Electricity Notes

Electricity

Class 10 Notes | Science Chapter 11

Introduction&Origins

Electricity plays a crucial role in our modern world, powering everything from homes and schools to hospitals and industries. But what exactly is electricity? How does it travel through an electric circuit? And what factors influence the flow of current? In this chapter, we'll explore these questions, uncover the principles behind electric circuits, and delve into the heating effects of electric current and its many practical applications.

Historical Origin: The word electricity originates from the Greek word "Elektron" (amber). The Greek philosopher Thales was the first to observe the attracting capacity of certain materials when rubbed together.

1. Electric Current and Circuit

•  Electricity is one of the most convenient and widely used forms of energy in the world.

•  Conductors are those substances in which electric charge can flow freely. Metals are electric conductors. Copper and aluminium are the two most commonly used metals in electricity.

•  Electrical insulators are those materials in which electric charges do not move freely. Rubber, glass, and plastic are examples of insulators.

•  Current electricity is the electricity which deals with moving charges.

•  A continuous and closed path of an electric current is called an electric circuit.

•  Electric current is defined as the rate of flow of electric charge through a cross-section of a conductor.

Electric Current (I) = Charge Flown (Q) / Time (t)   ⇒   I = Q / t

Flow of Electric Current

• SI Unit:  The S.I. unit of electric current is ampere (A). One-ampere current is equivalent to the flow of 1 coulomb of charge per second through a section of a conductor. Thus, 1 A = 1 C s⁻¹.

• Direction:  The direction of the flow of electric current is taken as opposite to the direction of the flow of negatively charged electrons. In other terms, the direction of conventional current is taken as the direction along which positive charges will move if allowed to do so.

• Electron Flow:  In conductors, the flow of electric current is due to the flow of electrons. The flow of 6.25 × 10¹⁸ electrons per second across a cross-section of a conductor constitutes a current of 1 ampere.

Solved Example 1.1 Question: A current of 0.75 A flows through a heater for 8 minutes. Calculate the amount of electric charge that passes through the circuit. Solution: Given: Current (I) = 0.75 A, Time (t) = 8 min = 8 × 60 s = 480 s Formula: Q = I × t Q = 0.75 A × 480 s = 360 C Answer: The electric charge that flows through the circuit is 360 C.

2. Electric Potential&Potential Difference

The flow of electric charge in a circuit is similar to water flow in a pipe:

•  The water pipe is analogous to the electric conductor, and the amount of water flowing across a section per second corresponds to electric current.

•  As water flows from higher pressure (level) to lower pressure (level), the electric current flows from higher electric potential to lower electric potential.

•  The potential difference across a conductor is maintained by using a cell or battery.

•  The chemical action within a cell generates the potential difference between its electrodes, due to which current flows through the circuit.

•  Potential difference between two points in an electric circuit carrying current is defined as the work done to move a unit positive charge from one point to another:

Potential Difference (V_A - V_B) = Work Done (W) / Charge (Q)

• SI Unit:  The SI unit of electric potential difference is volt (V), where 1 volt (V) = 1 joule (J) / 1 coulomb (C) = 1 J C⁻¹.

Solved Example 2.1 Question: How much work is done in moving a charge of 3 C between two points with a potential difference of 9 V? Solution: Given: Charge (Q) = 3 C, Potential Difference (V) = 9 V Formula: Work Done (W) = V × Q W = 9 V × 3 C = 27 J Answer: 27 J of work is done in moving the charge.

Components of Circuit Diagram

3. Circuit Diagram

•  A circuit diagram is a simplified representation of the components of an electrical circuit using either images of distinct parts or standard symbols.

•  It shows the relative positions of all the elements and their connections to one another.

•  It is often used to provide a visual representation of the circuit to an electrician.

4. Ohm's Law

Georg Simon Ohm established a relationship between the electric current flowing through a conductor and the potential difference across its ends, due to which current flows.

Statement of Ohm's Law: The temperature remaining constant, the current passing through a conductor is directly proportional to the potential difference across its ends, i.e.,

V ∝ I   ⇒   V = I R

•  Here, the constant R is known as the resistance of the given conductor. For a given conductor, its resistance is constant at a given temperature.

