रतिबाधा बदलती है, जिससे धारा में उल्लेखनीय परिवर्तन होता है। इलेक्ट्रॉनिक परिपथ यह परिवर्तन पकड़कर अलार्म बजाता है।ac circuits mein resonance. Darwaze mein kai turns ki coil hoti hai jo capacitor se judi hoti hai aur circuit resonance par tuned rehta hai. Coil se metal guzarne par iski impedance badalti hai, jisse current mein bada badlaav hota hai. Electronic circuit ye badlaav pakad kar alarm bajata hai.Resonance in ac circuits. The doorway has a coil with many turns joined to a capacitor. The circuit is tuned so it is at resonance. When metal passes through the coil, it changes the coil’s impedance. This causes a big change in the current. An electronic circuit notices this change and sounds the alarm.
Didn’t get this part?समझ नहीं आया?Samajh nahi aaya?
7.8Transformersट्रांसफ़ॉर्मरTransformers
To change an alternating voltage to a larger or smaller value we use a transformer, which works on the principle of mutual induction. It has two coils, insulated from each other, wound on a soft-iron core — either one on top of the other, or on separate limbs of the core. The primary coil (Np turns) is usually the input; the secondary coil (Ns turns) is the output.प्रत्यावर्ती वोल्टता को बड़े या छोटे मान में बदलने के लिए हम ट्रांसफ़ॉर्मर का उपयोग करते हैं, जो अन्योन्य प्रेरण के सिद्धांत पर कार्य करता है। इसमें एक-दूसरे से विद्युतरोधी दो कुंडलियाँ नर्म लौह क्रोड पर लिपटी होती हैं — या तो एक के ऊपर दूसरी, या क्रोड की अलग भुजाओं पर। प्राथमिक कुंडली (Np फेरे) प्रायः निवेश है; द्वितीयक कुंडली (Ns फेरे) निर्गत है।Alternating voltage ko badi ya chhoti value mein badalne ke liye hum transformer use karte hain, jo mutual induction ke principle par kaam karta hai. Ismein ek-doosre se insulated do coils soft-iron core par lipti hoti hain — ya to ek ke upar doosri, ya core ki alag limbs par. Primary coil (Np turns) aksar input hai; secondary coil (Ns turns) output hai.To change an ac voltage to a higher or lower value, we use a transformer. It works on the principle of mutual induction. It has two coils, insulated from each other, wound on a soft-iron core. The coils are either wound one on top of the other, or on separate arms (limbs) of the core. The primary coil (Np turns) is usually the input. The secondary coil (Ns turns) is the output.
Tap to zoom · 6 labels to exploreज़ूम करने के लिए टैप करें · 6 लेबलTap kijiye → zoom · 6 labels
Tap to zoom · 6 labels to exploreज़ूम करने के लिए टैप करें · 6 लेबलTap kijiye → zoom · 6 labelsAn alternating voltage on the primary drives a current that produces an alternating magnetic flux in the core. This flux links the secondary and induces an emf in it. Take an ideal transformer: the primary has negligible resistance and all the flux links both coils. If φ is the flux in each turn at time t:प्राथमिक पर प्रत्यावर्ती वोल्टता एक धारा चलाती है जो क्रोड में प्रत्यावर्ती चुंबकीय फ्लक्स बनाती है। यह फ्लक्स द्वितीयक से गुज़रकर उसमें emf प्रेरित करता है। एक आदर्श ट्रांसफ़ॉर्मर लीजिए: प्राथमिक का प्रतिरोध नगण्य है और पूरा फ्लक्स दोनों कुंडलियों से गुज़रता है। यदि समय t पर प्रत्येक फेरे में फ्लक्स φ हो:Primary par alternating voltage ek current chalata hai jo core mein alternating magnetic flux banata hai. Ye flux secondary se guzar kar usmein emf induce karta hai. Ek ideal transformer lijiye: primary ka resistance negligible hai aur poora flux dono coils se guzarta hai. Agar time t par har turn mein flux φ ho:An ac voltage on the primary coil drives a current. This current makes an alternating (changing) magnetic flux in the core. The flux passes through the secondary coil and induces an emf in it. Consider an ideal transformer: the primary has almost no resistance, and all the flux passes through both coils. Let φ be the flux in each turn at time t:
