CLASS 12 PHYSICS · CHAPTER 3 · READ
Current Electricityविद्युत धाराCurrent Electricity
Current → why charges flow in conductors → Ohm’s law → drift of electrons and mobility → where Ohm’s law fails → resistivity and temperature → power → cells, emf, internal resistance → Kirchhoff’s rules → Wheatstone bridge.धारा → चालकों में आवेश क्यों बहते हैं → ओम नियम → इलेक्ट्रॉनों का अपवाह और गतिशीलता → ओम नियम कहाँ विफल → प्रतिरोधकता और ताप → शक्ति → सेल, विद्युत वाहक बल, आंतरिक प्रतिरोध → किरखोफ़ के नियम → व्हीटस्टोन सेतु।Current → conductors mein charges kyun behte hain → Ohm’s law → electrons ka drift aur mobility → Ohm’s law kahan fail → resistivity aur temperature → power → cells, emf, internal resistance → Kirchhoff’s rules → Wheatstone bridge.
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 haiDidn’t understand a paragraph? Tap it (or its Explain button) → Hindi / Hinglish / Simple English.कोई पैराग्राफ़ समझ नहीं आया? उस पर (या समझें बटन पर) टैप करें → अंग्रेज़ी / हिंग्लिश / सरल अंग्रेज़ी।Koi paragraph samajh nahi aaya? Us par (ya Samjho button par) tap kijiye → Hindi / English / Simple English.
Try it yourself — live simulationsख़ुद करके देखें — लाइव सिमुलेशनKhud karke dekhiye — live simulations
Diagrams in this chapterइस अध्याय के चित्रIs chapter ke diagrams
▶ Animate: electrons drifting in a wire▶ एनिमेट: तार में अपवाहित होते इलेक्ट्रॉन▶ Animate: wire mein drift karte electronsAnimated · 3D
▶ Animate: balancing a Wheatstone bridge▶ एनिमेट: व्हीटस्टोन सेतु को संतुलित करना▶ Animate: Wheatstone bridge balance karnaAnimated
Resistance, length and area: R = ρl/Aप्रतिरोध, लंबाई और क्षेत्रफल: R = ρl/AResistance, length and area: R = ρl/AFig 3.2
Drift of electrons and electric currentइलेक्ट्रॉनों का अपवाह और विद्युत धाराDrift of electrons and electric currentFig 3.3–3.4
Where Ohm’s law holds — and where it failsओम नियम कहाँ लागू होता है — और कहाँ नहींWhere Ohm’s law holds — and where it failsFig 3.5–3.7
How resistivity changes with temperatureताप के साथ प्रतिरोधकता कैसे बदलती हैHow resistivity changes with temperatureFig 3.8–3.10
A cell: emf and internal resistanceसेल: विद्युत वाहक बल और आंतरिक प्रतिरोधA cell: emf and internal resistanceFig 3.12
Cells in series and in parallelश्रेणी और पार्श्व क्रम में सेलCells in series and in parallelFig 3.13–3.14
Kirchhoff’s junction rule and loop ruleकिरखोफ़ के संधि नियम और लूप नियमKirchhoff’s junction rule and loop ruleFig 3.15
Kirchhoff’s rules at work: the network of NCERT Example 3.6किरखोफ़ के नियमों का उपयोग: NCERT उदाहरण 3.6 का परिपथKirchhoff’s rules at work: the network of NCERT Example 3.6Fig 3.17
The Wheatstone bridgeव्हीटस्टोन सेतुThe Wheatstone bridgeFig 3.18In this chapter (NCERT sections)इस अध्याय में (NCERT खंड)Is chapter mein (NCERT sections)
- 3.1IntroductionभूमिकाIntroduction
- 3.2Electric Currentविद्युत धाराElectric Current
- 3.3Electric Currents in Conductorsचालकों में विद्युत धाराConductors mein Electric Current
- 3.4Ohm’s Lawओम नियमOhm’s Law
- 3.5Drift of Electrons and the Origin of Resistivityइलेक्ट्रॉनों का अपवाह तथा प्रतिरोधकता का मूलElectrons ka Drift aur Resistivity ka Origin
- 3.5.1MobilityगतिशीलताMobility
- 3.6Limitations of Ohm’s Lawओम नियम की सीमाएँOhm’s Law ki Limitations
- 3.7Resistivity of Various Materialsविभिन्न पदार्थों की प्रतिरोधकताAlag Materials ki Resistivity
- 3.8Temperature Dependence of Resistivityप्रतिरोधकता की ताप पर निर्भरताResistivity ki Temperature par Dependence
- 3.9Electrical Energy, Powerविद्युत ऊर्जा, शक्तिElectrical Energy, Power
- 3.10Cells, emf, Internal Resistanceसेल, विद्युत वाहक बल, आंतरिक प्रतिरोधCells, emf, Internal Resistance
- 3.11Cells in Series and in Parallelश्रेणी तथा पार्श्व क्रम में सेलSeries aur Parallel mein Cells
- 3.12Kirchhoff’s Rulesकिरखोफ़ के नियमKirchhoff’s Rules
- 3.13Wheatstone Bridgeव्हीटस्टोन सेतुWheatstone Bridge
3.1IntroductionभूमिकाIntroduction
In Chapters 1 and 2 the charges were at rest. Now they move. A flow of charge is an electric current — lightning is a huge, short burst of it; the current in a torch or a phone charger is steady. This chapter is about steady currents in wires and the laws that govern them.अध्याय 1 और 2 में आवेश विराम में थे। अब वे गति करते हैं। आवेश का प्रवाह विद्युत धारा है — तड़ित इसका विशाल, क्षणिक झटका है; टॉर्च या फ़ोन चार्जर की धारा स्थायी होती है। यह अध्याय तारों में स्थायी धाराओं और उनके नियमों के बारे में है।Chapter 1 aur 2 mein charges rest mein the. Ab wo move karte hain. Charge ka flow electric current hai — bijli iska bahut bada, short jhatka hai; torch ya phone charger ka current steady hota hai. Ye chapter wires mein steady currents aur unke laws ke baare mein hai.In Chapters 1 and 2, the charges were at rest. Now they move. A flow of charge is called an electric current. Lightning is a huge current that lasts a very short time. The current in a torch or a phone charger is steady (it does not change). This chapter is about steady currents in wires, and the laws they follow.
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3.2Electric Currentविद्युत धाराElectric Current
If a net charge q flows across an area in time t, the current is the rate of flow of charge. If the current changes with time, use a very small interval: I(t) = lim Δq/Δt. By convention the direction of current is the direction in which positive charge would flow.यदि समय t में किसी क्षेत्रफल से नेट आवेश q बहता है, तो धारा आवेश प्रवाह की दर है। यदि धारा समय के साथ बदले, तो बहुत छोटा अंतराल लें: I(t) = lim Δq/Δt। परिपाटी से धारा की दिशा वह है जिसमें धन आवेश बहता।Agar time t mein kisi area se net charge q behta hai, to current charge flow ki rate hai. Agar current time ke saath badle, to bahut chhota interval lo: I(t) = lim Δq/Δt. Convention se current ki direction wo hai jismein positive charge behta.Suppose a net charge q flows across an area in time t. The current is the rate of flow of charge (charge flowing each second). If the current changes with time, we take a very small time interval: I(t) = lim Δq/Δt. By convention (an agreed rule), the direction of current is the direction in which positive charge would flow.
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3.3Electric Currents in Conductorsचालकों में विद्युत धाराConductors mein Electric Current
A metal has free electrons moving randomly. Without a field, as many cross an area one way as the other, so there is no net current. Apply a field E and the electrons are pushed opposite to E — a net flow, i.e., a current. To keep the current steady, a source (a cell) must keep the ends at different potentials by continually moving charge round a closed circuit. In electrolytes, both positive and negative ions carry current.धातु में मुक्त इलेक्ट्रॉन यादृच्छिक गति करते हैं। क्षेत्र के बिना जितने एक ओर से किसी क्षेत्रफल को पार करते हैं उतने ही दूसरी ओर से, इसलिए कोई नेट धारा नहीं। क्षेत्र E लगाएँ तो इलेक्ट्रॉन E के विपरीत धकेले जाते हैं — नेट प्रवाह, यानी धारा। धारा को स्थायी रखने के लिए एक स्रोत (सेल) को बंद परिपथ में आवेश लगातार घुमाकर सिरों को अलग विभव पर रखना होता है। विद्युत अपघट्यों में धन और ऋण दोनों आयन धारा ले जाते हैं।Metal mein free electrons random motion karte hain. Field ke bina jitne ek taraf se kisi area ko cross karte hain utne hi doosri taraf se, isliye koi net current nahi. Field E lagaiye to electrons E ke opposite dhakele jaate hain — net flow, yaani current. Current ko steady rakhne ke liye ek source (cell) ko closed circuit mein charge lagataar ghuma kar ends ko alag potential par rakhna hota hai. Electrolytes mein positive aur negative dono ions current le jaate hain.A metal has free electrons that move about randomly. With no field, as many electrons cross an area one way as the other way. So there is no net current. Now apply a field E. The electrons are pushed opposite to E. This gives a net flow — a current. To keep the current steady, we need a source (like a cell). It keeps the two ends at different potentials by moving charge round and round a closed circuit (a complete loop). In electrolytes (liquids that conduct), both positive and negative ions carry the current.