•  Electrical resistance of a conductor may be considered as a measure of the opposition offered by it for the flow of electric charge through it. Mathematically, R = V / I.

•  S.I. unit of electrical resistance is ohm (Ω), where 1 Ω = 1 V / 1 A.

Activity: Understanding Ohm's Law

•  Objective: To study the relationship between potential difference (V) across a nichrome wire and current (I) passing through it by varying the number of cells.

•  Materials Needed: Nichrome wire (length 0.5 m), Ammeter, Voltmeter, Four 1.5 V cells, Connecting wires, Switch, Circuit board.

•  Procedure: Connect nichrome wire (XY) in series with ammeter and in parallel with voltmeter using 1 cell. Record V and I. Repeat by adding 2nd, 3rd, and 4th cells.

•  Observation & Conclusion: The ratio V/I remains approximately constant in each case. The V-I graph is a straight line passing through the origin. Thus, V/I is a constant ratio (R).

Important Note: The current through a resistor is inversely proportional to its resistance. If the resistance is doubled, the current gets halved.

5. Factors Affecting Resistance of a Conductor

At a given temperature, the resistance (R) of a conductor depends on:

•  (i) Length of the conductor (L)

•  (ii) Cross-sectional area of the conductor (A)

•  (iii) Nature of the material of the conductor

R ∝ L   and   R ∝ 1/A   ⇒   R = ρ L / A

• Resistivity (ρ):  Where ρ (rho) is a constant known as the resistivity of the material. Its value depends only on the nature of the material and temperature, independent of dimensions.

• Definition & Unit:  Resistivity is defined as the resistance offered by a cube of that material of side 1 m when current flows perpendicular to opposite faces. Unit: ohm-metre (Ω m).

• Range:  Metals and alloys have low resistivity (10⁻⁸ to 10⁻⁶ Ω m) and are good conductors. Insulators like rubber and glass have resistivity of 10¹² to 10¹⁷ Ω m.

• Temperature Effect:  For metallic conductors, resistance and resistivity increase with an increase in temperature.

Applications Based on Resistivity

•  Pure Metals (Copper & Aluminium): Very low resistivity, used for electrical transmission lines.

•  Alloys (Nichrome): Resistivity is higher than pure metals, do not oxidise easily at high temperatures. Used as heating elements in electric iron, heater, radiator, geyser, etc.

•  Tungsten Filaments: Used in electric lamps because tungsten has an extremely high melting point.

•  Rheostat: Variable resistance device used in electrical circuits to regulate current without changing voltage source.

•  Ammeter: Instrument measuring electric current directly in amperes (connected in series).

•  Voltmeter: Instrument measuring potential difference directly in volts (connected in parallel).

Solved Example 5.1 Question: The resistance of a metal wire of length 1.5 m is 30 Ω at 25°C. If the diameter of the wire is 0.4 mm, what will be the resistivity of the metal at that temperature? Solution: Given: Resistance (R) = 30 Ω, Length (l) = 1.5 m, Diameter (d) = 0.4 mm = 4 × 10⁻⁴ m Formula: ρ = R × A / l = R × (π d² / 4) / l Calculation: ρ = 30 × (π × (4 × 10⁻⁴)² / 4) / 1.5 = 2.51 × 10⁻⁶ Ω m Answer: The resistivity of the metal at 25°C is 2.51 × 10⁻⁶ Ω m.

Solved Example 5.2 Question: A wire made of a specific material has a length l and cross-sectional area A, with a resistance of 6 Ω. What would be the resistance of another wire made of the same material but with length l/3 and cross-sectional area 3A? Solution: For initial wire: R₁ = ρ (l / A) = 6 Ω For new wire: R₂ = ρ ((l / 3) / 3A) = ρ (l / 9A) = (1 / 9) × R₁ Calculation: R₂ = (1 / 9) × 6 Ω = 1 Ω Answer: The resistance of the new wire is 1 Ω.

6. Resistance of System of Resistors

In an electric circuit, resistances may be connected in series or parallel arrangement.

A) Series Grouping of Resistances

•  The current flowing through all the resistances is the same.

•  The total voltage across the combination is equal to the sum of voltage drops across individual resistors (V = V₁ + V₂ + V₃).

