Class 10 • Science

Chapter 11: Electricity

Uncover the laws governing the flow of electric charges. Learn about current, electric potential, Ohm's law, series and parallel circuits, heating effects, and electric power calculations that run modern appliances. Electric charges के बहाव (flow) को govern करने वाले नियमों (laws) को समझें। Current, electric potential, Ohm's law, series and parallel circuits, heating effects, और modern appliances को चलाने वाले electric power calculations के बारे में सीखें।

🌐 Language / भाषा:

💡 Remember from before:

In Chapter 9 and 10, we learned how light acts as an energy source. Now we explore electrical energy. Chapter 9 और 10 में हमने सीखा था कि light कैसे एक energy source की तरह काम करती है। अब हम electrical energy को explore करेंगे।
Matter is composed of atoms, which contain positively charged protons and negatively charged electrons. सारे matter atoms से मिलकर बने हैं, जिनमें positively charged protons और negatively charged electrons होते हैं।
Electric charge is a fundamental property of matter, and the flow of electrons produces electric current. Electric charge matter की एक fundamental property (मूल गुण) है, और electrons का flow ही electric current produce करता है।

1. Electric Current and Circuit (विद्युत धारा और परिपथ)

Section Goal: Define electric charge and electric current, identify their SI units, perform calculations using I = Q/t, and describe the construction of an electric circuit (विद्युत परिपथ).

Analogy: 🌊 The Water Pipe Analogy. Think of electric current like water flowing through a pipe. The water molecules are like electrons. The rate at which water flows (e.g. 5 litres per second) is like the current (measured in amperes). Just like you need a pump to keep water moving and a closed pipeline to prevent leaks, electrons need a battery and a closed pathway (circuit) to flow continuously!

Electric charge (charge) is a fundamental property of particles. The SI unit of electric charge is the coulomb (C). One coulomb is equivalent to the charge contained in nearly 6 × 10¹⁸ electrons. An electron possesses a negative charge of 1.6 × 10^-19 C [NCERT p.186]. Electric charge (charge) particles की एक fundamental property है। Electric charge का SI unit coulomb (C) होता है। एक coulomb लगभग 6 × 10¹⁸ electrons में मौजूद charge के बराबर होता है। एक electron पर 1.6 × 10^-19 C का negative charge होता है [NCERT p.186]।

Electric current (electric current) is defined as the rate of flow of electric charge through a cross-section. If a net charge Q flows through any cross-section of a conductor in time t, then the current I is [NCERT p.186]: Electric current (electric current) को किसी cross-section से बहने वाले electric charge के flow के rate (दर) के रूप में define किया जाता है। यदि conductor के cross-section से समय t में net charge Q बहता है, तो current I होगा [NCERT p.186]:

I = Q / t

The SI unit of electric current is the ampere (A), named after the French scientist André-Marie Ampère. One ampere is constituted by the flow of one coulomb of charge per second, i.e., 1 A = 1 C/s [NCERT p.186]. Small quantities of current are expressed in milliampere (1 mA = 10^-3 A) or microampere (1 μA = 10^-6 A). Electric current का SI unit ampere (A) है, जिसे French scientist André-Marie Ampère के नाम पर रखा गया है। एक ampere current प्रति second एक coulomb charge के flow से बनता है, यानी 1 A = 1 C/s [NCERT p.186]। Current की छोटी quantities को milliampere (1 mA = 10^-3 A) या microampere (1 μA = 10^-6 A) में express किया जाता है।

Conventional Current Direction: Historically, when electricity was discovered, electrons were not known. Current was considered to be the flow of positive charges, and the direction of flow of positive charges was taken to be the direction of electric current. Conventionally, in an electric circuit, the direction of electric current is taken as opposite to the direction of the flow of electrons (which are negative charges) [NCERT p.186]. Conventional Current Direction: Historically, जब electricity की खोज हुई, तब electrons के बारे में पता नहीं था। इसलिए, current को positive charges का flow माना गया और उनके flow की direction को ही electric current की direction मान लिया गया। Conventionally, किसी electric circuit में electric current की direction को electrons के flow की direction के opposite (जो कि negative charges हैं) माना जाता है [NCERT p.186]।

An electric circuit (electric circuit) is a closed and continuous path of an electric current. If the path is broken anywhere (or the switch is turned OFF), the current stops flowing and the appliances do not work [NCERT p.186]. An instrument called an ammeter (ammeter) is used to measure electric current in a circuit. It is always connected in series in a circuit because it has a very low resistance and needs all current to pass through it [NCERT p.187]. एक electric circuit (electric circuit) electric current का एक closed (बंद) और continuous path (सतत मार्ग) होता है। यदि path कहीं से भी टूट जाए (या switch बंद हो जाए), तो current बहना बंद हो जाता है और appliances काम नहीं करते [NCERT p.186]। Circuit में current measure करने के लिए ammeter (ammeter) नामक instrument का उपयोग किया जाता है। इसे circuit में हमेशा series (series) में जोड़ा जाता है क्योंकि इसका resistance बहुत कम होता है और इसे पूरे current को अपने अंदर से गुजारना होता है [NCERT p.187]।

⚡ Checkpoint 1: Electric Current & Charge

Section SummarySection Summary (अनुभाग सारांश)

Charge SI Unit: Coulomb (C). 1 electron = -1.6 × 10^-19 C. Charge SI Unit: Coulomb (C). 1 electron = -1.6 × 10^-19 C.
Current (I = Q/t): Flow of charge per unit time. Unit: Ampere (A). 1 A = 1 C/s. Current (I = Q/t): प्रति unit time में charge का बहाव। Unit: Ampere (A). 1 A = 1 C/s.
Direction: Opposite to the flow of electrons (from positive to negative terminal). Direction: Electrons के flow के opposite (positive terminal से negative terminal की ओर)।
Ammeter: Connected in series to measure current; has low resistance. Ammeter: Current measure करने के लिए series में जोड़ा जाता है; इसका resistance कम होता है।

2. Electric Potential and Potential Difference (विद्युत विभव और विभवांतर)

Section Goal: Define electric potential difference, state its SI unit (volt), perform calculations using V = W/Q, and explain how a voltmeter is connected. Electric potential difference (potential difference) को define करना, इसके SI unit (volt) को समझना, V = W/Q से calculations करना, और voltmeter के connection को समझाना।

Analogy: ⛰️ The Gravity Slope Analogy. Water doesn't flow in a flat, level pipe unless there is a pressure difference (like a slope). A water molecule at the top of a hill has high potential energy and flows naturally to the bottom (low potential). In a wire, electrons will not flow unless there is an "electrical slope" or pressure difference. This electrical pressure is called potential difference. A cell or battery acts like a pump that lifts electrons to the top of the hill, creating this pressure!