Here εp = vp (otherwise the zero-resistance primary would carry an infinite current), and if the secondary is open or draws little current, εs ≈ vs. Dividing the two equations:यहाँ εp = vp (वरना शून्य प्रतिरोध वाली प्राथमिक में अनंत धारा बहती), और यदि द्वितीयक खुली हो या कम धारा ले, तो εs ≈ vs। दोनों समीकरणों को भाग देने पर:Yahan εp = vp (warna zero resistance wali primary mein infinite current behta), aur agar secondary open ho ya kam current le, to εs ≈ vs. Dono equations ko divide karne par:Here εp = vp (if not, the zero-resistance primary would carry an infinite current). If the secondary circuit is open, or takes very little current, then εs ≈ vs. Now divide one equation by the other:
This used three assumptions: (i) the primary resistance and current are small; (ii) the same flux links both coils (very little flux escapes the core); (iii) the secondary current is small. If the transformer is 100% efficient (no energy loss), power in = power out: ip vp = is vs. A well-designed transformer can have an efficiency of more than 95%. Combining:इसमें तीन मान्यताएँ थीं: (i) प्राथमिक का प्रतिरोध और धारा कम है; (ii) दोनों कुंडलियों से वही फ्लक्स गुज़रता है (क्रोड से बहुत कम फ्लक्स बाहर जाता है); (iii) द्वितीयक धारा कम है। यदि ट्रांसफ़ॉर्मर 100% दक्ष हो (कोई ऊर्जा हानि नहीं), तो निवेशी शक्ति = निर्गत शक्ति: ip vp = is vs। अच्छी तरह बने ट्रांसफ़ॉर्मर की दक्षता 95% से अधिक हो सकती है। मिलाने पर:Ismein teen assumptions thi: (i) primary ka resistance aur current kam hai; (ii) dono coils se same flux guzarta hai (core se bahut kam flux bahar jaata hai); (iii) secondary current kam hai. Agar transformer 100% efficient ho (koi energy loss nahi), to input power = output power: ip vp = is vs. Achhe se bane transformer ki efficiency 95% se zyada ho sakti hai. Milane par:We made three assumptions here: (i) the primary’s resistance and current are small; (ii) the same flux passes through both coils (very little flux leaks out of the core); (iii) the secondary current is small. If the transformer is 100% efficient (no energy is lost), then power in = power out: ip vp = is vs. A well-made transformer can have an efficiency of more than 95%. Putting these together:
If Ns > Np, the voltage is stepped up (Vs > Vp): a step-up transformer — but the secondary current is smaller (Is < Ip). Example from the book: Np = 100, Ns = 200, so Ns/Np = 2; a 220 V input at 10 A becomes 440 V at 5.0 A. If Ns < Np, it is a step-down transformer: Vs < Vp and Is > Ip.यदि Ns > Np, तो वोल्टता बढ़ती है (Vs > Vp): उच्चायी (step-up) ट्रांसफ़ॉर्मर — पर द्वितीयक धारा कम होती है (Is < Ip)। पुस्तक का उदाहरण: Np = 100, Ns = 200, अतः Ns/Np = 2; 10 A पर 220 V निवेश 5.0 A पर 440 V बन जाता है। यदि Ns < Np, तो यह अपचायी (step-down) ट्रांसफ़ॉर्मर है: Vs < Vp और Is > Ip।Agar Ns > Np, to voltage badhta hai (Vs > Vp): step-up transformer — par secondary current kam hota hai (Is < Ip). Book ka example: Np = 100, Ns = 200, isliye Ns/Np = 2; 10 A par 220 V input 5.0 A par 440 V ban jaata hai. Agar Ns < Np, to ye step-down transformer hai: Vs < Vp aur Is > Ip.More turns on the output side → more voltage but less current. Fewer turns → less voltage but more current.