•  The total resistance of the combination is equal to the sum of individual resistances:

R_s = R₁ + R₂ + R₃ + ...

B) Parallel Grouping of Resistances

•  The voltage across each resistor is the same and equal to the voltage across the whole combination.

•  The currents in various resistors are inversely proportional to their resistances, and total current is the sum of individual currents (I = I₁ + I₂ + I₃).

•  The sum of reciprocals of separate resistances is equal to the reciprocal of equivalent resistance:

1 / R_p = 1 / R₁ + 1 / R₂ + 1 / R₃ + ...

C) Household Electric Circuits

•  In household circuits, series circuit is NOT followed because if one component fails, the circuit breaks and all components stop working. Also, different appliances require different currents.

•  A parallel circuit divides current according to component requirements, allows separate on/off switches for each appliance, and reduces overall resistance. Hence, household appliances are connected in parallel.

Solved Example 6.1 Question: An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery in series. Calculate: (a) Total resistance of circuit    (b) Current through circuit (c) Potential difference across (i) electric lamp and (ii) conductor    (d) Power of lamp Solution: (a) Total Resistance R = R₁ + R₂ = 20 Ω + 4 Ω = 24 Ω (b) Current I = V / R = 6 V / 24 Ω = 0.25 A (c) (i) Voltage across lamp V₁ = I × R₁ = 0.25 A × 20 Ω = 5 V     (ii) Voltage across conductor V₂ = I × R₂ = 0.25 A × 4 Ω = 1 V (d) Power of lamp P = V₁ × I = 5 V × 0.25 A = 1.25 W

7. Heating Effect of Electric Current

If a potential difference V is applied across a conductor of resistance R and current I flows for time t, then according to Joule's law of heating, the electric energy consumed (and heat produced H) is given by:

H = W = Q V = V I t = I² R t = (V² t) / R  (in Joules)

•  Use W = I² R t for series circuits (since current I is constant).

•  Use W = (V² t) / R for parallel circuits (since voltage V is constant across branches).

Practical Applications of Heating Effect

•  Electric Bulb: Electric heating produces light. Filaments are made of high melting point tungsten. Bulbs are filled with chemically inactive nitrogen and argon gases to prolong filament life.

•  Electric Fuse: Protects circuits/appliances by stopping unduly high current. Made of a metal/alloy wire with appropriate melting point placed in series. If current exceeds rated capacity, temperature rises, fuse melts, and breaks the circuit.

Solved Example 7.1 Question: An electric heater uses 900 W of power at maximum heat and 400 W at minimum heat operating at 240 V. Calculate current and resistance in each case. Solution: (a) Maximum Heating (P = 900 W, V = 240 V):     Current I = P / V = 900 / 240 = 3.75 A     Resistance R = V / I = 240 / 3.75 = 64 Ω (b) Minimum Heating (P = 400 W, V = 240 V):     Current I = P / V = 400 / 240 = 1.67 A     Resistance R = V / I = 240 / 1.67 = 144 Ω

8. Electric Power

The time rate of doing electric work is called electric power:

P = W / t = V I = I² R = V² / R

• SI Unit:  SI Unit: Watt (W), where 1 Watt = 1 Volt × 1 Ampere = 1 V A.

• Commercial Unit:  Practical units of electric energy are watt-hour and kilowatt-hour. Commercial unit is kilowatt-hour (kW h).

1 kW h = 1000 W h = 3.6 × 10⁶ J

 

Chemical Reactions And Equations Notes

 Chemical Reactions and Equations

Class 10 Science — Chapter 1 Complete Chapter Notes

In our daily lives, physical changes involve alterations in appearance or state without creating new substances, like melting ice. On the other hand, chemical changes result in the formation of entirely new substances, such as cooking food or rusting iron. Recognizing these changes helps us understand when a chemical reaction occurs.

1. What is a Chemical Reaction?

A chemical reaction is a process in which one or more substances, the Reactants, are converted to one or more different substances, the Products.

2. Chemical Equations

A chemical equation is a symbolic representation of a chemical reaction using the symbols and formulas of the substances involved.