What makes the electric charge flow? For flow of charges in a conducting metallic wire, the gravity itself has no role to play; the electrons move only if there is a difference of electric pressure – called the electric potential difference (potential difference) along the conductor [NCERT p.187]. Electric charge को क्या बहने पर मजबूर करता है? किसी conducting metallic wire में charges के flow के लिए gravity का कोई role नहीं होता; electrons तभी move करते हैं जब वहाँ electric pressure में अंतर हो — जिसे conductor के साथ electric potential difference (potential difference) कहा जाता है [NCERT p.187]।

We define the electric potential difference between two points in an electric circuit carrying some current as the work done to move a unit charge from one point to the other [NCERT p.188]: हम current-carrying electric circuit के दो points के बीच electric potential difference को एक point से दूसरे point तक unit charge को ले जाने में किए गए work done के रूप में define करते हैं [NCERT p.188]:

V = W / Q

The SI unit of electric potential difference is the volt (V), named after Alessandro Volta. One volt is defined as the potential difference between two points in a current-carrying conductor when one joule of work is done to move a charge of one coulomb from one point to the other, i.e., 1 V = 1 J/C [NCERT p.188]. Electric potential difference का SI unit volt (V) है, जिसे Alessandro Volta के नाम पर रखा गया है। जब 1 coulomb charge को एक point से दूसरे point तक ले जाने में 1 joule का work done होता है, तो उन दो points के बीच potential difference को 1 volt कहा जाता है, यानी 1 V = 1 J/C [NCERT p.188]।

The potential difference is measured by an instrument called the voltmeter (voltmeter). The voltmeter is always connected in parallel across the points between which the potential difference is to be measured because it has a very high resistance and takes negligible current from the circuit [NCERT p.188]. Potential difference को voltmeter (voltmeter) नामक instrument से measure किया जाता है। Voltmeter को हमेशा उन points के cross parallel (parallel) में जोड़ा जाता है जिनके बीच potential difference measure करना हो, क्योंकि इसका resistance बहुत high होता है और यह circuit से न के बराबर current लेता है [NCERT p.188]।

⚡ Checkpoint 2: Potential DifferenceCheckpoint 2: potential difference

Section SummarySection Summary (अनुभाग सारांश)

Potential Difference (V = W/Q): Work done per unit charge. Unit: Volt (V). 1 V = 1 J/C. Potential Difference (V = W/Q): प्रति unit charge पर किया गया work done। Unit: Volt (V)। 1 V = 1 J/C।
Source: Chemical action inside a cell or battery maintains the potential difference. Source: Cell या battery के अंदर की chemical action potential difference को maintain रखती है।
Voltmeter: Connected in parallel; has high resistance. Voltmeter: Parallel में जोड़ा जाता है; इसका resistance बहुत high होता है।

Figure 11.2 — 3D render of Ohm's law experimental circuit containing battery, ammeter, voltmeter, and resistance coil.

3. Ohm's Law (ओम का नियम)

Section Goal: Explain Ohm's law, define resistance and its unit (ohm), interpret the V-I linear graph, and calculate V, I, or R. Ohm's law को समझाना, resistance और उसके unit (ohm) को define करना, V-I linear graph को interpret करना, और V, I, या R calculate करना।

Analogy: 🚴 The Wind Resistance Analogy. Imagine you are riding a bicycle. The force you pedal with is like Voltage. How fast you go is like the Current. The strong headwind pushing against you is the Resistance. If you pedal harder (increase Voltage), you go faster (increase Current). But if the wind gets stronger (increase Resistance), you go slower (decrease Current). This relationship is what Ohm's law describes!

In 1827, a German physicist Georg Simon Ohm found a relationship between the current I flowing in a metallic wire and the potential difference V across its terminals. According to Ohm's Law (Ohm's law), the electric current flowing through a metallic conductor is directly proportional to the potential difference across its ends, provided its temperature remains constant [NCERT p.189]. 1827 में, एक German physicist Georg Simon Ohm ने wire में बहने वाले current I और उसके ends के potential difference V के बीच relation खोजा। Ohm's Law (Ohm's law) के अनुसार, किसी metallic conductor में बहने वाला current उसके ends के cross potential difference के directly proportional होता है, बशर्ते उसका temperature constant रहे [NCERT p.189]।

V ∝ I ⇒ V = I R

where R is a constant for the given metallic wire at a given temperature and is called its resistance (resistance). It is the property of a conductor to resist the flow of charges through it [NCERT p.189]. जहाँ R दिए गए metallic wire के लिए constant है और इसे इसका resistance (resistance) कहते हैं। यह किसी conductor की वह property है जिससे वह अपने अंदर बहने वाले charges के flow का विरोध (resist) करता है [NCERT p.189]।

The SI unit of resistance is the ohm (Ω). According to Ohm's law, R = V / I. If the potential difference across the two ends of a conductor is 1 V and the current through it is 1 A, then the resistance of the conductor is 1 Ω, i.e., 1 Ω = 1 V/A [NCERT p.189]. Also, the current through a resistor is inversely proportional to its resistance (I = V/R) [NCERT p.189]. Resistance का SI unit ohm (Ω) है। Ohm's law के अनुसार, R = V / I। यदि ends के cross potential difference 1 V हो और बहने वाला current 1 A हो, तो conductor का resistance 1 Ω होगा, यानी 1 Ω = 1 V/A [NCERT p.189]। साथ ही, current resistance के inversely proportional होता है (I = V/R) [NCERT p.189]।

V-I Graph: For an ohmic conductor, the graph of potential difference V versus current I is a straight line passing through the origin. The slope of this line represents the resistance of the conductor [NCERT p.189]. V-I Graph: Ohmic conductor के लिए potential difference V और current I का graph origin से pass होने वाली एक straight line होती है। इस line का slope conductor के resistance को represent करता है [NCERT p.189]।

⚡ Lab 1: Ohm's Law V-I Graph Builder

Voltageविभव (Voltage) (V): 6 V

Resistance (R): 30 Ω

Calculated Current (I = V/R): 0.20 A

⚡ Checkpoint 3: Ohm's LawCheckpoint 3: ओम का नियम

Section SummarySection Summary (अनुभाग सारांश)

Ohm's Law: V = IR (at constant temperature).
Resistance (R): Opposition to current. Unit: Ohm (Ω). 1 Ω = 1 V / 1 A.
Current: Inversely proportional to resistance. Doubling R halves I.
V-I Graph: A straight line; steeper slope indicates higher resistance.