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3.4Ohm’s Lawओम नियमOhm’s Law
Ohm’s law (G.S. Ohm, 1828): the potential difference V across a conductor is proportional to the current I through it. The constant is the resistance R.ओम नियम (जी.एस. ओम, 1828): किसी चालक के सिरों के बीच विभवांतर V उसमें बहने वाली धारा I के अनुक्रमानुपाती है। स्थिरांक प्रतिरोध R है।Ohm’s law (G.S. Ohm, 1828): kisi conductor ke ends ke beech potential difference V usmein behne wale current I ke proportional hai. Constant resistance R hai.Double the voltage and the current doubles. R tells how hard it is for current to flow.
Tap to zoom · 3 labels to exploreज़ूम करने के लिए टैप करें · 3 लेबलTap kijiye → zoom · 3 labelsR depends on the material and on the shape. Put two identical slabs end to end: the same current, double the voltage, so R ∝ l. Put them side by side: double the current for the same voltage, so R ∝ 1/A (see the slabs). Together:R पदार्थ और आकार दोनों पर निर्भर है। दो समान पट्टिकाएँ सिरे-से-सिरा रखें: वही धारा, दोगुनी वोल्टता, इसलिए R ∝ l। साथ-साथ रखें: उसी वोल्टता पर दोगुनी धारा, इसलिए R ∝ 1/A (पट्टिकाएँ देखें)। मिलाकर:R material aur shape dono par depend karta hai. Do same slabs end-to-end rakhiye: wahi current, double voltage, isliye R ∝ l. Side-by-side rakhiye: usi voltage par double current, isliye R ∝ 1/A (slabs dekhiye). Milakar:R depends on the material and on the shape of the conductor. Join two identical slabs end to end. The current is the same, but the voltage needed is double. So R ∝ l (resistance grows with length). Now put them side by side. The current doubles for the same voltage. So R ∝ 1/A (resistance falls as the area grows). Putting both together: 👉 (see the slabs)
Ohm’s law can also be written for a point inside the material. With current density j = I/A (A m⁻²) and E = V/l, V = IR becomes E = ρj, or j = σE, where σ = 1/ρ is the conductivity (S m⁻¹).ओम नियम को पदार्थ के अंदर किसी बिंदु के लिए भी लिखा जा सकता है। धारा घनत्व j = I/A (A m⁻²) और E = V/l लेने पर V = IR बन जाता है E = ρj, या j = σE, जहाँ σ = 1/ρ चालकता है (S m⁻¹)।Ohm’s law ko material ke andar kisi point ke liye bhi likh sakte hain. Current density j = I/A (A m⁻²) aur E = V/l lene par V = IR ban jaata hai E = ρj, ya j = σE, jahan σ = 1/ρ conductivity hai (S m⁻¹).Ohm’s law can also be written for one point inside the material. Use the current density j = I/A (current per unit area, in A m⁻²) and E = V/l. Then V = IR becomes E = ρj, or j = σE. Here σ = 1/ρ is called the conductivity (unit S m⁻¹).
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3.5Drift of Electrons and the Origin of Resistivityइलेक्ट्रॉनों का अपवाह तथा प्रतिरोधकता का मूलElectrons ka Drift aur Resistivity ka Origin
In a field, each electron accelerates at a = −eE/m — until it collides with an ion and its velocity is randomised again. The average time between collisions is the relaxation time τ. Averaged over all electrons, the random velocities cancel and what remains is a small steady drift velocity (see the drift):क्षेत्र में हर इलेक्ट्रॉन a = −eE/m से त्वरित होता है — जब तक किसी आयन से टकरा न जाए और उसका वेग फिर यादृच्छिक न हो जाए। टक्करों के बीच औसत समय विश्रांति काल τ है। सभी इलेक्ट्रॉनों का औसत लेने पर यादृच्छिक वेग निरस्त हो जाते हैं और एक छोटा स्थायी अपवाह वेग बचता है (अपवाह देखें):Field mein har electron a = −eE/m se accelerate hota hai — jab tak kisi ion se takra na jaaye aur uski velocity phir random na ho jaaye. Collisions ke beech average time relaxation time τ hai. Saare electrons ka average lene par random velocities cancel ho jaati hain aur ek chhoti steady drift velocity bachti hai (drift dekhiye):In a field, each electron speeds up with acceleration a = −eE/m. This goes on until it hits (collides with) an ion. After the hit, its velocity becomes random again. The average time between two collisions is called the relaxation time τ. When we average over all the electrons, the random velocities cancel out. What is left is a small, steady drift velocity (a slow overall movement in one direction): 👉 (see the drift)
Tap to zoom · 6 labels to exploreज़ूम करने के लिए टैप करें · 6 लेबलTap kijiye → zoom · 6 labelsWatch the electrons zigzag. Switch the field on: on top of the random motion they slowly drift against E. Change V and see the drift speed and the current change.इलेक्ट्रॉनों को टेढ़ा-मेढ़ा चलते देखें। क्षेत्र चालू करें: यादृच्छिक गति के ऊपर वे धीरे-धीरे E के विपरीत खिसकते हैं। V बदलें और अपवाह चाल व धारा बदलती देखें।Electrons ko zigzag chalte dekhiye. Field on kijiye: random motion ke upar wo dheere-dheere E ke against khiskte hain. V badliye aur drift speed aur current badalte dekhiye.
3.5.1MobilityगतिशीलताMobility
Mobility μ is the drift speed per unit field: μ = |vd|/E = eτ/m. Its SI unit is m² V⁻¹ s⁻¹ (often quoted in cm² V⁻¹ s⁻¹). High mobility means charges move easily — important for semiconductors (Chapter 14).गतिशीलता μ प्रति एकांक क्षेत्र अपवाह चाल है: μ = |vd|/E = eτ/m। इसका SI मात्रक m² V⁻¹ s⁻¹ है (प्रायः cm² V⁻¹ s⁻¹ में)। अधिक गतिशीलता यानी आवेश आसानी से चलते हैं — अर्धचालकों के लिए महत्वपूर्ण (अध्याय 14)।Mobility μ per unit field drift speed hai: μ = |vd|/E = eτ/m. Iski SI unit m² V⁻¹ s⁻¹ hai (aksar cm² V⁻¹ s⁻¹ mein). Zyada mobility matlab charges aasaani se chalte hain — semiconductors ke liye important (Chapter 14).Mobility μ is the drift speed for each unit of field: μ = |vd|/E = eτ/m. Its SI unit is m² V⁻¹ s⁻¹ (it is often given in cm² V⁻¹ s⁻¹). High mobility means the charges move easily. This is important for semiconductors (Chapter 14).
(a) Estimate the drift speed of electrons in a copper wire of cross-section 1.0 × 10⁻⁷ m² carrying 1.5 A. Take one free electron per atom; density of copper 9.0 × 10³ kg m⁻³, atomic mass 63.5 u. (b) Compare it with the thermal speed of copper atoms and with the speed at which the field travels along the wire.(a) 1.0 × 10⁻⁷ m² काट वाले ताँबे के तार में 1.5 A धारा पर इलेक्ट्रॉनों की अपवाह चाल का अनुमान लगाएँ। प्रति परमाणु एक मुक्त इलेक्ट्रॉन; ताँबे का घनत्व 9.0 × 10³ kg m⁻³, परमाणु द्रव्यमान 63.5 u। (b) इसकी तुलना ताँबे के परमाणुओं की ऊष्मीय चाल और तार में क्षेत्र के चलने की चाल से करें।(a) 1.0 × 10⁻⁷ m² cross-section wale copper wire mein 1.5 A current par electrons ki drift speed estimate kijiye. Har atom ek free electron; copper ki density 9.0 × 10³ kg m⁻³, atomic mass 63.5 u. (b) Iski tulna copper atoms ki thermal speed aur wire mein field ke chalne ki speed se kijiye.(a) Find roughly the drift speed of electrons in a copper wire. The wire has a cross-section of 1.0 × 10⁻⁷ m² and carries 1.5 A. Take one free electron for each atom. The density of copper is 9.0 × 10³ kg m⁻³ and its atomic mass is 63.5 u. (b) Compare this drift speed with the thermal speed (heat motion speed) of copper atoms, and with the speed at which the electric field travels along the wire.
Show solutionहल देखेंSolution dekhiye
(a) n = (6.0 × 10²³/63.5 g) × 9.0 × 10⁶ g m⁻³ ≈ 8.5 × 10²⁸ m⁻³. vd = I/(neA) = 1.5/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁷) ≈ 1.1 mm s⁻¹. (b) Thermal speeds of the atoms are about 2 × 10² m s⁻¹ (10⁵ times larger); the field travels at about 3 × 10⁸ m s⁻¹ (10¹¹ times larger). That is why a bulb lights instantly although the electrons crawl.(a) n = (6.0 × 10²³/63.5 g) × 9.0 × 10⁶ g m⁻³ ≈ 8.5 × 10²⁸ m⁻³। vd = I/(neA) = 1.5/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁷) ≈ 1.1 mm s⁻¹। (b) परमाणुओं की ऊष्मीय चाल लगभग 2 × 10² m s⁻¹ (10⁵ गुना अधिक); क्षेत्र लगभग 3 × 10⁸ m s⁻¹ से चलता है (10¹¹ गुना अधिक)। इसीलिए बल्ब तुरंत जलता है जबकि इलेक्ट्रॉन रेंगते हैं।(a) n = (6.0 × 10²³/63.5 g) × 9.0 × 10⁶ g m⁻³ ≈ 8.5 × 10²⁸ m⁻³. vd = I/(neA) = 1.5/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁷) ≈ 1.1 mm s⁻¹. (b) Atoms ki thermal speed lagbhag 2 × 10² m s⁻¹ (10⁵ guna zyada); field lagbhag 3 × 10⁸ m s⁻¹ se chalta hai (10¹¹ guna zyada). Isiliye bulb turant jalta hai jabki electrons rengte hain.(a) First find n, the number of free electrons per m³: n = (6.0 × 10²³/63.5 g) × 9.0 × 10⁶ g m⁻³ ≈ 8.5 × 10²⁸ m⁻³. Then vd = I/(neA) = 1.5/(8.5 × 10²⁸ × 1.6 × 10⁻¹⁹ × 1.0 × 10⁻⁷) ≈ 1.1 mm s⁻¹. (b) The atoms move due to heat at about 2 × 10² m s⁻¹, which is 10⁵ times faster. The field travels along the wire at about 3 × 10⁸ m s⁻¹, which is 10¹¹ times faster. This is why a bulb lights up at once, even though the electrons move very slowly.