A step-up transformer does not create energy. Voltage goes up by the same factor that current goes down, so power in ≈ power out. And a transformer does not work on steady dc: a constant current gives a constant flux, so nothing is induced in the secondary.उच्चायी ट्रांसफ़ॉर्मर ऊर्जा नहीं बनाता। वोल्टता जितने गुना बढ़ती है, धारा उतने ही गुना घटती है, इसलिए निवेशी शक्ति ≈ निर्गत शक्ति। और ट्रांसफ़ॉर्मर स्थिर dc पर काम नहीं करता: स्थिर धारा से फ्लक्स स्थिर रहता है, इसलिए द्वितीयक में कुछ प्रेरित नहीं होता।Step-up transformer energy nahi banata. Voltage jitne guna badhta hai, current utne hi guna ghatta hai, isliye input power ≈ output power. Aur transformer steady dc par kaam nahi karta: constant current se flux constant rehta hai, isliye secondary mein kuch induce nahi hota.A step-up transformer does not make extra energy. The voltage goes up by the same factor that the current goes down, so power in ≈ power out. Also, a transformer does not work on steady dc. A constant current gives a constant flux, so nothing is induced in the secondary.
Choose the primary and secondary turns, the input voltage and the load. The ideal-transformer relations give Vs, Is and Ip live — check that power in = power out, and switch to dc to see why the output vanishes.प्राथमिक और द्वितीयक के फेरे, निवेशी वोल्टता और भार चुनिए। आदर्श ट्रांसफ़ॉर्मर के संबंध Vs, Is और Ip लाइव देते हैं — जाँचिए कि निवेशी शक्ति = निर्गत शक्ति, और dc पर बदलकर देखिए कि निर्गत क्यों गायब हो जाता है।Primary aur secondary ke turns, input voltage aur load chuniye. Ideal transformer ke relations Vs, Is aur Ip live dete hain — check kijiye ki input power = output power, aur dc par switch karke dekhiye ki output kyun gaayab ho jaata hai.
Rotate a real-looking transformer: laminated soft-iron core, a primary coil with fewer turns and a secondary with more (step-up). Tap any part to see its name and job; the glowing flux path shows the alternating flux going round the core.एक असली जैसा ट्रांसफ़ॉर्मर घुमाइए: पटलित नर्म लौह क्रोड, कम फेरों वाली प्राथमिक और अधिक फेरों वाली द्वितीयक कुंडली (उच्चायी)। किसी भी भाग पर टैप करके उसका नाम और काम देखिए; चमकता फ्लक्स पथ क्रोड में घूमता प्रत्यावर्ती फ्लक्स दिखाता है।Ek asli jaisa transformer ghumaiye: laminated soft-iron core, kam turns wali primary aur zyada turns wali secondary coil (step-up). Kisi bhi part par tap karke uska naam aur kaam dekhiye; chamakta flux path core mein ghoomta alternating flux dikhata hai.
Energy losses in real transformers:वास्तविक ट्रांसफ़ॉर्मरों में ऊर्जा हानियाँ:Real transformers mein energy losses:Energy losses in real transformers (where the lost energy goes):
- (i) Flux leakage: not all the primary’s flux passes through the secondary (poor core design, air gaps). Reduced by winding the primary and secondary one over the other.(i) फ्लक्स क्षरण: प्राथमिक का पूरा फ्लक्स द्वितीयक से नहीं गुज़रता (क्रोड की खराब बनावट, वायु अंतराल)। प्राथमिक और द्वितीयक को एक के ऊपर दूसरी लपेटकर घटाया जाता है।(i) Flux leakage: primary ka poora flux secondary se nahi guzarta (core ka kharab design, air gaps). Primary aur secondary ko ek ke upar doosri lapet kar kam kiya jaata hai.(i) Flux leakage: not all the flux of the primary passes through the secondary (because of poor core design or air gaps). This is reduced by winding the primary and secondary coils one over the other.