⚡ Try-It Lab 1b: 3D Wire Electron Flow Sandbox

Rotate the 3D wire segment by dragging. Adjust the Voltage slider to speed up the drifting electrons (small blue spheres). Increase the Resistance slider to add impurity ions (orange spheres) which cause collisions, vibrating the copper atoms (large purple spheres) and generating heat (glow)! 3D wire segment को drag करके rotate करें। Drifting electrons (छोटे नीले गोले) की speed बढ़ाने के लिए Voltage slider को adjust करें। Impurity ions (नारंगी गोले) जोड़ने के लिए Resistance slider बढ़ाएं, जो collisions (टक्कर) का कारण बनते हैं, जिससे copper atoms (बड़े बैंगनी गोले) vibrate होते हैं और heat (चमक) पैदा होती है!

Initializing 3D WebGL...

Voltage (V): 6 V

Resistance (R): 20 Ω

Calculated Flow Details:

                                Drift Current (I = V/R): 0.30 A

                                Thermal Heat Output (P = I²R): 1.80 W

                                Lattice Vibrations: Medium

4. Factors on which Resistance Depends (प्रतिरोध को प्रभावित करने वाले कारक)

Section Goal: Explain the factors affecting resistance, define resistivity (प्रतिरोधकता) (ρ), and compare conductors, alloys, and insulators.

Analogy: 🚶 The Crowded Corridor Analogy. Imagine walking down a corridor. 1. If the corridor is longer, you encounter more people and it takes longer (Resistance increases with length: R ∝ l). 2. If the corridor is wider, it is much easier to squeeze through (Resistance decreases with wider area: R ∝ 1/A). 3. If the floor is made of sticky mud instead of smooth tiles, your speed decreases (corresponds to the nature of the material, or resistivity).

The resistance of a uniform metallic conductor depends on [NCERT p.190]: एक uniform metallic conductor का resistance इन बातों पर depend करता है [NCERT p.190]:

its length (l) — directly proportional: R ∝ l इसकी length (l) पर — directly proportional: R ∝ l
its area of cross-section (A) — inversely proportional: R ∝ 1/A इसके cross-section के area (A) पर — inversely proportional: R ∝ 1/A
the temperature of the conductor conductor के temperature पर
the nature of the material conductor के material के nature (प्रकृति) पर

Combining these factors, we get [NCERT p.191]: इन factors को combine करने पर हमें मिलता है [NCERT p.191]:

R = ρ ( l / A )

where ρ (rho) is a constant of proportionality called the electrical resistivity (resistivity) of the material of the conductor. The SI unit of resistivity is the ohm-meter (Ω·m) [NCERT p.191]. जहाँ ρ (rho) proportionality constant है जिसे material की electrical resistivity (resistivity) कहते हैं। Resistivity का SI unit ohm-meter (Ω·m) होता है [NCERT p.191]।

Resistivity is a characteristic property of the material. Metals and alloys have very low resistivity in the range of 10^-8 Ω·m to 10^-6 Ω·m, making them good conductors. Insulators like rubber and glass have very high resistivity in the range of 10¹² to 10¹⁷ Ω·m [NCERT p.191]. Resistivity material की एक characteristic property (विशिष्ट गुण) है। Metals और alloys की resistivity बहुत कम (10^-8 Ω·m से 10^-6 Ω·m) होती है, जो इन्हें good conductors बनाती है। Rubber और glass जैसे insulators की resistivity बहुत high (10¹² से 10¹⁷ Ω·m) होती है [NCERT p.191]।

Alloys (like nichrome, manganin, constantan) have higher resistivity than their constituent metals. Alloys do not oxidise (burn) readily at high temperatures, which is why they are commonly used in heating elements of electrical appliances like electric irons, heaters, and toasters [NCERT p.191]. Silver is the best electrical conductor, followed by copper and aluminium. Alloys (जैसे nichrome, manganin, constantan) की resistivity उनके constituent pure metals से अधिक होती है। Alloys high temperature पर आसानी से oxidise (जलते) नहीं हैं, यही कारण है कि इनका उपयोग electric irons, heaters, और toasters जैसे electrical heating appliances के heating elements में किया जाता है [NCERT p.191]। Silver सबसे अच्छा electrical conductor है, जिसके बाद copper और aluminium आते हैं।

⚡ Checkpoint 4: Factors of ResistanceCheckpoint 4: resistance को प्रभावित करने वाले कारक

Section SummarySection Summary (अनुभाग सारांश)

Formula: R = ρl/A. Resistivity (ρ) depends ONLY on material and temperature. Formula: R = ρl/A. Resistivity (ρ) केवल material के nature और temperature पर depend करती है।
Length & Area: Long and thin wires have high resistance; short and thick wires have low resistance. Length & Area: लंबे और पतले wires का resistance अधिक होता है; छोटे और मोटे wires का resistance कम होता है।
Alloys: High resistivity, do not melt/oxidize easily, used in heating appliances. Alloys: High resistivity, आसानी से melt/oxidize नहीं होते, heating appliances में उपयोग होते हैं।

Figure 11.3 — 3D schematic comparison of resistor components wired in series vs. parallel configurations.