(a) In Example 3.1 the drift speed is only about a mm per second for a current of a few amperes. How then is a current set up almost the instant a circuit is closed? (b) The drift is caused by the force of the electric field on the electrons, and a force should cause acceleration. Why then do electrons have a steady average drift speed? (c) If the drift speed is so small and the electron’s charge is so small, how can we still get large currents in a conductor? (d) When electrons drift in a metal from lower to higher potential, are all the free electrons moving in the same direction? (e) Are the paths of electrons between successive collisions (with the positive ions of the metal) straight lines (i) when there is no electric field, (ii) when there is an electric field?(a) उदाहरण 3.1 में कुछ ऐम्पियर धारा के लिए अपवाह चाल केवल लगभग एक mm प्रति सेकंड है। तब परिपथ बंद करते ही लगभग तुरंत धारा कैसे स्थापित हो जाती है? (b) अपवाह इलेक्ट्रॉनों पर विद्युत क्षेत्र के बल के कारण होता है, और बल से त्वरण होना चाहिए। फिर इलेक्ट्रॉनों की औसत अपवाह चाल स्थिर क्यों होती है? (c) यदि अपवाह चाल इतनी कम है और इलेक्ट्रॉन का आवेश भी इतना कम है, तो चालक में बड़ी धारा कैसे मिलती है? (d) जब इलेक्ट्रॉन धातु में निम्न विभव से उच्च विभव की ओर अपवाहित होते हैं, तो क्या सभी मुक्त इलेक्ट्रॉन एक ही दिशा में चल रहे होते हैं? (e) क्रमागत टक्करों (धातु के धन आयनों से) के बीच इलेक्ट्रॉनों के पथ क्या सरल रेखाएँ होते हैं — (i) विद्युत क्षेत्र की अनुपस्थिति में, (ii) विद्युत क्षेत्र की उपस्थिति में?(a) Example 3.1 mein kuch ampere current ke liye drift speed sirf lagbhag ek mm per second hai. To circuit band karte hi current lagbhag turant kaise shuru ho jaata hai? (b) Drift electrons par electric field ke force se hota hai, aur force se acceleration hona chahiye. Phir electrons ki average drift speed steady kyun rehti hai? (c) Agar drift speed itni kam hai aur electron ka charge bhi itna chhota hai, to conductor mein bada current kaise milta hai? (d) Jab electrons metal mein lower se higher potential ki taraf drift karte hain, to kya saare free electrons ek hi direction mein chal rahe hote hain? (e) Do collisions ke beech (metal ke positive ions se) electrons ke paths kya seedhi lines hote hain — (i) electric field na ho tab, (ii) electric field ho tab?(a) In Example 3.1, the drift speed is only about a mm per second, for a current of a few amperes. Then how does a current start almost at the very moment we close a circuit? (b) The drift is caused by the force of the electric field on the electrons. A force should cause acceleration. Then why do electrons have a steady average drift speed? (c) The drift speed is very small, and the charge of an electron is very small. How can we still get large currents in a conductor? (d) When electrons drift in a metal from lower to higher potential, do all the free electrons move in the same direction? (e) Are the paths of electrons between two collisions (with the positive ions of the metal) straight lines (i) when there is no electric field, (ii) when there is an electric field?
Show solutionहल देखेंSolution dekhiye
(a) The electric field is set up throughout the circuit almost instantly (at nearly the speed of light), and it starts a local electron drift at every point at once. The current does not wait for electrons to travel from one end of the wire to the other. (It does take a short while for the current to reach its steady value.)(a) विद्युत क्षेत्र पूरे परिपथ में लगभग तुरंत (लगभग प्रकाश की चाल से) स्थापित हो जाता है और हर बिंदु पर एक साथ स्थानीय इलेक्ट्रॉन अपवाह शुरू कर देता है। धारा को इलेक्ट्रॉनों के तार के एक सिरे से दूसरे सिरे तक पहुँचने की प्रतीक्षा नहीं करनी पड़ती। (हाँ, धारा को अपने स्थायी मान तक पहुँचने में थोड़ा समय लगता है।)(a) Electric field poore circuit mein lagbhag turant (lagbhag light ki speed se) ban jaata hai aur har point par ek saath local electron drift shuru kar deta hai. Current ko electrons ke wire ke ek end se doosre end tak pahunchne ka intezaar nahi karna padta. (Haan, current ko apni steady value tak pahunchne mein thoda time lagta hai.)(a) When we close the circuit, the electric field is set up in the whole circuit almost at once (at nearly the speed of light). It starts electrons drifting at every point at the same time. So the current does not wait for electrons to travel from one end of the wire to the other. (It does take a short time for the current to reach its steady value.)
(b) Each free electron does accelerate, and its drift speed grows — until it collides with a positive ion. After the collision it loses its drift speed, accelerates again, collides again, and so on. On the average, therefore, the electrons have only a steady drift speed.(b) प्रत्येक मुक्त इलेक्ट्रॉन सचमुच त्वरित होता है और उसकी अपवाह चाल बढ़ती है — जब तक वह किसी धन आयन से टकरा नहीं जाता। टक्कर के बाद वह अपनी अपवाह चाल खो देता है, फिर त्वरित होता है, फिर टकराता है, और यही चलता रहता है। इसलिए औसतन इलेक्ट्रॉनों की केवल एक स्थिर अपवाह चाल होती है।(b) Har free electron sach mein accelerate hota hai aur uski drift speed badhti hai — jab tak wo kisi positive ion se takra nahi jaata. Collision ke baad wo apni drift speed kho deta hai, phir accelerate hota hai, phir takrata hai, aur ye chalta rehta hai. Isliye average mein electrons ki sirf ek steady drift speed hoti hai.(b) Each free electron does speed up, and its drift speed grows — until it hits a positive ion. After the hit, it loses its drift speed. Then it speeds up again, hits again, and so on. So on average, the electrons have only a steady drift speed.
(c) Because the number density of free electrons is enormous, about 10²⁹ m⁻³. In I = neAvd, a huge n makes up for the tiny e and vd.(c) क्योंकि मुक्त इलेक्ट्रॉनों का संख्या घनत्व बहुत विशाल है, लगभग 10²⁹ m⁻³। I = neAvd में विशाल n, छोटे e और vd की भरपाई कर देता है।(c) Kyunki free electrons ki number density bahut zyada hai, lagbhag 10²⁹ m⁻³. I = neAvd mein bahut bada n, chhote e aur vd ki kami poori kar deta hai.(c) Because the number of free electrons in each m³ is huge — about 10²⁹ m⁻³. In I = neAvd, the very large n makes up for the tiny e and the tiny vd.
(d) By no means. The small drift velocity is only superposed on the large random (thermal) velocities of the electrons, which point in all directions.(d) बिल्कुल नहीं। छोटा अपवाह वेग तो इलेक्ट्रॉनों के बड़े यादृच्छिक (ऊष्मीय) वेगों पर केवल अध्यारोपित होता है, जो सभी दिशाओं में होते हैं।(d) Bilkul nahi. Chhota drift velocity to electrons ke bade random (thermal) velocities par sirf superpose hota hai, jo har direction mein hote hain.(d) Not at all. The small drift velocity is only added on top of (superposed on) the large random (thermal) velocities of the electrons. These random velocities point in all directions.
(e) (i) With no field, the paths between collisions are straight lines (no force acts). (ii) With a field, the paths are in general curved — the constant force −eE bends them, just as gravity bends the path of a thrown ball.(e) (i) क्षेत्र न होने पर टक्करों के बीच पथ सरल रेखाएँ होते हैं (कोई बल नहीं लगता)। (ii) क्षेत्र होने पर पथ सामान्यतः वक्र होते हैं — नियत बल −eE उन्हें मोड़ देता है, ठीक वैसे जैसे गुरुत्व फेंकी गई गेंद का पथ मोड़ देता है।(e) (i) Field na ho to collisions ke beech paths seedhi lines hote hain (koi force nahi lagta). (ii) Field ho to paths aam taur par curved hote hain — constant force −eE unhe mod deta hai, bilkul waise jaise gravity phenki gayi ball ka path mod deti hai.(e) (i) With no field, the paths between collisions are straight lines, because no force acts on the electron. (ii) With a field, the paths are usually curved. The steady force −eE bends them, just as gravity bends the path of a ball you throw.