- (ii) Resistance of the windings: the wire heats up (I²R). In high-current, low-voltage windings this is reduced by using thick wire.(ii) कुंडलियों का प्रतिरोध: तार गरम होता है (I²R)। उच्च धारा, निम्न वोल्टता वाली कुंडलियों में मोटे तार से इसे घटाया जाता है।(ii) Windings ka resistance: taar garam hota hai (I²R). High-current, low-voltage windings mein mote taar se ise kam kiya jaata hai.(ii) Resistance of the windings: the wire gets hot (I²R loss). In windings that carry high current at low voltage, this is reduced by using thick wire.
- (iii) Eddy currents: the alternating flux induces eddy currents in the iron core and heats it. Reduced by using a laminated core.(iii) भँवर धाराएँ: प्रत्यावर्ती फ्लक्स लौह क्रोड में भँवर धाराएँ प्रेरित कर उसे गरम करता है। पटलित क्रोड से इसे घटाया जाता है।(iii) Eddy currents: alternating flux iron core mein eddy currents induce karke use garam karta hai. Laminated core se ise kam kiya jaata hai.(iii) Eddy currents: the changing flux induces swirling currents (eddy currents) in the iron core, and these heat it. This is reduced by using a laminated core (made of thin insulated sheets).
- (iv) Hysteresis: the core’s magnetisation is reversed again and again; the energy spent appears as heat. Kept small by using a magnetic material with low hysteresis loss.(iv) शैथिल्य: क्रोड का चुंबकन बार-बार उलटता है; व्यय ऊर्जा ऊष्मा के रूप में प्रकट होती है। कम शैथिल्य हानि वाले चुंबकीय पदार्थ से इसे कम रखा जाता है।(iv) Hysteresis: core ka magnetisation baar-baar palatta hai; kharch hui energy heat ban jaati hai. Kam hysteresis loss wale magnetic material se ise kam rakha jaata hai.(iv) Hysteresis: the core’s magnetisation is reversed again and again, and the energy used for this turns into heat. It is kept small by using a magnetic material with low hysteresis loss.
Long-distance transmission: the generator’s output voltage is stepped up, so the current and hence the I²R loss in the lines is cut down. It is transmitted to an area sub-station near the consumers, where the voltage is stepped down, and stepped down again at distribution sub-stations and utility poles before a supply of 240 V reaches our homes.लंबी दूरी का संचरण: जनित्र की निर्गत वोल्टता बढ़ाई जाती है, जिससे धारा और इसलिए लाइनों में I²R हानि घटती है। इसे उपभोक्ताओं के पास के क्षेत्रीय उपकेंद्र तक भेजा जाता है, जहाँ वोल्टता घटाई जाती है, और फिर वितरण उपकेंद्रों और खंभों पर और घटाई जाती है, तब जाकर 240 V की आपूर्ति हमारे घरों तक पहुँचती है।Lambi doori ka transmission: generator ka output voltage badhaya jaata hai, jisse current aur isliye lines mein I²R loss kam hota hai. Ise consumers ke paas ke area sub-station tak bheja jaata hai, jahan voltage ghataya jaata hai, aur phir distribution sub-stations aur poles par aur ghataya jaata hai, tab jaakar 240 V ki supply hamare gharon tak pahunchti hai.Sending power over long distances: the output voltage of the generator is stepped up (raised). This lowers the current, and so lowers the I²R loss in the wires. The power is sent to an area sub-station near the users. There the voltage is stepped down. It is stepped down again at distribution sub-stations and on electric poles, until a supply of 240 V reaches our homes.