5. Resistance of a System of Resistors (प्रतिरोधकों का संयोजन)

Section Goal: Contrast series and parallel resistor combinations, derive total resistance formulas, and explain why parallel wiring is used in homes. Series (series) और parallel (पार्श्वक्रम) combinations की तुलना करना, total resistance formulas derive करना, और समझाना कि घरों में parallel wiring क्यों की जाती है।

Analogy: 🛤️ Toll Booths on a Highway. 1. Series combination (श्रेणीक्रम संयोजन) is like having three toll booths one after another on a single highway lane. Every car has to stop at all three. The total delay (resistance) is the sum of all individual delays: R_total = R₁ + R₂ + R₃. 2. Parallel combination (पार्श्वक्रम संयोजन) is like opening three toll booths side-by-side. The traffic can split and pass through different booths simultaneously. This speeds up flow and reduces total resistance to less than even the smallest individual booth: 1/R_total = 1/R₁ + 1/R₂ + 1/R₃.

Series CombinationSeries Combination (series combination)

When resistors are joined end-to-end, they are said to be connected in series. In a series circuit [NCERT p.193]: जब resistors को सिरे-से-सिरा (end-to-end) मिलाकर जोड़ा जाता है, तो उन्हें series में connected कहा जाता है। एक series circuit में [NCERT p.193]:

The current (I) remains the same through all resistors. सभी resistors में से बहने वाला current (I) समान रहता है।
The total potential difference (V) divides across the resistors: V = V₁ + V₂ + V₃. Total potential difference (V) सभी resistors के cross बंट जाता है: V = V₁ + V₂ + V₃।
The equivalent resistance is the sum of individual resistances: Equivalent resistance सभी individual resistances का sum होता है:

R_s = R₁ + R₂ + R₃

Parallel CombinationParallel Combination (पार्श्वक्रम combination)

When resistors are connected across the same two nodes, they are in parallel. In a parallel circuit [NCERT p.197]: जब resistors को समान दो points के cross जोड़ा जाता है, तो वे parallel में होते हैं। एक parallel circuit में [NCERT p.197]:

The potential difference (V) remains the same across all resistors. सभी resistors के cross potential difference (V) समान रहता है।
The total current (I) divides through the branches: I = I₁ + I₂ + I₃. Total current (I) सभी branches में विभाजित (divide) हो जाता है: I = I₁ + I₂ + I₃।
The reciprocal of the equivalent resistance is: Equivalent resistance का reciprocal (व्युत्क्रम) होता है:

1 / R_p = 1 / R₁ + 1 / R₂ + 1 / R₃

Domestic Wiring: Why Parallel?Domestic Wiring: Parallel Connection क्यों?

We do not connect domestic appliances in series because [NCERT p.201]: हम घरेलू उपकरणों (domestic appliances) को series में नहीं जोड़ते हैं क्योंकि [NCERT p.201]:

If one appliance fails or is turned off, the circuit is broken and all other appliances stop working. यदि कोई एक appliance खराब हो जाए या बंद कर दिया जाए, तो circuit टूट जाता है और बाकी सभी appliances काम करना बंद कर देते हैं।
Each appliance needs a different current (e.g. an AC needs 15 A while a bulb needs 0.1 A), but in series, current is identical. हर appliance को अलग-अलग current चाहिए (जैसे AC को 15 A चाहिए जबकि bulb को 0.1 A), लेकिन series में current identical (एक समान) रहता है।
In series, the voltage divides, so devices do not get their rated 220 V. In parallel, every device gets the full 220 V. Series में voltage divide हो जाता है, जिससे devices को अपना rated 220 V नहीं मिल पाता। Parallel में, हर device को पूरा 220 V मिलता है।

⚡ Lab 2: Series-Parallel Resistance Calculator

Circuit Typeपरिपथ का प्रकार (Circuit Type)

Resistor 1resistanceक 1 (R₁): 10 Ω

Resistor 2 (R₂): 10 Ω

Equivalent Resistance (R_eq): 20.0 Ω

🔍 Spot the Mistake: Parallel Resistance Trap

Students often make a critical algebraic mistake when calculating parallel resistance: forgetting to invert the final fraction! Look at the calculation steps below. Find the step where the mistake was made and click on it. Students parallel resistance calculate करते समय अक्सर एक critical algebraic mistake करते हैं: अंतिम fraction को पलटना (invert करना) भूल जाना! नीचे दिए गए calculation steps को देखें, गलती पहचानें और उस पर click करें।

Step 1: Write down the parallel formula and substitute values
For two parallel resistors of 3 Ω and 6 Ω:

1/R_p = 1/3 + 1/6

Step 2: Find the common denominator and add the fractions

1/R_p = 2/6 + 1/6 = 3/6 = 1/2

Step 3: State the final resistance

R_p = 1/2 Ω

⚡ Checkpoint 5: Series and Parallel ResistorsCheckpoint 5: series और parallel resistanceक

Section SummarySection Summary (अनुभाग सारांश)

Series (V divides, I identical): R_total = R₁ + R₂ + R₃. Total resistance increases.
Parallel (I divides, V identical): 1/R_total = 1/R₁ + 1/R₂ + 1/R₃. Total resistance is smaller than the smallest branch.
Domestic Parallel: Provides independent switches, identical voltage (220V), and safety from whole-circuit failure.

Figure 11.4 — 3D render of a resistive heating wire glowing red-hot from current flow inside a toaster.

6. Heating Effect of Electric Current (विद्युत धारा का तापीय प्रभाव)

Section Goal: Explain the heating effect of current, state Joule's law of heating (H = I²Rt), and list safety and household applications. Current के heating effect को समझाना, Joule's law of heating (H = I²Rt) का statement देना, और इसके household व safety applications बताना।

Analogy: 摩擦 Friction on a Rope. Imagine pulling a rough rope through your hands quickly. The friction of the rope rubbing against your skin creates heat. When current flows, electrons collide with the atoms of the metal conductor. These microscopic collisions cause the atoms to vibrate faster, creating heat. In a high-resistance wire like Nichrome, the friction is massive, converting electrical energy into heat quickly!