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3.6Limitations of Ohm’s Lawओम नियम की सीमाएँOhm’s Law ki Limitations
Ohm’s law works for many materials, but it is not a fundamental law of nature. Three kinds of deviation are seen (see the graphs): (1) V stops being proportional to I; (2) the relation depends on the sign of V — reversing V does not give the same current (a diode); (3) the relation is not unique — the same I occurs at more than one V (e.g. GaAs). Materials that obey Ohm’s law are called ohmic.ओम नियम बहुत से पदार्थों पर लागू है, पर यह प्रकृति का मूल नियम नहीं है। तीन प्रकार के विचलन दिखते हैं (ग्राफ़ देखें): (1) V, I के अनुक्रमानुपाती नहीं रहता; (2) संबंध V के चिह्न पर निर्भर करता है — V उलटने पर वही धारा नहीं मिलती (डायोड); (3) संबंध अद्वितीय नहीं — एक ही I एक से अधिक V पर मिलता है (जैसे GaAs)। जो पदार्थ ओम नियम मानते हैं वे ओमीय कहलाते हैं।Ohm’s law bahut se materials par lagta hai, par ye nature ka fundamental law nahi hai. Teen tarah ke deviations dikhte hain (graphs dekhiye): (1) V, I ke proportional nahi rehta; (2) relation V ke sign par depend karta hai — V ulatne par wahi current nahi milta (diode); (3) relation unique nahi — ek hi I ek se zyada V par milta hai (jaise GaAs). Jo materials Ohm’s law maante hain wo ohmic kehlaate hain.Ohm’s law works for many materials, but it is not a basic law of nature. We see three kinds of departure from it: (1) V is no longer proportional to I (the graph is not a straight line); (2) the result depends on the sign of V — if we reverse V, we do not get the same current (for example, a diode); (3) the relation is not unique — the same I can happen at more than one V (for example, GaAs). Materials that obey Ohm’s law are called ohmic. 👉 (see the graphs)
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3.7Resistivity of Various Materialsविभिन्न पदार्थों की प्रतिरोधकताAlag Materials ki Resistivity
Resistivities span an enormous range. Metals have low ρ, about 10⁻⁸ to 10⁻⁶ Ω m. Insulators like glass and rubber have 10²² to 10²⁴ times more. Semiconductors (Si, Ge) lie in between on a logarithmic scale, and their resistivity falls as temperature rises or with added impurities.प्रतिरोधकताएँ बहुत बड़े परास में फैली हैं। धातुओं की ρ कम, लगभग 10⁻⁸ से 10⁻⁶ Ω m। विद्युतरोधी जैसे काँच और रबर की 10²² से 10²⁴ गुना अधिक। अर्धचालक (Si, Ge) लघुगणकीय पैमाने पर बीच में हैं, और ताप बढ़ने या अशुद्धियाँ मिलाने पर इनकी प्रतिरोधकता घटती है।Resistivities bahut bade range mein phaili hain. Metals ki ρ kam, lagbhag 10⁻⁸ se 10⁻⁶ Ω m. Insulators jaise glass aur rubber ki 10²² se 10²⁴ guna zyada. Semiconductors (Si, Ge) log scale par beech mein hain, aur temperature badhne ya impurities milaane par inki resistivity ghatati hai.Resistivities vary over a huge range. Metals have low ρ, about 10⁻⁸ to 10⁻⁶ Ω m. Insulators like glass and rubber have 10²² to 10²⁴ times more. Semiconductors (Si, Ge) come in between, on a logarithmic scale (a scale that goes up in powers of ten). Their resistivity falls when temperature rises, or when we add impurities.
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3.8Temperature Dependence of Resistivityप्रतिरोधकता की ताप पर निर्भरताResistivity ki Temperature par Dependence
For a metal over a limited range of temperature, resistivity increases almost linearly:किसी धातु के लिए सीमित ताप परास में प्रतिरोधकता लगभग रैखिक रूप से बढ़ती है:Kisi metal ke liye limited temperature range mein resistivity lagbhag linearly badhti hai:For a metal, over a limited range of temperature, resistivity increases almost in a straight line with temperature:
Tap to zoom · 3 labels to exploreज़ूम करने के लिए टैप करें · 3 लेबलTap kijiye → zoom · 3 labelsWhy? From ρ = m/(ne²τ): in a metal, n hardly changes, but hotter ions vibrate more, collisions are more frequent, τ falls, so ρ rises (copper). Alloys like nichrome, manganin and constantan have high ρ that hardly changes with T (nichrome) — ideal for heaters and standard resistors. In a semiconductor, heating frees many more carriers: n rises faster than τ falls, so ρ falls (semiconductor).क्यों? ρ = m/(ne²τ) से: धातु में n लगभग नहीं बदलता, पर गर्म आयन अधिक कंपन करते हैं, टक्करें अधिक होती हैं, τ घटता है, इसलिए ρ बढ़ती है (ताँबा)। नाइक्रोम, मैंगनिन, कॉन्सटेंटन जैसी मिश्रधातुओं की ρ अधिक और ताप से लगभग अपरिवर्तित (नाइक्रोम) — हीटर और मानक प्रतिरोधों के लिए उत्तम। अर्धचालक में गर्म करने पर बहुत अधिक वाहक मुक्त होते हैं: n, τ के घटने से तेज़ बढ़ता है, इसलिए ρ घटती है (अर्धचालक)।Kyun? ρ = m/(ne²τ) se: metal mein n lagbhag nahi badalta, par garam ions zyada vibrate karte hain, collisions zyada hote hain, τ ghatata hai, isliye ρ badhti hai (copper). Nichrome, manganin, constantan jaisi alloys ki ρ zyada aur temperature se lagbhag same (nichrome) — heaters aur standard resistors ke liye best. Semiconductor mein garam karne par bahut zyada carriers free hote hain: n, τ ke ghatne se tez badhta hai, isliye ρ ghatati hai (semiconductor).Why? Look at ρ = m/(ne²τ). In a metal, n hardly changes. But hotter ions shake (vibrate) more, so collisions happen more often. So τ becomes smaller, and ρ becomes bigger. Alloys like nichrome, manganin and constantan have a high ρ that hardly changes with T. This makes them ideal for heaters and standard resistors. In a semiconductor, heating frees many more charge carriers. n rises faster than τ falls, so ρ becomes smaller. 👉 (copper) 👉 (nichrome) 👉 (semiconductor)
Pick a material, change the wire’s length, thickness and temperature, and apply a voltage. R = ρl/A and the current update live.पदार्थ चुनें, तार की लंबाई, मोटाई और ताप बदलें, और वोल्टता लगाएँ। R = ρl/A और धारा तुरंत बदलती हैं।Material chuniye, wire ki length, motaai aur temperature badliye, aur voltage lagaiye. R = ρl/A aur current live badalte hain.
A nichrome toaster element has resistance 75.3 Ω at room temperature (27.0 °C), measured with a tiny current. On a 230 V supply the current settles at 2.68 A. Find the steady temperature of the element (α for nichrome = 1.70 × 10⁻⁴ °C⁻¹).टोस्टर के नाइक्रोम तत्व का प्रतिरोध कमरे के ताप (27.0 °C) पर, बहुत छोटी धारा से मापने पर, 75.3 Ω है। 230 V पर धारा 2.68 A पर स्थिर होती है। तत्व का स्थायी ताप ज्ञात करें (नाइक्रोम का α = 1.70 × 10⁻⁴ °C⁻¹)।Toaster ke nichrome element ka resistance room temperature (27.0 °C) par, bahut chhote current se naapne par, 75.3 Ω hai. 230 V par current 2.68 A par steady hota hai. Element ka steady temperature nikaalo (nichrome ka α = 1.70 × 10⁻⁴ °C⁻¹).The heating element of a nichrome toaster has a resistance of 75.3 Ω at room temperature (27.0 °C), measured with a very small current. On a 230 V supply, the current settles at 2.68 A. Find the steady temperature of the element. (α for nichrome = 1.70 × 10⁻⁴ °C⁻¹.)
Show solutionहल देखेंSolution dekhiye
R₂ = 230/2.68 = 85.8 Ω. From R₂ = R₁[1 + α(T₂ − T₁)]: T₂ − T₁ = (85.8 − 75.3)/(75.3 × 1.70 × 10⁻⁴) ≈ 820 °C. So T₂ ≈ 847 °C.R₂ = 230/2.68 = 85.8 Ω। R₂ = R₁[1 + α(T₂ − T₁)] से: T₂ − T₁ = (85.8 − 75.3)/(75.3 × 1.70 × 10⁻⁴) ≈ 820 °C। अतः T₂ ≈ 847 °C।R₂ = 230/2.68 = 85.8 Ω. R₂ = R₁[1 + α(T₂ − T₁)] se: T₂ − T₁ = (85.8 − 75.3)/(75.3 × 1.70 × 10⁻⁴) ≈ 820 °C. Isliye T₂ ≈ 847 °C.First find the hot resistance: R₂ = 230/2.68 = 85.8 Ω. Now use R₂ = R₁[1 + α(T₂ − T₁)]. So T₂ − T₁ = (85.8 − 75.3)/(75.3 × 1.70 × 10⁻⁴) ≈ 820 °C. Add the room temperature: T₂ ≈ 847 °C.
A platinum resistance thermometer reads 5 Ω at the ice point, 5.23 Ω at the steam point and 5.795 Ω in a hot bath. Find the temperature of the bath.प्लैटिनम प्रतिरोध थर्मामीटर का प्रतिरोध हिमांक पर 5 Ω, भाप बिंदु पर 5.23 Ω और एक गर्म कुंड में 5.795 Ω है। कुंड का ताप ज्ञात करें।Platinum resistance thermometer ka resistance ice point par 5 Ω, steam point par 5.23 Ω aur ek garam bath mein 5.795 Ω hai. Bath ka temperature nikaalo.A platinum resistance thermometer shows 5 Ω at the ice point (0 °C), 5.23 Ω at the steam point (100 °C), and 5.795 Ω in a hot bath. Find the temperature of the bath.