Tap to zoom · 6 labels to exploreज़ूम करने के लिए टैप करें · 6 लेबलTap kijiye → zoom · 6 labelsAround you: your phone charger contains a small step-down transformer (inside its switch-mode circuit); the grey drum on an electricity pole near your house is a step-down distribution transformer; and its gentle hum comes from the core’s magnetisation reversing 100 times every second on 50 Hz mains.आपके आस-पास: आपके फ़ोन चार्जर में एक छोटा अपचायी ट्रांसफ़ॉर्मर होता है (उसके स्विच-मोड परिपथ में); घर के पास बिजली के खंभे पर लगा स्लेटी ड्रम अपचायी वितरण ट्रांसफ़ॉर्मर है; और इसकी हल्की गुनगुनाहट 50 Hz मेन्स पर क्रोड के चुंबकन के हर सेकंड 100 बार उलटने से आती है।Aapke aas-paas: aapke phone charger mein ek chhota step-down transformer hota hai (uske switch-mode circuit mein); ghar ke paas bijli ke khambe par laga grey drum step-down distribution transformer hai; aur iski halki hum 50 Hz mains par core ke magnetisation ke har second 100 baar palatne se aati hai.Around you: your phone charger has a small step-down transformer (inside its switch-mode circuit). The grey drum on an electric pole near your house is a step-down distribution transformer. Its soft humming sound comes from the core’s magnetisation reversing 100 times every second on 50 Hz mains.
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- AC through R: i = (vm/R) sin ωt, in phase with v. Average current over a cycle = 0, but average power = ½ im²R.R में ac: i = (vm/R) sin ωt, v के साथ समान कला में। एक चक्र में औसत धारा = 0, पर औसत शक्ति = ½ im²R।R mein ac: i = (vm/R) sin ωt, v ke saath in phase. Ek cycle mein average current = 0, par average power = ½ im²R.AC through R: i = (vm/R) sin ωt, in step (in phase) with v. The average current over a cycle = 0, but the average power = ½ im²R.
- rms value = 0.707 × peak (sine wave). 220 V mains (rms) has a peak of about 311 V. Meters read rms.rms मान = 0.707 × शिखर (ज्या तरंग)। 220 V मेन्स (rms) का शिखर लगभग 311 V। मीटर rms पढ़ते हैं।rms value = 0.707 × peak (sine wave). 220 V mains (rms) ka peak lagbhag 311 V. Meters rms padhte hain.rms value = 0.707 × peak value (for a sine wave). The 220 V mains (rms) has a peak of about 311 V. Meters show rms values.
- A phasor is a rotating arrow (anticlockwise, ω); length = amplitude, vertical projection = instantaneous value. Voltage and current are scalars.फ़ेज़र घूमता तीर है (वामावर्त, ω); लंबाई = आयाम, ऊर्ध्वाधर प्रक्षेप = तात्क्षणिक मान। वोल्टता और धारा अदिश हैं।Phasor ghoomta teer hai (anticlockwise, ω); length = amplitude, vertical projection = instantaneous value. Voltage aur current scalars hain.A phasor is a rotating arrow (turning anticlockwise at ω). Its length = the amplitude, and its vertical shadow (projection) = the value at that instant. Voltage and current are scalars.
- Pure L: XL = ωL, current LAGS voltage by π/2. Pure C: XC = 1/ωC, current LEADS voltage by π/2. Average power in both = 0.शुद्ध L: XL = ωL, धारा वोल्टता से π/2 पीछे। शुद्ध C: XC = 1/ωC, धारा वोल्टता से π/2 आगे। दोनों में औसत शक्ति = 0।Pure L: XL = ωL, current voltage se π/2 PEECHE. Pure C: XC = 1/ωC, current voltage se π/2 AAGE. Dono mein average power = 0.Pure L: XL = ωL, and current LAGS behind voltage by π/2. Pure C: XC = 1/ωC, and current LEADS voltage by π/2. Average power in both = 0.
- Series LCR: VR ∥ I, VL ahead of I by π/2, VC behind I by π/2. Z = √(R² + (XC − XL)²), tan φ = (XC − XL)/R.श्रेणी LCR: VR ∥ I, VL, I से π/2 आगे, VC, I से π/2 पीछे। Z = √(R² + (XC − XL)²), tan φ = (XC − XL)/R।Series LCR: VR ∥ I, VL, I se π/2 aage, VC, I se π/2 peeche. Z = √(R² + (XC − XL)²), tan φ = (XC − XL)/R.Series LCR: VR ∥ I (in step), VL is ahead of I by π/2, VC is behind I by π/2. Z = √(R² + (XC − XL)²), and tan φ = (XC − XL)/R.
