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