The substances that undergo chemical change in the reaction, Magnesium and Oxygen, are the Reactants. When magnesium burns in air, it combines with oxygen to form magnesium oxide.
Chemical equations can be made more concise and useful if we use chemical formulae instead of words. Example: The chemical formula for water is H₂O, which indicates that it is made up of two hydrogen atoms and one oxygen atom.

1. Writing a Chemical Equation

Representation of a chemical reaction in terms of symbols and chemical formulae of the reactants and products is known as a chemical equation.
Physical state notation:

• For solids, the symbol is "(s)".

• For liquids, it is "(l)".

• For gases, it is "(g)".

• For aqueous solutions, it is "(aq)".

• For gas produced in the reaction, it is represented by "(↑)".

• For precipitate formed in the reaction, it is represented by "(↓)".

The reactants are on the left (LHS) of the arrow, while the products are on the right (RHS). A plus sign (+) links the different reactants and products together.

2. Balanced Chemical Equations

A balanced account of a chemical reaction is a complete chemical equation, which symbolically depicts the reactants, products, and their physical states.
The Law of Conservation of Mass states that in a chemical reaction, atoms can't be created or destroyed. This means that the total number of atoms for each element in the starting materials (reactants) must be the same as in the end products, keeping the overall mass the same.

Balanced chemical equation definition: A chemical equation is considered balanced when the number of atoms for each element on the reactant side is identical to the product side.

Steps for Balancing Chemical Equations

• To balance chemical equations, coefficients (numeric values preceding chemical symbols or formulas) are utilized. These coefficients represent the number of atoms or molecules involved. Adjust the coefficients as needed to make sure that the number of each type of atom is the same on both sides of the equation.
• For instance, in the equation Zn + HCl → ZnCl₂ + H₂, balancing involves adding coefficients to achieve equilibrium: Zn + 2HCl → ZnCl₂ + H₂. This balancing process often involves trial and error adjustments to ensure the equality of atoms on both sides.

Detailed Step-by-Step Example: Fe + H₂O → Fe₃O₄ + H₂

Step	Procedure&Reaction State
Step 1	Write the number of atoms of elements present in reactants and in products in a table.
Step 2	Balance the atom which is maximum in number on either side of a chemical equation. Start by balancing oxygen atoms (4 on RHS). Multiply H₂O on LHS by 4: Fe + 4 × H₂O → Fe₃O₄ + H₂
Step 3	Balancing oxygen creates an imbalance in hydrogen (8 on LHS, 2 on RHS). Multiply H₂ on RHS by 4: Fe + 4 × H₂O → Fe₃O₄ + 4 × H₂
Step 4	Balance iron atoms (1 on LHS, 3 on RHS). Multiply Fe on LHS by 3: 3 × Fe + 4 × H₂O → Fe₃O₄ + 4 × H₂ Balanced Equation: 3Fe + 4H₂O → Fe₃O₄ + 4H₂

Informative Notation and Reaction Conditions:
• Including physical states makes chemical equations more informative: (s) for solid, (l) for liquid, (g) for gas, and (aq) for aqueous solution.
• Reaction conditions (temperature, pressure, catalyst, sunlight, chlorophyll) are indicated above and/or below the arrow in the chemical equation.
• Example (Photosynthesis):
  6CO₂(g) + 6H₂O(l) —[Sunlight / Chlorophyll]→ C₆H₁₂O₆(aq) + 6O₂(g)

3. Types of Chemical Reactions

Chemical reactions are classified into several categories based on various factors:

1. Combination Reaction

A reaction in which a single product is formed from two or more reactants is known as a combination reaction.

Examples:
(i) Calcium oxide reacts vigorously with water to produce slaked lime (calcium hydroxide), releasing a large amount of heat:
    CaO(s) + H₂O(l) → Ca(OH)₂(aq) + Heat

(ii) White-washing of walls: A solution of slaked lime, Ca(OH)₂,reacts slowly with carbon dioxide in air to form a thin layer of calcium carbonate (CaCO₃) on walls, giving a shiny finish after 2–3 days. (Note: Marble also has the chemical formula CaCO₃):
    Ca(OH)₂(aq) + CO₂(g) → CaCO₃(s) + H₂O(l)