If the electric circuit is purely resistive, that is, a configuration of resistors practically connected to a battery, the energy source is continuously dissipated entirely in the form of heat. This is known as the heating effect of electric current (heating effect) [NCERT p.202]. यदि कोई electric circuit purely resistive है, यानी उसमें केवल resistors ही battery से जुड़े हैं, तो source की energy लगातार पूरी तरह से heat के रूप में lost होती रहती है। इसे heating effect of electric current (heating effect) कहते हैं [NCERT p.202]।

Consider a current I flowing through a resistor of resistance R. Let the potential difference across it be V. The work done in moving charge Q is V × Q, and the power input is P = V(Q/t) = VI. The energy supplied to the circuit by the source in time t is H = P × t = VIt. Applying Ohm's law (V=IR), we obtain Joule's Law of Heating (Joule's law) [NCERT p.202]: मान लीजिए resistance R वाले resistor से current I बह रहा है और potential difference V है। Charge Q को move करने में किया गया work done V × Q होगा, और input power P = V(Q/t) = VI होगी। Source द्वारा समय t में supplied energy H = P × t = VIt होगी। Ohm's law (V=IR) apply करने पर हमें Joule's Law of Heating (Joule's law) प्राप्त होता है [NCERT p.202]:

H = I² R t

This law implies that heat produced in a resistor is [NCERT p.202]: इस नियम के अनुसार, किसी resistor में produced heat [NCERT p.202]:

directly proportional to the square of current for a given resistance, दिए गए resistance के लिए current के square के directly proportional होती है,
directly proportional to resistance for a given current, and दिए गए current के लिए resistance के directly proportional होती है, और
directly proportional to the time for which the current flows. उस समय (time) के directly proportional होती है जिसके लिए current बहता है।

Practical Applications:व्यावहारिक अनुप्रयोग (Practical Applications):

Heating Appliances: Electric iron, toaster, oven, kettle, heater rely on high-resistance elements (like nichrome) to convert current to heat. Heating Appliances: Electric iron, toaster, oven, kettle, heater आदि high-resistance elements (जैसे nichrome) पर rely करते हैं ताकि वे current को heat में बदल सकें।
Electric Bulb: The filament is made of tungsten because it has a very high melting point (3380 °C) and does not melt when glowing white-hot. Bulbs are filled with chemically inactive nitrogen or argon to prevent oxidation [NCERT p.202]. Electric Bulb: इसका filament tungsten का बना होता है क्योंकि इसका melting point (गलनांक) बहुत high (3380 °C) होता है और यह सफेद-गर्म (white-hot) होने पर भी पिघलता नहीं है। Bulbs को chemically inactive nitrogen या argon गैसों से भरा जाता है ताकि filament का oxidation न हो [NCERT p.202]।
Electric Fuse: A safety device connected in series with the live wire. It consists of a wire made of a metal/alloy with a low melting point (e.g. tin-lead alloy). If the current exceeds a safe limit (due to overloading or short-circuiting), the fuse wire melts and breaks the circuit, protecting appliances [NCERT p.202]. Electric Fuse: यह live wire के साथ series में जुड़ा एक safety device है। इसमें कम melting point वाले metal/alloy (जैसे tin-lead alloy) का wire होता है। यदि overloading या short-circuit के कारण current सुरक्षित सीमा से अधिक हो जाता है, तो fuse wire पिघलकर circuit को break कर देता है, जिससे appliances सुरक्षित रहते हैं [NCERT p.202]।

⚡ Lab 3: Joule's Heating Simulator

Currentelectric current (Current) (I): 2 A

Resistanceresistance (Resistance) (R): 5 Ω

Timeसमय (Time) (t): 10 s

Generated Heat (H = I²Rt): 200 J

Temp: 25°C

⚡ Checkpoint 6: Heating EffectCheckpoint 6: electric current का तापीय प्रभाव

Section SummarySection Summary (अनुभाग सारांश)

Joule's Law (H = I²Rt): Energy converted to heat in a resistor.
Bulb: Uses high melting point Tungsten filament and inert gases.
Fuse: Safety device with a low melting point wire; connected in series with the live wire.

7. Electric Power (विद्युत शक्ति)

Section Goal: Define electric power, state its formulas (P = VI = I²R = V²/R), define commercial unit of energy (kWh), and solve bill calculations. Electric power को define करना, इसके formulas (P = VI = I²R = V²/R) को समझना, energy के commercial unit (kWh) को समझना, और bills calculations करना।

Analogy: 🔋 The Hourly Water Tank Consumption. Imagine your house water usage. If you run a shower, you consume water at a specific rate (Litres per minute - like Power). The total water you drain from your tank over the entire day depends on how long you run the shower (Litres per minute × hours = total Litres - like Energy). In electricity, the rate of energy consumption is Power (watts), and the total electricity used is Energy (kilowatt-hours), which you are billed for! 🔋 The Hourly Water Tank Consumption: पानी की टंकी का उपभोग। अपने घर में पानी के उपयोग की कल्पना करें। यदि आप शॉवर चलाते हैं, तो आप एक विशिष्ट दर (Litres per minute - जैसे Power) पर पानी का उपभोग करते हैं। पूरे दिन में आप टंकी से कुल कितना पानी बहाते हैं, यह इस बात पर निर्भर करता है कि आपने शॉवर कितनी देर तक चलाया (Litres per minute × hours = total Litres - जैसे Energy)। Electricity में, energy consumption का rate Power (watts) कहलाता है, और उपयोग की गई कुल बिजली Energy (kilowatt-hours) कहलाती है, जिसके लिए आपको बिल भेजा जाता है!