Show solutionहल देखेंSolution dekhiye
t = (Rₜ − R₀)/(R₁₀₀ − R₀) × 100 = (5.795 − 5)/(5.23 − 5) × 100 = (0.795/0.23) × 100 ≈ 345.65 °C.t = (Rₜ − R₀)/(R₁₀₀ − R₀) × 100 = (5.795 − 5)/(5.23 − 5) × 100 = (0.795/0.23) × 100 ≈ 345.65 °C।t = (Rₜ − R₀)/(R₁₀₀ − R₀) × 100 = (5.795 − 5)/(5.23 − 5) × 100 = (0.795/0.23) × 100 ≈ 345.65 °C.The resistance changes evenly with temperature, so: t = (Rₜ − R₀)/(R₁₀₀ − R₀) × 100 = (5.795 − 5)/(5.23 − 5) × 100 = (0.795/0.23) × 100 ≈ 345.65 °C.
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3.9Electrical Energy, Powerविद्युत ऊर्जा, शक्तिElectrical Energy, Power
When charge ΔQ = IΔt moves through a potential drop V, it loses potential energy IVΔt. In a conductor the electrons do not speed up — the energy goes into collisions and heats the conductor. The rate is the power:जब आवेश ΔQ = IΔt विभवपात V से गुज़रता है, तो वह IVΔt स्थितिज ऊर्जा खोता है। चालक में इलेक्ट्रॉन तेज़ नहीं होते — ऊर्जा टक्करों में जाकर चालक को गर्म करती है। इसकी दर शक्ति है:Jab charge ΔQ = IΔt potential drop V se guzarta hai, to wo IVΔt potential energy khota hai. Conductor mein electrons tez nahi hote — energy collisions mein jaakar conductor ko garam karti hai. Iski rate power hai:When a charge ΔQ = IΔt moves through a potential drop V, it loses potential energy IVΔt. In a conductor, the electrons do not speed up. Instead, the energy goes into collisions and heats the conductor. The energy used each second is the power:
Why power lines use very high voltage: to deliver power P through cables of resistance Rc, the current is I = P/V and the wasted power is Pc = I²Rc = P²Rc/V². Raising V cuts the loss as 1/V². That is why transmission lines carry dangerous voltages, and transformers step the voltage down near our homes.बिजली की लाइनें बहुत ऊँची वोल्टता क्यों: Rc प्रतिरोध वाले केबलों से शक्ति P पहुँचाने में धारा I = P/V और व्यर्थ शक्ति Pc = I²Rc = P²Rc/V²। V बढ़ाने से हानि 1/V² से घटती है। इसीलिए पारेषण लाइनें खतरनाक वोल्टता ले जाती हैं, और ट्रांसफ़ॉर्मर हमारे घरों के पास वोल्टता घटाते हैं।Bijli ki lines bahut high voltage kyun: Rc resistance wale cables se power P pahunchaane mein current I = P/V aur wasted power Pc = I²Rc = P²Rc/V². V badhaane se loss 1/V² se ghatta hai. Isiliye transmission lines khatarnaak voltage le jaati hain, aur transformers hamare gharon ke paas voltage ghataate hain.Why power lines use very high voltage: suppose we send power P through cables with resistance Rc. The current is I = P/V, and the power wasted as heat is Pc = I²Rc = P²Rc/V². So if we raise V, the loss falls as 1/V². That is why transmission lines carry dangerous high voltages. Transformers then step the voltage down near our homes.
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3.10Cells, emf, Internal Resistanceसेल, विद्युत वाहक बल, आंतरिक प्रतिरोधCells, emf, Internal Resistance
A cell has two electrodes, P (positive) and N (negative), in an electrolyte (see the cell). When no current flows, the potential difference between P and N is the emf ε of the cell. (Despite the name "electromotive force", ε is a potential difference, not a force.) The electrolyte itself has a resistance — the internal resistance r.सेल में विद्युत अपघट्य में दो इलेक्ट्रोड होते हैं, P (धन) और N (ऋण) (सेल देखें)। जब कोई धारा नहीं बहती, तब P और N के बीच विभवांतर सेल का विद्युत वाहक बल (emf) ε है। (नाम "बल" होने पर भी ε एक विभवांतर है, बल नहीं।) विद्युत अपघट्य का अपना प्रतिरोध है — आंतरिक प्रतिरोध r।Cell mein electrolyte mein do electrodes hote hain, P (positive) aur N (negative) (cell dekhiye). Jab koi current nahi behta, tab P aur N ke beech potential difference cell ka emf ε hai. (Naam "force" hone par bhi ε ek potential difference hai, force nahi.) Electrolyte ka apna resistance hai — internal resistance r.A cell has two electrodes, P (positive) and N (negative), dipped in an electrolyte. When no current flows, the potential difference between P and N is called the emf ε of the cell. (Its full name is “electromotive force”, but ε is really a potential difference, not a force.) The electrolyte itself has some resistance. This is called the internal resistance r. 👉 (see the cell)
Tap to zoom · 6 labels to exploreज़ूम करने के लिए टैप करें · 6 लेबलTap kijiye → zoom · 6 labelsWhen the cell drives a current I through an external resistor R, part of the emf is used up inside the cell, so the terminal voltage is less than ε:जब सेल बाह्य प्रतिरोधक R में धारा I बहाता है, तो emf का कुछ भाग सेल के अंदर ही खर्च हो जाता है, इसलिए टर्मिनल वोल्टता ε से कम होती है:Jab cell external resistor R mein current I bahata hai, to emf ka kuch part cell ke andar hi kharch ho jaata hai, isliye terminal voltage ε se kam hoti hai:When the cell sends a current I through an outside resistor R, part of the emf is used up inside the cell itself. So the voltage at the terminals (the terminal voltage) is less than ε:
Change the load R, the emf and the internal resistance r. See how the current, the terminal voltage V = ε − Ir and the power in R change. Try R = 0 (short circuit).भार R, emf और आंतरिक प्रतिरोध r बदलें। देखें धारा, टर्मिनल वोल्टता V = ε − Ir और R में शक्ति कैसे बदलती हैं। R = 0 (लघुपथ) आज़माएँ।Load R, emf aur internal resistance r badliye. Dekhiye current, terminal voltage V = ε − Ir aur R mein power kaise badalte hain. R = 0 (short circuit) try kijiye.
A car battery has emf 12 V and internal resistance 0.4 Ω. What is the maximum current it can supply?कार की बैटरी का emf 12 V और आंतरिक प्रतिरोध 0.4 Ω है। यह अधिकतम कितनी धारा दे सकती है?Car battery ka emf 12 V aur internal resistance 0.4 Ω hai. Ye maximum kitna current de sakti hai?A car battery has an emf of 12 V and an internal resistance of 0.4 Ω. What is the largest current it can give?
Show solutionहल देखेंSolution dekhiye
Maximum current flows when the terminals are joined directly (R = 0): I = ε/r = 12/0.4 = 30 A.अधिकतम धारा तब बहती है जब टर्मिनल सीधे जोड़ दिए जाएँ (R = 0): I = ε/r = 12/0.4 = 30 A।Maximum current tab behta hai jab terminals seedhe jod diye jaayein (R = 0): I = ε/r = 12/0.4 = 30 A.The largest current flows when the two terminals are joined directly by a wire (R = 0). Then I = ε/r = 12/0.4 = 30 A.
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3.11Cells in Series and in Parallelश्रेणी तथा पार्श्व क्रम में सेलSeries aur Parallel mein Cells
Tap to zoom · 8 labels to exploreज़ूम करने के लिए टैप करें · 8 लेबलTap kijiye → zoom · 8 labelsSeries (positive of one joined to negative of the next): the emfs add and the internal resistances add: ε(eq) = ε₁ + ε₂, r(eq) = r₁ + r₂. If one cell is connected the other way round (negative to negative), its emf is subtracted: ε(eq) = ε₁ − ε₂.श्रेणी (एक का धन अगले के ऋण से): emf जुड़ते हैं और आंतरिक प्रतिरोध जुड़ते हैं: ε(eq) = ε₁ + ε₂, r(eq) = r₁ + r₂। यदि एक सेल उलटा जुड़ा हो (ऋण से ऋण), तो उसका emf घटता है: ε(eq) = ε₁ − ε₂।Series (ek ka positive agle ke negative se): emfs judte hain aur internal resistances judte hain: ε(eq) = ε₁ + ε₂, r(eq) = r₁ + r₂. Agar ek cell ulta juda ho (negative se negative), to uska emf ghatta hai: ε(eq) = ε₁ − ε₂.Series (the positive of one cell joined to the negative of the next): the emfs add up, and the internal resistances add up: ε(eq) = ε₁ + ε₂, r(eq) = r₁ + r₂. If one cell is joined the wrong way round (negative to negative), its emf is subtracted: ε(eq) = ε₁ − ε₂.