Exothermic Chemical Reactions: Reactions that release heat along with the formation of products are called exothermic chemical reactions.
Other Examples of Exothermic Reactions:
• Burning of Coal: C(s) + O₂(g) → CO₂(g)
• Formation of Water: 2H₂(g) + O₂(g) → 2H₂O(l)
• Burning of Natural Gas: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)
• Respiration: Digestion breaks carbohydrates (rice, potatoes, bread) into glucose. Glucose combines with oxygen in cells to yield energy:
  C₆H₁₂O₆(aq) + 6O₂(g) → 6CO₂(g) + 6H₂O(l) + Energy

2. Decomposition Reaction

A reaction in which a single reactant breaks down when exposed to heat, light, or electricity to give two or more simpler products is called a decomposition reaction.

There are three types of decomposition reactions:

(a) Thermal Decomposition

A decomposition reaction carried out by heating is called thermal decomposition.

Example 1: Heating of Ferrous Sulphate
Green ferrous sulphate crystals (FeSO₄·7H₂O) lose water when heated and turn white. On further heating, it decomposes into ferric oxide (Fe₂O₃ - solid), sulphur dioxide (SO₂ - gas), and sulphur trioxide (SO₃ - gas). The characteristic smell of burning sulphur is observed.
    2FeSO₄(s) —[Heat]→ Fe₂O₃(s) + SO₂(g) + SO₃(g)

Example 2: Decomposition of Calcium Carbonate
Heating calcium carbonate breaks it down into calcium oxide (quick lime) and carbon dioxide gas. Calcium oxide is widely used in cement manufacturing.
    CaCO₃(s) —[Heat]→ CaO(s) + CO₂(g)

Example 3: Heating of Lead Nitrate Powder
Heating lead nitrate powder produces brown fumes of nitrogen dioxide (NO₂) along with lead oxide and oxygen gas.
    2Pb(NO₃)₂(s) —[Heat]→ 2PbO(s) + 4NO₂(g) + O₂(g)

(b) Photolytic Decomposition

A decomposition reaction carried out in the presence of light energy.

Example 1: Decomposition of Hydrogen Peroxide
Hydrogen peroxide (H₂O₂) breaks down into water and oxygen gas when exposed to sunlight.

Example 2: Effect of Sunlight on Silver Chloride and Silver Bromide
White silver chloride turns grey in sunlight due to decomposition into silver metal and chlorine gas:
    2AgCl(s) —[Sunlight]→ 2Ag(s) + Cl₂(g)
Similarly, silver bromide decomposes:
    2AgBr(s) —[Sunlight]→ 2Ag(s) + Br₂(g)
Note: These reactions of silver halides are used in black and white photography.

(c) Electrolytic Decomposition

A decomposition reaction carried out by passing an electric current through an aqueous solution or molten state of a compound.
Example: Electrolysis of Water — Decomposition of water into hydrogen gas and oxygen gas by passing electric current through acidulated water:
    2H₂O(l) —[Electric Current]→ 2H₂(g) + O₂(g)

Endothermic Reactions: Decomposition reactions require energy in the form of heat, light, or electricity to break bonds. Reactions in which energy is absorbed are called endothermic reactions.

3. Displacement Reaction

A displacement reaction is a chemical reaction in which a more reactive element displaces a less reactive element from its salt solution. Both metals and non-metals take part in displacement reactions.

Important Reaction: Iron Nails in Copper Sulphate Solution
    Fe(s) + CuSO₄(aq) → FeSO₄(aq) + Cu(s)
Observation: The iron nail becomes brownish due to copper deposition, and the blue colour of the copper sulphate solution fades to pale green (FeSO₄).

Other Examples:
• Zinc displaces copper: Zn(s) + CuSO₄(aq) → ZnSO₄(aq) + Cu(s)
• Lead displaces copper: Pb(s) + CuCl₂(aq) → PbCl₂(aq) + Cu(s)
Zinc and lead are more reactive elements than copper.

4. Double Displacement Reaction

A double displacement reaction is a chemical reaction where two compounds react by exchanging their ions (cations and anions) to form two new compounds.