We know that the rate of doing work is power. This is also the rate of consumption of energy. The electric power (P) is the rate at which electrical energy is dissipated or consumed in an electric circuit [NCERT p.205]: हम जानते हैं कि work done करने का rate power है। यह energy consumption का rate भी है। किसी electric circuit में electrical energy के consume या dissipate होने की rate को electric power (P) कहा जाता है [NCERT p.205]:

P = V I = I² R = V² / R

The SI unit of electric power is the watt (W). It is the power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V, i.e., 1 W = 1 V × 1 A [NCERT p.205]. Since W is small, we use kilowatt (1 kW = 1000 W). Electric power का SI unit watt (W) है। यह उस device द्वारा consumed power है जो 1 V potential difference पर 1 A current लेता है, यानी 1 W = 1 V × 1 A [NCERT p.205]। चूंकि watt एक छोटा unit है, इसलिए हम kilowatt (1 kW = 1000 W) का उपयोग करते हैं।

Commercial Unit of Electrical Energy: The product of power and time gives electrical energy. The SI unit of electrical energy is the Joule (J). However, the commercial unit of electrical energy is the kilowatt-hour (kWh), commonly known as a 'unit' of electricity [NCERT p.206]. Commercial Unit of Electrical Energy: power और time का product (गुणनफल) electrical energy देता है। Electrical energy का SI unit Joule (J) है। लेकिन electrical energy का commercial unit kilowatt-hour (kWh) है, जिसे आमतौर पर बिजली का 'unit' (यूनिट) कहा जाता है [NCERT p.206]।

1 kWh = 1 kW × 1 hour = 1000 W × 3600 s = 3.6 × 10⁶ Joules (J)

⚡ Lab 4: House Load & Bill CalculatorLab 4: घरेलू उपकरण और बिजली बिल कैलकुलेटर

Select Deviceउपकरण चुनें (Select Device)

Hours used per day:प्रतिदिन उपयोग के घंटे: 8 hours

Electricity Rate (per kWh): ₹बिजली की दर (प्रति kWh): ₹ 6

Monthly Consumption:मासिक उपभोग (Monthly Consumption): 14.4 kWh (units)

Estimated Cost (30 Days):अनुमानित लागत (30 दिन): ₹ 86.40

Interactive Load Visualizerलोड विज़ुअलाइज़र (Interactive Load Visualizer)

See how high wattage appliances drastically increase your energy bill compared to low power bulbs. देखें कि कैसे high wattage वाले उपकरण low power bulbs की तुलना में आपके बिजली के बिल को तेजी से बढ़ाते हैं।

Device Power :उपकरण की शक्ति (Power): 10 W
Monthly Hours: 240 hrs
Energy (P×t)  : 2.4 kWh

⚡ Checkpoint 7: Electric PowerCheckpoint 7: electric शक्ति

Section SummarySection Summary (अनुभाग सारांश)

Power (P = VI = I²R = V²/R): Rate of electrical energy consumption. Unit: Watt (W).
Commercial Energy Unit: Kilowatt-hour (kWh) or 'Unit'.
Conversion: 1 kWh = 3.6 × 10⁶ J.
Cost: Total units (kWh) × cost per unit.

NCERT Solutions

In-Text Questions (पाठ्यपुस्तक के प्रश्न)

Q1. What does an electric circuit mean? [NCERT p.188]

Answer: An electric circuit is a closed, continuous, and conducting path through which an electric current flows. A typical electric circuit consists of a source of electric current (a cell or battery), connecting wires, an electrical load (like a bulb), and a switch to turn the current on or off.

Q2. Define the unit of current. [NCERT p.188]

Answer: The SI unit of current is the ampere (A). One ampere is defined as the current constituted by the flow of one coulomb of charge per second through any cross-section of a conductor.
1 A = 1 C / 1 s

Q3. Calculate the number of electrons constituting one coulomb of charge. [NCERT p.188]

Answer: The charge on one electron (e) is 1.6 × 10^-19 C.
Let n be the number of electrons constituting a charge Q = 1 C.
Using the quantization of charge:
Q = n × e
n = Q / e = 1 C / (1.6 × 10^-19 C)
n = 6.25 × 10¹⁸ electrons.

Q4. Name a device that helps to maintain a potential difference across a conductor. [NCERT p.188]

Answer: A cell or a battery (a combination of cells) helps to maintain a potential difference across a conductor by chemical reaction inside it.

Q5. What is meant by saying that the potential difference between two points is 1 V? [NCERT p.188]

Answer: The potential difference between two points is said to be 1 volt (V) if 1 joule (J) of work is done to move a charge of 1 coulomb (C) from one point to the other in a current-carrying circuit.
1 V = 1 J / 1 C

Q6. How much energy is given to each coulomb of charge passing through a 6 V battery? [NCERT p.188]

Answer: Given: Potential difference V = 6 V, Charge Q = 1 C (each coulomb).
The potential difference is defined as W/Q, where W is the work done (which is equal to the energy given).
Therefore, Energy W = V × Q = 6 V × 1 C = 6 Joules (J).

Q7. On what factors does the resistance of a conductor depend? [NCERT p.200]

Answer: The resistance (R) of a conductor depends on the following factors:

Length of the conductor (l): R is directly proportional to length (R ∝ l).
Area of cross-section (A): R is inversely proportional to cross-sectional area (R ∝ 1/A).
Nature of the material: Different materials have different electrical resistivities (ρ).
Temperature: For metallic conductors, resistance increases with temperature.

Q8. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why? [NCERT p.200]

Answer: Current will flow more easily through the thick wire.
Reason: The resistance (R) of a wire is inversely proportional to its cross-sectional area (R ∝ 1/A). A thick wire has a larger cross-sectional area, meaning it has lower resistance than a thin wire of the same material and length. Since current I = V/R, for the same voltage source, lower resistance allows more current to flow.

Q9. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it? [NCERT p.200]

Answer: According to Ohm's law, current is directly proportional to the potential difference across the component (I = V/R). If the resistance (R) remains constant and the voltage (V) decreases to half (V/2), the current will also decrease to half of its original value (I/2).

Q10. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal? [NCERT p.200]

Answer: Coils of electric toasters and irons are made of alloys (like nichrome) rather than pure metals because:

Alloys have much higher electrical resistivity than pure metals, generating more heat.
Alloys do not readily oxidise (burn) or melt at high operating temperatures.

Exercise Questions (अध्याय के अभ्यास प्रश्न)

Q1. A piece of wire of resistance R is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is R', then the ratio R/R' is: (a) 1/25 (b) 1/5 (c) 5 (d) 25 [NCERT p.212]

Answer: (d) 25
Explanation: When the wire is cut into 5 equal parts, the resistance of each part becomes R/5 (since resistance is directly proportional to length).
When these 5 parts are connected in parallel, the equivalent resistance R' is:
1/R' = 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5) + 1/(R/5)
1/R' = 5/R + 5/R + 5/R + 5/R + 5/R = 25/R
⇒ R' = R / 25
⇒ Ratio R / R' = 25.