Parallel (positives together, negatives together) (see parallel):पार्श्व (धन साथ, ऋण साथ) (पार्श्व देखें):Parallel (positives saath, negatives saath) (parallel dekhiye):Parallel (all positives joined together, and all negatives joined together): 👉 (see parallel)
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3.12Kirchhoff’s Rulesकिरखोफ़ के नियमKirchhoff’s Rules
Real circuits are networks of cells and resistors that cannot always be reduced to series and parallel. Two rules solve any of them (see the rules):वास्तविक परिपथ सेलों और प्रतिरोधकों के ऐसे जाल होते हैं जिन्हें हमेशा श्रेणी और पार्श्व में नहीं बदला जा सकता। दो नियम इनमें से किसी को भी हल कर देते हैं (नियम देखें):Real circuits cells aur resistors ke aise networks hote hain jinhe hamesha series aur parallel mein reduce nahi kar sakte. Do rules inmein se kisi ko bhi solve kar dete hain (rules dekhiye):Real circuits are networks of cells and resistors. We cannot always reduce them to simple series and parallel parts. Two rules can solve any such circuit: 👉 (see the rules)
Tap to zoom · 6 labels to exploreज़ूम करने के लिए टैप करें · 6 लेबलTap kijiye → zoom · 6 labels- (a) Junction rule: at any junction, the sum of currents entering = sum of currents leaving. This is conservation of charge.(a) संधि नियम: किसी भी संधि पर आने वाली धाराओं का योग = जाने वाली धाराओं का योग। यह आवेश संरक्षण है।(a) Junction rule: kisi bhi junction par aane wale currents ka sum = jaane wale currents ka sum. Ye charge conservation hai.(a) Junction rule: at any junction (a point where wires meet), the total current coming in = the total current going out. This is because charge is conserved (it is not lost or made).
- (b) Loop rule: the algebraic sum of changes in potential around any closed loop is zero (Σε = ΣIR). This is because potential is single-valued — energy is conserved.(b) लूप (पाश) नियम: किसी भी बंद लूप में विभव परिवर्तनों का बीजगणितीय योग शून्य है (Σε = ΣIR)। क्योंकि विभव एक-मानी है — ऊर्जा संरक्षित है।(b) Loop rule: kisi bhi closed loop mein potential changes ka algebraic sum zero hai (Σε = ΣIR). Kyunki potential single-valued hai — energy conserved hai.(b) Loop rule: if you go round any closed loop, the algebraic sum (sum with signs) of the changes in potential is zero (Σε = ΣIR). This is because each point has only one value of potential — energy is conserved.
Sign rules: going through a resistor along the current, the potential drops (−IR); against the current, it rises (+IR). Going through a cell from − to +, the potential rises (+ε). Choose a direction for each unknown current; a negative answer simply means the current flows the other way.चिह्न नियम: प्रतिरोधक से धारा की दिशा में जाने पर विभव घटता है (−IR); धारा के विरुद्ध बढ़ता है (+IR)। सेल में − से + जाने पर विभव बढ़ता है (+ε)। हर अज्ञात धारा की एक दिशा मान लें; ऋणात्मक उत्तर का अर्थ बस यह है कि धारा उलटी दिशा में है।Sign rules: resistor se current ki direction mein jaane par potential ghatta hai (−IR); current ke against badhta hai (+IR). Cell mein − se + jaane par potential badhta hai (+ε). Har unknown current ki ek direction maan lo; negative answer ka matlab bas itna hai ki current ulti direction mein hai.Sign rules: when you cross a resistor in the direction of the current, the potential drops (−IR). When you cross it against the current, the potential rises (+IR). When you cross a cell from − to +, the potential rises (+ε). Choose a direction for each unknown current. If an answer comes out negative, it just means the current flows the other way.
Twelve 1 Ω resistors form the edges of a cube. A 10 V battery (no internal resistance) is connected across two diagonally opposite corners. Find the equivalent resistance and the current in each edge.बारह 1 Ω प्रतिरोधक एक घन के किनारे बनाते हैं। 10 V बैटरी (आंतरिक प्रतिरोध नहीं) विकर्णतः विपरीत दो कोनों के बीच जुड़ी है। तुल्य प्रतिरोध और हर किनारे में धारा ज्ञात करें।Barah 1 Ω resistors ek cube ke edges banaate hain. 10 V battery (internal resistance nahi) diagonally opposite do corners ke beech judi hai. Equivalent resistance aur har edge mein current nikaalo.Twelve resistors, each of 1 Ω, form the edges of a cube. A 10 V battery (with no internal resistance) is joined across two corners that are diagonally opposite. Find the equivalent resistance and the current in each edge.
Show solutionहल देखेंSolution dekhiye
By symmetry the current I in each of the 3 edges leaving the first corner is the same; at each next corner it splits equally into I/2 and I/2; the 3 edges entering the last corner again carry I each. Loop rule along one path: ε = IR + (I/2)R + IR = (5/2)IR. Total current = 3I, so R(eq) = ε/3I = (5/6) Ω. With ε = 10 V: 3I = 12 A, so I = 4 A in the 6 edges at the two ends and 2 A in the 6 middle edges.सममिति से पहले कोने से निकलने वाले 3 किनारों में धारा I समान है; अगले हर कोने पर यह I/2 और I/2 में बराबर बँटती है; अंतिम कोने में आने वाले 3 किनारों में फिर I। एक पथ पर लूप नियम: ε = IR + (I/2)R + IR = (5/2)IR। कुल धारा = 3I, अतः R(eq) = ε/3I = (5/6) Ω। ε = 10 V से: 3I = 12 A, इसलिए दोनों सिरों के 6 किनारों में I = 4 A और बीच के 6 किनारों में 2 A।Symmetry se pehle corner se nikalne wale 3 edges mein current I same hai; agle har corner par ye I/2 aur I/2 mein barabar bant-ta hai; last corner mein aane wale 3 edges mein phir I. Ek path par loop rule: ε = IR + (I/2)R + IR = (5/2)IR. Total current = 3I, isliye R(eq) = ε/3I = (5/6) Ω. ε = 10 V se: 3I = 12 A, isliye dono ends ke 6 edges mein I = 4 A aur beech ke 6 edges mein 2 A.By symmetry, the current I in each of the 3 edges leaving the first corner is the same. At each next corner, it splits equally into I/2 and I/2. The 3 edges going into the last corner again carry I each. Use the loop rule along one path: ε = IR + (I/2)R + IR = (5/2)IR. The total current is 3I. So R(eq) = ε/3I = (5/6) Ω. With ε = 10 V: 3I = 12 A. So I = 4 A in the 6 edges at the two ends, and 2 A in the 6 middle edges.
Tap to zoom · 13 labels to exploreज़ूम करने के लिए टैप करें · 13 लेबलTap kijiye → zoom · 13 labelsDetermine the current in each branch of the network shown in Fig. 3.17: a 10 V cell with a 1 Ω resistor in branch CA, 4 Ω in AB, 4 Ω in AD, 2 Ω in BC, 2 Ω in CD, and a 5 V cell (negligible internal resistance) in branch DEB.चित्र 3.17 में दिखाए गए परिपथ की प्रत्येक शाखा में धारा ज्ञात कीजिए: शाखा CA में 10 V का सेल और 1 Ω प्रतिरोध, AB में 4 Ω, AD में 4 Ω, BC में 2 Ω, CD में 2 Ω, और शाखा DEB में 5 V का सेल (आंतरिक प्रतिरोध नगण्य)।Fig. 3.17 ke network ki har branch mein current nikaliye: branch CA mein 10 V cell aur 1 Ω resistor, AB mein 4 Ω, AD mein 4 Ω, BC mein 2 Ω, CD mein 2 Ω, aur branch DEB mein 5 V cell (internal resistance negligible).Find the current in each branch of the network in Fig. 3.17. Branch CA has a 10 V cell and a 1 Ω resistor. AB has 4 Ω, AD has 4 Ω, BC has 2 Ω and CD has 2 Ω. Branch DEB has a 5 V cell (its internal resistance is so small we can ignore it).
Show solutionहल देखेंSolution dekhiye
Step 1 — junction rule first. Give unknown names to as few currents as possible: I₁ in CA (from C to A through the 10 V cell), I₂ in AB (A → B) and I₃ in DEB (D → E → B). The junction rule then fixes the rest: at A, AD carries I₁ − I₂; at B, BC carries I₂ + I₃; at C, CD carries I₂ + I₃ − I₁ (from C to D). Only 3 unknowns are left.चरण 1 — पहले संधि नियम। जितनी कम हो सकें उतनी अज्ञात धाराओं को नाम दीजिए: CA में I₁ (C से A, 10 V सेल से होकर), AB में I₂ (A → B) और DEB में I₃ (D → E → B)। फिर संधि नियम बाकी तय कर देता है: A पर AD में I₁ − I₂; B पर BC में I₂ + I₃; C पर CD में I₂ + I₃ − I₁ (C से D की ओर)। अब केवल 3 अज्ञात बचे।Step 1 — pehle junction rule. Jitne kam ho sakein utne unknown currents ko naam dijiye: CA mein I₁ (C se A, 10 V cell se hokar), AB mein I₂ (A → B) aur DEB mein I₃ (D → E → B). Ab junction rule baaki tay kar deta hai: A par AD mein I₁ − I₂; B par BC mein I₂ + I₃; C par CD mein I₂ + I₃ − I₁ (C se D ki taraf). Ab sirf 3 unknowns bache.Step 1 — use the junction rule first. Give names to as few unknown currents as possible: I₁ in CA (from C to A through the 10 V cell), I₂ in AB (from A to B) and I₃ in DEB (from D to E to B). The junction rule then gives all the others. At A, AD carries I₁ − I₂. At B, BC carries I₂ + I₃. At C, CD carries I₂ + I₃ − I₁ (from C to D). Now only 3 unknowns are left.