- XC > XL → capacitive, current leads (φ > 0). XC < XL → inductive, current lags (φ < 0).XC > XL → धारितीय, धारा आगे (φ > 0)। XC < XL → प्रेरकीय, धारा पीछे (φ < 0)।XC > XL → capacitive, current aage (φ > 0). XC < XL → inductive, current peeche (φ < 0).If XC > XL, the circuit is capacitive and the current leads (φ > 0). If XC < XL, the circuit is inductive and the current lags (φ < 0).
- Resonance: XL = XC at ω0 = 1/√(LC); Z = R (minimum), current maximum = vm/R, power factor 1. Needs both L and C. Used in radio tuning and metal detectors.अनुनाद: ω0 = 1/√(LC) पर XL = XC; Z = R (न्यूनतम), धारा अधिकतम = vm/R, शक्ति गुणांक 1। L और C दोनों चाहिए। रेडियो ट्यूनिंग और धातु संसूचक में उपयोग।Resonance: ω0 = 1/√(LC) par XL = XC; Z = R (minimum), current maximum = vm/R, power factor 1. L aur C dono chahiye. Radio tuning aur metal detectors mein use.Resonance: XL = XC at ω0 = 1/√(LC). Then Z = R (smallest), the current is largest = vm/R, and the power factor is 1. It needs both L and C. It is used in radio tuning and metal detectors.
- P = VI cos φ. cos φ = 1 for pure R, 0 for pure L or C (wattless current). Only R dissipates power. A parallel capacitor improves a lagging power factor.P = VI cos φ। शुद्ध R के लिए cos φ = 1, शुद्ध L या C के लिए 0 (वाटहीन धारा)। केवल R शक्ति क्षय करता है। पार्श्व संधारित्र पिछड़े शक्ति गुणांक को सुधारता है।P = VI cos φ. Pure R ke liye cos φ = 1, pure L ya C ke liye 0 (wattless current). Sirf R power dissipate karta hai. Parallel capacitor lagging power factor sudharta hai.P = VI cos φ. cos φ = 1 for pure R, and 0 for pure L or pure C (wattless current). Only R uses up power. A capacitor joined in parallel improves a lagging power factor.
- Transformer (mutual induction, ac only): Vs/Vp = Ns/Np, Is/Ip = Np/Ns. Step-up: Ns > Np (V up, I down). Step-down: Ns < Np.ट्रांसफ़ॉर्मर (अन्योन्य प्रेरण, केवल ac): Vs/Vp = Ns/Np, Is/Ip = Np/Ns। उच्चायी: Ns > Np (V बढ़े, I घटे)। अपचायी: Ns < Np।Transformer (mutual induction, sirf ac): Vs/Vp = Ns/Np, Is/Ip = Np/Ns. Step-up: Ns > Np (V upar, I neeche). Step-down: Ns < Np.Transformer (works by mutual induction, ac only): Vs/Vp = Ns/Np and Is/Ip = Np/Ns. Step-up: Ns > Np (voltage up, current down). Step-down: Ns < Np.
- Transformer losses and cures: flux leakage (wind coils one over the other), winding resistance (thick wire), eddy currents (laminated core), hysteresis (low-loss core material).ट्रांसफ़ॉर्मर हानियाँ और उपाय: फ्लक्स क्षरण (कुंडलियाँ एक के ऊपर दूसरी), कुंडली प्रतिरोध (मोटा तार), भँवर धाराएँ (पटलित क्रोड), शैथिल्य (कम हानि वाला क्रोड पदार्थ)।Transformer losses aur upaay: flux leakage (coils ek ke upar doosri), winding resistance (mota taar), eddy currents (laminated core), hysteresis (kam loss wala core material).Transformer losses and their cures: flux leakage (wind coils one over the other), winding resistance (use thick wire), eddy currents (use a laminated core), hysteresis (use a low-loss core material).
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These are Simple Learning Hub’s own study notes written to follow the current (rationalised) NCERT Class 12 Physics textbook section by section; read them with the official NCERT chapter. Diagrams and simulations are original SLH work based on the NCERT figures they name. Report a mistake