Q2. Which of the following terms does not represent electrical power in a circuit? (a) I²R (b) IR² (c) VI (d) V²/R [NCERT p.212]

Answer: (b) IR²
Explanation: Power P is given by VI. Using Ohm's law V = IR, we get P = I²R. Alternatively, using I = V/R, we get P = V²/R. Thus, IR² is not a valid expression for electrical power.

Q3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be: (a) 100 W (b) 75 W (c) 50 W (d) 25 W [NCERT p.212]

Answer: (d) 25 W
Explanation:
Step 1: Calculate the resistance of the bulb.
P = V² / R ⇒ R = V² / P = (220)² / 100 = 48400 / 100 = 484 Ω.
Step 2: Calculate power consumed at 110 V.
P_new = V_new² / R = (110)² / 484 = 12100 / 484 = 25 W.

Q4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be: (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1 [NCERT p.212]

Answer: (c) 1:4
Explanation: Let the resistance of each wire be R.
1. Series: R_s = R + R = 2R.
Heat produced H_s = V²t / R_s = V²t / 2R.
2. Parallel: 1/R_p = 1/R + 1/R = 2/R ⇒ R_p = R/2.
Heat produced H_p = V²t / R_p = V²t / (R/2) = 2V²t / R.
3. Ratio: H_s / H_p = (V²t / 2R) / (2V²t / R) = 1/4 or 1:4.

Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points? [NCERT p.212]

Answer: A voltmeter is always connected in parallel across the two points between which the potential difference is to be measured. This is done because a voltmeter has a very high electrical resistance, drawing negligible current from the main circuit and ensuring the measured voltage is accurate.

Solved CBSE Previous Year Questions (PYQs)

Q1. Draw the V-I graph for an ohmic conductor. What does its slope represent? [CBSE 2019, 1 Mark] Q1. एक ohmic conductor के लिए V-I graph draw करें। इसका slope क्या represent करता है? [CBSE 2019, 1 Mark]

Answer:
The V-I graph for an ohmic conductor is a straight line passing through the origin. The slope of the V-I graph (V/I) represents the electrical resistance (R) of the conductor. Ohmic conductor के लिए V-I graph origin से pass होने वाली एक straight line होती है। V-I graph का slope (V/I) conductor के electrical resistance (R) को represent करता है।

Q2. Out of silver, copper, and constantan, which has the highest electrical resistivity? Mention its utility. [CBSE 2021, 2 Marks] Q2. Silver, copper, और constantan में से किसका electrical resistivity सबसे अधिक है? इसकी utility बताइए। [CBSE 2021, 2 Marks]

Answer:
Among silver, copper, and constantan, constantan (an alloy) has the highest electrical resistivity. Utility: It is used in making standard resistors and heating elements because its resistance is stable and doesn't change easily with temperature. Silver, copper, और constantan में से constantan (एक alloy) की resistivity सबसे अधिक होती है। Utility: इसका उपयोग standard resistors और heating elements बनाने में किया जाता है क्योंकि इसका resistance temperature के साथ आसानी से नहीं बदलता।

Q3. A wire of resistance 20 Ω is bent in the form of a closed circle. What is the equivalent resistance between two points along its diameter? [CBSE 2020, 3 Marks] Q3. 20 Ω resistance वाले एक wire को एक closed circle के रूप में मोड़ा जाता है। इसके diameter के two points के बीच equivalent resistance क्या होगा? [CBSE 2020, 3 Marks]

Answer:
The diameter divides the circular wire into two equal semicircular halves. The resistance of each half is R₁ = R₂ = 20 / 2 = 10 Ω. Since these two halves are in parallel, 1/R_eq = 1/10 + 1/10 = 2/10 = 1/5 ⇒ R_eq = 5 Ω. Diameter circle को दो equal semicircular halves में बांटता है। हर आधे भाग का resistance R₁ = R₂ = 20 / 2 = 10 Ω होगा। चूँकि ये दोनों parallel में जुड़े हैं, 1/R_eq = 1/10 + 1/10 = 2/10 = 1/5 ⇒ R_eq = 5 Ω.

Q4. (a) State Joule's law of heating and write its mathematical equation. (b) An electric heater of 1000 W works on 220 V. Calculate (i) the current drawn, (ii) resistance of the heater element, and (iii) the energy consumed in 2 hours in kWh. [CBSE 2022, 5 Marks] Q4. (a) Joule's law of heating का statement दें और इसका mathematical equation लिखें। (b) 1000 W का एक electric heater 220 V पर काम करता है। Calculate करें: (i) drawn current, (ii) heater element का resistance, और (iii) 2 hours में consumed energy (kWh में)। [CBSE 2022, 5 Marks]

Answer:
(a) Joule's Law states that heat produced in a resistor is directly proportional to the square of current (I²), resistance (R), and time (t): H = I²Rt.
(b) Given: P = 1000 W = 1 kW, V = 220 V, t = 2 h.
(i) Current I = P / V = 1000 / 220 = 4.55 A.
(ii) Resistance R = V / I = 220 / 4.55 = 48.4 Ω.
(iii) Energy E = P × t = 1 kW × 2 h = 2 kWh. (a) Joule's Law के अनुसार, किसी resistor में produced heat current के square (I²), resistance (R), और time (t) के directly proportional होती है: H = I²Rt.
(b) Given: P = 1000 W = 1 kW, V = 220 V, t = 2 h.
(i) Current I = P / V = 1000 / 220 = 4.55 A.
(ii) Resistance R = V / I = 220 / 4.55 = 48.4 Ω.
(iii) Energy E = P × t = 1 kW × 2 h = 2 kWh.