Example: Reaction of Sodium Sulphate and Barium Chloride
    Na₂SO₄(aq) + BaCl₂(aq) → BaSO₄(s)↓ + 2NaCl(aq)
Observation: A white insoluble precipitate of barium sulphate (BaSO₄) is formed.

Precipitation&Neutralisation Reactions:
• Any reaction that produces an insoluble precipitate is called a precipitation reaction.
• Neutralisation reactions (Acid + Base → Salt + Water) are also examples of double displacement reactions.

5. Endothermic and Exothermic Reactions Summary

(i) Exothermic Reaction: Energy is released into the surroundings, usually as heat. Example: C(s) + O₂(g) → CO₂(g) + Heat

(ii) Endothermic Reaction: Energy is absorbed from the surroundings in the form of heat, light, or electricity. Example (Photosynthesis): 6CO₂ + 6H₂O → C₆H₁₂O₆ + 6O₂

6. Oxidation and Reduction (Redox Reactions)

Oxidation and Reduction occur simultaneously in chemical reactions, known as redox reactions.

Definitions:
• Oxidation: Gain of oxygen OR loss of hydrogen / loss of electrons (increase in oxidation state).
• Reduction: Loss of oxygen OR gain of hydrogen / gain of electrons (decrease in oxidation state).

Oxidation of Copper to Copper Oxide:
When copper powder is heated in air, it turns black due to the formation of copper(II) oxide:
    2Cu + O₂ —[Heat]→ 2CuO
If hydrogen gas is passed over heated CuO, the black coating turns brown as the reverse reaction occurs:
    CuO + H₂ —[Heat]→ Cu + H₂O
Here, CuO loses oxygen (reduced) and H₂ gains oxygen (oxidized).

Other Examples of Redox Reactions:
• ZnO + C → Zn + CO (ZnO is reduced to Zn; C is oxidized to CO)
• MnO₂ + 4HCl → MnCl₂ + 2H₂O + Cl₂ (MnO₂ is reduced; HCl is oxidized to Cl₂)

Oxidizing Agent and Reducing Agent:
• Reducing Agent: The substance that gets oxidized (causes reduction of the other substance by donating electrons or adding hydrogen/removing oxygen).
• Oxidizing Agent: The substance that gets reduced (causes oxidation of the other substance by accepting electrons or supplying oxygen/removing hydrogen).
• Example: In Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g), Zinc is oxidized (reducing agent) and HCl is reduced (oxidizing agent).

4. Effects of Oxidation Reaction in Everyday Life

1. Corrosion

When a metal is attacked by substances around it such as moisture, air, oxygen, and acids, it is said to corrode, and this process is called corrosion. Corrosion causes flaking, structural weakening, and damage to bridges, iron railings, ships, and automobile bodies.

Chemical Equations of Corrosion:
• Rusting of Iron: 4Fe(s) + 3O₂(from air) + xH₂O(moisture) → 2Fe₂O₃·xH₂O (reddish-brown rust)
• Corrosion of Copper: Cu(s) + H₂O(moisture) + CO₂(air) → CuCO₃·Cu(OH)₂ (green coating)
• Corrosion of Silver: 2Ag(s) + H₂S(air) → Ag₂S(black coating) + H₂(g)

2. Rancidity

When fats and oils present in food materials are left exposed to air for a long time, they undergo oxidation, resulting in a change in taste and odour. This phenomenon is called rancidity.

Methods to Prevent Rancidity:

• Seal food materials in air-tight containers.

• Flush food packaging with inert gases like nitrogen (e.g., chips bags).

• Keep food in refrigerators to slow down oxidation.

• Add antioxidants or food preservatives.

Chapter Revision Questions&Solutions

REVISION QUESTION 1: Redox Roles in Zn + HCl Q: What roles do zinc and hydrochloric acid play in the redox reaction Zn(s) + 2HCl(aq) → ZnCl₂(aq) + H₂(g)? Solution: Zinc is oxidized to Zn²⁺ and acts as the REDUCING AGENT. Hydrochloric acid (H⁺) is reduced to H₂ gas and acts as the OXIDIZING AGENT.

REVISION QUESTION 2: Cause of Rancidity Q: What is the primary cause of rancidity in food containing fats and oils? Solution: The primary cause of rancidity is the oxidation of fats and oils when exposed to air over time.