Step 2 — loop rule on three loops. Going round a loop: a resistor crossed along the current is a drop (−IR), against it a rise (+IR); a cell crossed from − to + is a rise (+ε). The 10 V cell’s + terminal faces A and the 5 V cell’s + terminal faces B.चरण 2 — तीन लूपों पर लूप नियम। लूप में घूमते हुए: धारा की दिशा में प्रतिरोध पार करने पर विभवपात (−IR), विपरीत दिशा में वृद्धि (+IR); सेल को − से + की ओर पार करने पर वृद्धि (+ε)। 10 V सेल का + टर्मिनल A की ओर और 5 V सेल का + टर्मिनल B की ओर है।Step 2 — teen loops par loop rule. Loop mein ghoomte hue: current ki direction mein resistor paar kiya to drop (−IR), ulti direction mein rise (+IR); cell ko − se + ki taraf paar kiya to rise (+ε). 10 V cell ka + terminal A ki taraf aur 5 V cell ka + terminal B ki taraf hai.Step 2 — use the loop rule on three loops. As we go round a loop: crossing a resistor along the current is a drop in potential (−IR); crossing it against the current is a rise (+IR). Crossing a cell from − to + is a rise (+ε). In this circuit, the + terminal of the 10 V cell faces A, and the + terminal of the 5 V cell faces B.
Loop ADCA: 10 − 4(I1 − I2) + 2(I2 + I3 − I1) − I1 = 0 ⇒ 7I1 − 6I2 − 2I3 = 10
Loop ABCA: 10 − 4I2 − 2(I2 + I3) − I1 = 0 ⇒ I1 + 6I2 + 2I3 = 10
Loop BCDEB: 5 − 2(I2 + I3) − 2(I2 + I3 − I1) = 0 ⇒ 2I1 − 4I2 − 4I3 = −5लूप ADCA: 10 − 4(I1 − I2) + 2(I2 + I3 − I1) − I1 = 0 ⇒ 7I1 − 6I2 − 2I3 = 10
लूप ABCA: 10 − 4I2 − 2(I2 + I3) − I1 = 0 ⇒ I1 + 6I2 + 2I3 = 10
लूप BCDEB: 5 − 2(I2 + I3) − 2(I2 + I3 − I1) = 0 ⇒ 2I1 − 4I2 − 4I3 = −5Loop ADCA: 10 − 4(I1 − I2) + 2(I2 + I3 − I1) − I1 = 0 ⇒ 7I1 − 6I2 − 2I3 = 10
Loop ABCA: 10 − 4I2 − 2(I2 + I3) − I1 = 0 ⇒ I1 + 6I2 + 2I3 = 10
Loop BCDEB: 5 − 2(I2 + I3) − 2(I2 + I3 − I1) = 0 ⇒ 2I1 − 4I2 − 4I3 = −5Here are the three loop equations. On the left of the ⇒ sign is the loop rule written out; on the right is the same equation tidied up:
Loop ADCA: 10 − 4(I1 − I2) + 2(I2 + I3 − I1) − I1 = 0 ⇒ 7I1 − 6I2 − 2I3 = 10
Loop ABCA: 10 − 4I2 − 2(I2 + I3) − I1 = 0 ⇒ I1 + 6I2 + 2I3 = 10
Loop BCDEB: 5 − 2(I2 + I3) − 2(I2 + I3 − I1) = 0 ⇒ 2I1 − 4I2 − 4I3 = −5
Step 3 — solve. Adding the first two equations: 8I₁ = 20, so I₁ = 2.5 A. Put it in the second and third: 6I₂ + 2I₃ = 7.5 and 4I₂ + 4I₃ = 10. These give I₂ = 5/8 A = 0.625 A and I₃ = 15/8 A = 1.875 A.चरण 3 — हल। पहले दो समीकरण जोड़ने पर: 8I₁ = 20, अतः I₁ = 2.5 A। इसे दूसरे और तीसरे में रखिए: 6I₂ + 2I₃ = 7.5 और 4I₂ + 4I₃ = 10। इनसे I₂ = 5/8 A = 0.625 A और I₃ = 15/8 A = 1.875 A।Step 3 — solve. Pehli do equations jodne par: 8I₁ = 20, isliye I₁ = 2.5 A. Ise doosri aur teesri mein rakhiye: 6I₂ + 2I₃ = 7.5 aur 4I₂ + 4I₃ = 10. Inse I₂ = 5/8 A = 0.625 A aur I₃ = 15/8 A = 1.875 A.Step 3 — solve the equations. Add the first two equations. The I₂ and I₃ terms cancel, leaving 8I₁ = 20, so I₁ = 2.5 A. Put this value into the second and third equations. We get 6I₂ + 2I₃ = 7.5 and 4I₂ + 4I₃ = 10. Solving these two gives I₂ = 5/8 A = 0.625 A and I₃ = 15/8 A = 1.875 A.
Branch currents: AB: 5/8 A (A → B) · CA: 2.5 A (C → A) · DEB: 15/8 A (D → E → B) · AD: 15/8 A (A → D) · CD: 0 A · BC: 2.5 A (B → C).शाखा धाराएँ: AB: 5/8 A (A → B) · CA: 2.5 A (C → A) · DEB: 15/8 A (D → E → B) · AD: 15/8 A (A → D) · CD: 0 A · BC: 2.5 A (B → C)।Branch currents: AB: 5/8 A (A → B) · CA: 2.5 A (C → A) · DEB: 15/8 A (D → E → B) · AD: 15/8 A (A → D) · CD: 0 A · BC: 2.5 A (B → C).Current in each branch: AB: 5/8 A (from A to B) · CA: 2.5 A (from C to A) · DEB: 15/8 A (from D to E to B) · AD: 15/8 A (from A to D) · CD: 0 A (no current) · BC: 2.5 A (from B to C).
Check with a loop we did not use, BADEB: 5 V + (5/8 × 4) V − (15/8 × 4) V = 5 + 2.5 − 7.5 = 0 ✓. Any other loop gives no new equation — it is only a check. (Taking VC = 0: VA = 7.5 V, VB = 5 V, VD = 0 V — C and D are at the same potential, which is why no current flows in CD.)जिस लूप का उपयोग नहीं किया उससे जाँच, BADEB: 5 V + (5/8 × 4) V − (15/8 × 4) V = 5 + 2.5 − 7.5 = 0 ✓। कोई अन्य लूप नया समीकरण नहीं देता — वह केवल जाँच है। (VC = 0 लेने पर: VA = 7.5 V, VB = 5 V, VD = 0 V — C और D समान विभव पर हैं, इसीलिए CD में धारा नहीं बहती।)Jo loop use nahi kiya usse check, BADEB: 5 V + (5/8 × 4) V − (15/8 × 4) V = 5 + 2.5 − 7.5 = 0 ✓. Koi aur loop nayi equation nahi deta — wo sirf check hai. (VC = 0 lene par: VA = 7.5 V, VB = 5 V, VD = 0 V — C aur D same potential par hain, isiliye CD mein current nahi behta.)Check the answer using a loop we did not use, BADEB: 5 V + (5/8 × 4) V − (15/8 × 4) V = 5 + 2.5 − 7.5 = 0 ✓. Any other loop gives no new equation; it only checks the answer. (Take VC = 0. Then VA = 7.5 V, VB = 5 V and VD = 0 V. C and D are at the same potential. That is why no current flows in CD.)
The network of NCERT Example 3.6, solved live. Change the two emfs and the resistances; every branch current updates. Watch the junction rule (in = out) and tap a loop to check that the voltage changes round it add up to zero.NCERT उदाहरण 3.6 का परिपथ, लाइव हल। दोनों विद्युत वाहक बल और प्रतिरोध बदलिए; हर शाखा की धारा तुरंत बदलती है। संधि नियम (आने वाली = जाने वाली) देखिए और किसी लूप पर टैप करके जाँचिए कि उसके चारों ओर विभव परिवर्तनों का योग शून्य है।NCERT Example 3.6 ka network, live solve. Dono emf aur resistances badaliye; har branch ka current turant badalta hai. Junction rule (in = out) dekhiye aur kisi loop par tap karke check kijiye ki uske chaaron taraf voltage changes ka sum zero hai.
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3.13Wheatstone Bridgeव्हीटस्टोन सेतुWheatstone Bridge
Four resistors R₁, R₂, R₃, R₄ form a diamond ABCD; a battery is connected across A and C (the battery arm) and a galvanometer across B and D (the galvanometer arm) (see the bridge). When the bridge is balanced, no current flows through the galvanometer (Ig = 0). Applying the junction rule at B and D (I₁ = I₃, I₂ = I₄) and the loop rule to loops ADBA and CBDC gives the balance condition:चार प्रतिरोधक R₁, R₂, R₃, R₄ एक हीरा ABCD बनाते हैं; A और C के बीच बैटरी (बैटरी भुजा) और B और D के बीच गैल्वेनोमीटर (गैल्वेनोमीटर भुजा) (सेतु देखें)। सेतु संतुलित होने पर गैल्वेनोमीटर में कोई धारा नहीं (Ig = 0)। B और D पर संधि नियम (I₁ = I₃, I₂ = I₄) और लूप ADBA व CBDC पर लूप नियम लगाने से संतुलन प्रतिबंध मिलता है:Chaar resistors R₁, R₂, R₃, R₄ ek diamond ABCD banaate hain; A aur C ke beech battery (battery arm) aur B aur D ke beech galvanometer (galvanometer arm) (bridge dekhiye). Bridge balanced hone par galvanometer mein koi current nahi (Ig = 0). B aur D par junction rule (I₁ = I₃, I₂ = I₄) aur loops ADBA va CBDC par loop rule lagaane se balance condition milti hai:Four resistors R₁, R₂, R₃ and R₄ are joined to make a diamond shape ABCD. A battery is joined across A and C — this is the battery arm. A galvanometer (a meter that detects small currents) is joined across B and D — this is the galvanometer arm. When the bridge is balanced, no current flows through the galvanometer (Ig = 0). Use the junction rule at B and D (I₁ = I₃, I₂ = I₄). Then use the loop rule on loops ADBA and CBDC. This gives the balance condition: 👉 (see the bridge)
Tap to zoom · 6 labels to exploreज़ूम करने के लिए टैप करें · 6 लेबलTap kijiye → zoom · 6 labelsR₄ is the unknown. Adjust R₃ until the galvanometer needle reads zero, then check that R₄ = R₃ × R₂/R₁. The current through G is solved exactly from Kirchhoff’s rules.R₄ अज्ञात है। R₃ को तब तक बदलें जब तक गैल्वेनोमीटर की सुई शून्य न पढ़े, फिर जाँचें कि R₄ = R₃ × R₂/R₁। G में धारा किरखोफ़ नियमों से सटीक हल की जाती है।R₄ unknown hai. R₃ ko tab tak badliye jab tak galvanometer ki needle zero na padhe, phir check kijiye ki R₄ = R₃ × R₂/R₁. G mein current Kirchhoff’s rules se exactly solve hota hai.