Q5. Why are alloys commonly used in electrical heating devices like electric irons and heaters, rather than pure metals? [CBSE 2023, 2 Marks] Q5. Alloys का उपयोग electric irons और heaters जैसे heating devices में pure metals की तुलना में अधिक क्यों किया जाता है? [CBSE 2023, 2 Marks]

Answer:
Alloys are used because: (1) They have higher electrical resistivity than their constituent pure metals. (2) They do not oxidise (burn) or melt easily at high temperatures. Alloys का उपयोग इसलिए किया जाता है क्योंकि: (1) इनकी electrical resistivity इनकी pure metals की तुलना में अधिक होती है। (2) ये high temperatures पर आसानी से oxidise (जलते) या melt नहीं होते।

Q6. Compute the number of electrons constituting one coulomb of charge. [CBSE 2024, 2 Marks] Q6. एक coulomb charge बनाने वाले electrons की संख्या calculate करें। [CBSE 2024, 2 Marks]

Answer:
Given: Charge Q = 1 C, Charge of one electron e = 1.6 × 10⁻¹⁹ C. Using Q = ne ⇒ n = Q/e = 1 / (1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons. Given: Charge Q = 1 C, एक electron का charge e = 1.6 × 10⁻¹⁹ C. Q = ne का उपयोग करके ⇒ n = Q/e = 1 / (1.6 × 10⁻¹⁹) = 6.25 × 10¹⁸ electrons.

Q7. The V-I graphs for two resistors A and B are shown. If A makes a larger angle with the current axis than B, which of the two has higher resistance and why? [CBSE 2020/2023, 3 Marks] Q7. दो resistors A और B के लिए V-I graphs दिखाए गए हैं। यदि A current axis के साथ B की तुलना में बड़ा angle बनाता है, तो दोनों में से किसका resistance अधिक है और क्यों? [CBSE 2020/2023, 3 Marks]

Answer:
Resistance (R) is represented by the slope of the V-I graph (V/I). Since resistor A makes a larger angle (steeper slope) with the current axis (x-axis), it has a higher V/I ratio for any given current. Thus, resistor A has a higher resistance than resistor B. V-I graph का slope (V/I) resistance (R) को दर्शाता है। चूँकि resistor A, current axis (x-axis) के साथ बड़ा angle बनाता है, इसलिए किसी दिए गए current के लिए इसका V/I ratio अधिक है। अतः resistor A का resistance, B से अधिक है।

Q8. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, what will be the power consumed? [CBSE 2022/2025, 3 Marks] Q8. एक electric bulb पर 220 V और 100 W rated है। जब इसे 110 V पर चलाया जाता है, तो consumed power क्या होगी? [CBSE 2022/2025, 3 Marks]

Answer:
Resistance of the bulb: R = V² / P = (220)² / 100 = 484 Ω.
When operated on V' = 110 V, new power consumed P' = V'² / R = (110)² / 484 = 12100 / 484 = 25 W. Bulb का resistance: R = V² / P = (220)² / 100 = 484 Ω.
जब V' = 110 V पर operate किया जाता है, तो new power P' = V'² / R = (110)² / 484 = 12100 / 484 = 25 W.

Q9. Derive the expression for equivalent resistance of three resistors connected in parallel. [CBSE 2024, 3 Marks] Q9. Parallel में जुड़े तीन resistors के equivalent resistance का expression derive करें। [CBSE 2024, 3 Marks]

Answer:
Let three resistors R₁, R₂, and R₃ be connected in parallel across a potential difference V.
1. The potential difference V across each resistor is the same.
2. The total current I is the sum of currents in branches: I = I₁ + I₂ + I₃.
3. By Ohm's law: I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃, and total current I = V/R_p.
Substituting these: V/R_p = V/R₁ + V/R₂ + V/R₃ ⇒ 1/R_p = 1/R₁ + 1/R₂ + 1/R₃. मान लीजिए R₁, R₂, और R₃ parallel में potential difference V के साथ जुड़े हैं।
1. हर resistor के cross potential difference V समान रहता है।
2. Total current I सभी branches के currents का sum होता है: I = I₁ + I₂ + I₃.
3. Ohm's law के अनुसार: I₁ = V/R₁, I₂ = V/R₂, I₃ = V/R₃, और total current I = V/R_p.
इन्हें substitute करने पर: V/R_p = V/R₁ + V/R₂ + V/R₃ ⇒ 1/R_p = 1/R₁ + 1/R₂ + 1/R₃.

Q10. (a) Why is a series circuit not used for domestic electrical wiring? (b) An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it? [CBSE 2021/2025, 5 Marks] Q10. (a) घरेलू electrical wiring के लिए series circuit का उपयोग क्यों नहीं किया जाता? (b) 100 Ω का एक electric lamp, 50 Ω का toaster, और 500 Ω का water filter एक 220 V source से parallel in जुड़े हैं। उसी source से जुड़े electric iron का resistance क्या होगा जो इन तीनों के बराबर current लेता है, और इससे बहने वाला current क्या है? [CBSE 2021/2025, 5 Marks]

Answer:
(a) Series is not used because: (1) If one device fails, the circuit breaks and all devices turn off. (2) Different devices need different currents, which is impossible in series. (3) Devices don't get the full 220 V.
(b) Parallel resistance of lamp, toaster, filter:
1/R_p = 1/100 + 1/50 + 1/500 = (5 + 10 + 1) / 500 = 16 / 500
R_p = 500 / 16 = 31.25 Ω.
The electric iron takes as much current as all three, so its resistance is equal to the equivalent resistance: R_iron = 31.25 Ω.
Current through it: I = V / R_iron = 220 / 31.25 = 7.04 A. (a) Series का उपयोग इसलिए नहीं किया जाता क्योंकि: (1) यदि एक device खराब हो जाए, तो circuit टूट जाता है और सभी बंद हो जाते हैं। (2) अलग-अलग devices को अलग current चाहिए, जो series में मुमकिन नहीं। (3) Devices को पूरा 220 V नहीं मिलता।
(b) Lamp, toaster, और filter का parallel resistance:
1/R_p = 1/100 + 1/50 + 1/500 = (5 + 10 + 1) / 500 = 16 / 500
R_p = 500 / 16 = 31.25 Ω.
Electric iron इन तीनों के बराबर current लेता है, इसलिए इसका resistance equivalent resistance के बराबर होगा: R_iron = 31.25 Ω।
इसमें से बहने वाला current: I = V / R_iron = 220 / 31.25 = 7.04 A।

Sources:

NCERT Class 10 Science, Chapter 11: Electricity, pp. 186–215. (Reprint 2026-27).
CBSE Class 10 Science Syllabus & Board Exam Paper Guidelines (2025-26).