A Wheatstone bridge has AB = 100 Ω, BC = 10 Ω, CD = 5 Ω and DA = 60 Ω. A 15 Ω galvanometer is connected across BD and 10 V is maintained across AC. Find the galvanometer current.व्हीटस्टोन सेतु में AB = 100 Ω, BC = 10 Ω, CD = 5 Ω और DA = 60 Ω हैं। BD के बीच 15 Ω का गैल्वेनोमीटर है और AC पर 10 V बनाए रखा गया है। गैल्वेनोमीटर धारा ज्ञात करें।Wheatstone bridge mein AB = 100 Ω, BC = 10 Ω, CD = 5 Ω aur DA = 60 Ω hain. BD ke beech 15 Ω ka galvanometer hai aur AC par 10 V maintain hai. Galvanometer current nikaalo.In a Wheatstone bridge, AB = 100 Ω, BC = 10 Ω, CD = 5 Ω and DA = 60 Ω. A galvanometer of 15 Ω is joined across BD. A voltage of 10 V is kept across AC. Find the current through the galvanometer.
Show solutionहल देखेंSolution dekhiye
The bridge is not balanced (100/60 ≠ 10/5). Let I₁ flow in AB, I₂ in AD and Ig from B to D. Loop BADB: 100I₁ + 15Ig − 60I₂ = 0. Loop BCDB: 10(I₁ − Ig) − 15Ig − 5(I₂ + Ig) = 0. Loop ADCEA: 60I₂ + 5(I₂ + Ig) = 10. Solving: I₂ = 31.5 Ig and 13I₂ + Ig = 2, so Ig ≈ 4.87 mA.सेतु संतुलित नहीं है (100/60 ≠ 10/5)। AB में I₁, AD में I₂ और B से D की ओर Ig लें। लूप BADB: 100I₁ + 15Ig − 60I₂ = 0। लूप BCDB: 10(I₁ − Ig) − 15Ig − 5(I₂ + Ig) = 0। लूप ADCEA: 60I₂ + 5(I₂ + Ig) = 10। हल करने पर: I₂ = 31.5 Ig और 13I₂ + Ig = 2, अतः Ig ≈ 4.87 mA।Bridge balanced nahi hai (100/60 ≠ 10/5). AB mein I₁, AD mein I₂ aur B se D ki taraf Ig lo. Loop BADB: 100I₁ + 15Ig − 60I₂ = 0. Loop BCDB: 10(I₁ − Ig) − 15Ig − 5(I₂ + Ig) = 0. Loop ADCEA: 60I₂ + 5(I₂ + Ig) = 10. Solve karne par: I₂ = 31.5 Ig aur 13I₂ + Ig = 2, isliye Ig ≈ 4.87 mA.First, the bridge is not balanced, because 100/60 ≠ 10/5. So some current flows through the galvanometer. Let I₁ flow in AB, I₂ in AD, and Ig from B to D. Now write the loop rule for three loops. Loop BADB: 100I₁ + 15Ig − 60I₂ = 0. Loop BCDB: 10(I₁ − Ig) − 15Ig − 5(I₂ + Ig) = 0. Loop ADCEA: 60I₂ + 5(I₂ + Ig) = 10. Solving the first two gives I₂ = 31.5 Ig. The third gives 13I₂ + Ig = 2. Together, Ig ≈ 4.87 mA.
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★Quick revisionझटपट दोहरावQuick revision
- Current = rate of flow of charge; direction = that of positive charge flow. Steady current needs a closed circuit and a source.धारा = आवेश प्रवाह की दर; दिशा = धन आवेश प्रवाह की। स्थायी धारा के लिए बंद परिपथ और स्रोत चाहिए।Current = charge flow ki rate; direction = positive charge flow ki. Steady current ke liye closed circuit aur source chahiye.Current is the rate of flow of charge. Its direction is the direction in which positive charge flows. A steady current needs a closed circuit (a full loop) and a source, like a cell.
- Ohm’s law V = IR; R = ρl/A; j = σE. ρ depends on material and temperature, not shape.ओम नियम V = IR; R = ρl/A; j = σE। ρ पदार्थ और ताप पर निर्भर, आकार पर नहीं।Ohm’s law V = IR; R = ρl/A; j = σE. ρ material aur temperature par depend, shape par nahi.Ohm’s law: V = IR. Resistance R = ρl/A. Also j = σE. Resistivity ρ depends on the material and on temperature, but not on the shape.
- Electrons drift opposite to E with vd = eEτ/m (~mm/s); I = neAvd; ρ = m/(ne²τ); mobility μ = eτ/m.इलेक्ट्रॉन E के विपरीत vd = eEτ/m (~mm/s) से खिसकते हैं; I = neAvd; ρ = m/(ne²τ); गतिशीलता μ = eτ/m।Electrons E ke opposite vd = eEτ/m (~mm/s) se khiskte hain; I = neAvd; ρ = m/(ne²τ); mobility μ = eτ/m.Electrons drift opposite to E with drift speed vd = eEτ/m (only about mm/s). Current I = neAvd. Resistivity ρ = m/(ne²τ). Mobility μ = eτ/m.
- Ohm’s law is not universal: non-linear, direction-dependent (diode) and non-unique (GaAs) behaviours exist.ओम नियम सार्वत्रिक नहीं: अरैखिक, दिशा-निर्भर (डायोड) और अद्वितीय नहीं (GaAs) व्यवहार होते हैं।Ohm’s law universal nahi: non-linear, direction-dependent (diode) aur non-unique (GaAs) behaviours hote hain.Ohm’s law does not work for everything. Some materials are non-linear (V–I graph is not a straight line), some depend on direction (like a diode), and some are non-unique (like GaAs).
- Metals: ρ rises with T (α > 0). Alloys (nichrome, manganin): high ρ, nearly constant. Semiconductors: ρ falls with T (α < 0).धातुएँ: ताप के साथ ρ बढ़ती है (α > 0)। मिश्रधातुएँ (नाइक्रोम, मैंगनिन): ρ अधिक, लगभग स्थिर। अर्धचालक: ताप के साथ ρ घटती है (α < 0)।Metals: temperature ke saath ρ badhti hai (α > 0). Alloys (nichrome, manganin): ρ zyada, lagbhag constant. Semiconductors: temperature ke saath ρ ghatati hai (α < 0).Metals: ρ goes up when T goes up (α > 0). Alloys (nichrome, manganin): ρ is high and almost constant. Semiconductors: ρ goes down when T goes up (α < 0).
- Power P = VI = I²R = V²/R. High-voltage transmission cuts line loss (∝ 1/V²).शक्ति P = VI = I²R = V²/R। उच्च-वोल्टता पारेषण से लाइन हानि (∝ 1/V²) घटती है।Power P = VI = I²R = V²/R. High-voltage transmission se line loss (∝ 1/V²) ghatta hai.Power P = VI = I²R = V²/R. Sending power at high voltage reduces the loss in the lines, because the loss is ∝ 1/V².
- Cell: emf ε (open-circuit voltage), terminal V = ε − Ir, I = ε/(R + r).सेल: emf ε (खुले परिपथ की वोल्टता), टर्मिनल V = ε − Ir, I = ε/(R + r)।Cell: emf ε (open-circuit voltage), terminal V = ε − Ir, I = ε/(R + r).Cell: emf ε is its voltage when no current flows. Terminal voltage V = ε − Ir, and current I = ε/(R + r).
- Series cells: emfs and r add. Parallel cells: ε(eq)/r(eq) = Σεᵢ/rᵢ, 1/r(eq) = Σ1/rᵢ.श्रेणी सेल: emf और r जुड़ते हैं। पार्श्व सेल: ε(eq)/r(eq) = Σεᵢ/rᵢ, 1/r(eq) = Σ1/rᵢ।Series cells: emfs aur r judte hain. Parallel cells: ε(eq)/r(eq) = Σεᵢ/rᵢ, 1/r(eq) = Σ1/rᵢ.Cells in series: the emfs add, and the internal resistances r add. Cells in parallel: ε(eq)/r(eq) = Σεᵢ/rᵢ, and 1/r(eq) = Σ1/rᵢ.
- Kirchhoff: junction rule (charge conservation), loop rule (energy conservation). Wheatstone balance: R₂/R₁ = R₄/R₃.किरखोफ़: संधि नियम (आवेश संरक्षण), लूप नियम (ऊर्जा संरक्षण)। व्हीटस्टोन संतुलन: R₂/R₁ = R₄/R₃।Kirchhoff: junction rule (charge conservation), loop rule (energy conservation). Wheatstone balance: R₂/R₁ = R₄/R₃.Kirchhoff’s rules: the junction rule comes from conservation of charge, and the loop rule comes from conservation of energy. A Wheatstone bridge is balanced when R₂/R₁ = R₄/R₃.
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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