Electricity
Chapter 11 • Complete Study Notes, Important Formulae, Activities and Solved Numericals
Electricity plays a crucial role in our modern world, powering everything from homes and schools to hospitals and industries. In this chapter, we explore electric circuits, the factors that influence the flow of current, the heating effect of electric current and its 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: Substances in which electric charge can flow freely. Metals are conductors. Copper and aluminium are commonly used in electricity.
- Insulators: Materials in which electric charges do not move freely. Rubber, glass and plastic are examples.
- Current electricity: Deals with moving charges.
- Electric circuit: A continuous and closed path of an electric current.
Here, I = electric current, Q = charge flown and t = time.
Flow of Electric Current and Units
- The SI unit of electric current is ampere (A).
- One ampere is the flow of one coulomb of charge per second: 1 A = 1 C s⁻¹.
- The unit is named after the French scientist André-Marie Ampère.
- Conventional current is taken opposite to the direction of flow of negatively charged electrons. It is along the direction in which positive charges would move.
- In conductors, electric current is due to the flow of electrons. The flow of 6.25 × 10¹⁸ electrons per second across a cross-section constitutes a current of 1 ampere.
📌 Solved Example 1: Charge Calculation
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.
Given: I = 0.75 A; t = 8 min = 480 s
Formula: Q = I × t
Calculation: Q = 0.75 × 480 = 360 C
2. Electric Potential and 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.
- Just as water flows from higher pressure (level) to lower pressure (level), 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.
SI unit: volt (V), where 1 V = 1 J / 1 C = 1 J C⁻¹.
📌 Solved Example 2: Work Done Calculation
Question: How much work is done in moving a charge of 3 C between two points with a potential difference of 9 V?
Given: Q = 3 C; V = 9 V
Formula: W = V × Q
Calculation: W = 9 × 3 = 27 J
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 or engineer.
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.
- 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 is a measure of the opposition offered by a conductor to the flow of electric charge through it.
- The SI unit of electrical resistance is ohm (Ω), where 1 Ω = 1 V / 1 A.
Activity: Verifying 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: Nichrome wire (0.5 m), ammeter, voltmeter, four 1.5 V cells, connecting wires, switch and circuit board.
- Procedure: Connect the nichrome wire in series with an ammeter and in parallel with a voltmeter using one 1.5 V cell. Record current and potential difference. Increase the cells to 2, 3 and 4 successively, recording I and V each time. Plot V against I.
- Conclusion: The ratio V/I remains constant in each case. The V-I graph is a straight line passing through the origin, proving V ∝ I.
5. Factors Affecting Resistance of a Conductor
At a given temperature, the resistance of a conductor depends on:
- Length (L): R ∝ L
- Area of cross-section (A): R ∝ 1/A
- Nature of the material of the conductor.
Here, ρ (rho) is the resistivity of the material.
- Resistivity is the resistance offered by a cube of that material of side 1 m when current flows perpendicular to opposite faces.
- Unit of resistivity: ohm-metre (Ω·m).
- Its value depends only on the nature of the material and temperature, and is independent of dimensions.
- Metals and alloys have low resistivity (10⁻⁸ to 10⁻⁶ Ω·m) and are good conductors. Insulators such as rubber and glass have resistivity of 10¹² to 10¹⁷ Ω·m.
- Pure metals have very low resistivity. Copper and aluminium are used for transmission lines.
- Alloys have higher resistivity and do not oxidise easily. Nichrome is used in heaters, irons and geysers.
- Tungsten is used for lamp filaments due to its high melting point.
Rheostat
A variable resistance device used to regulate current without changing the voltage source.
Ammeter
An instrument connected in series to measure current.
Voltmeter
An instrument connected in parallel to measure potential difference.
📌 Solved Example 3: Resistivity Calculation
Question: The resistance of a metal wire of length 1.5 m is 30 Ω at 25°C. If its diameter is 0.4 mm, find resistivity at 25°C.
Given: R = 30 Ω; d = 0.4 mm = 4 × 10⁻⁴ m; l = 1.5 m
Formula: ρ = R × A / l = R × (πd²/4) / l
Calculation: ρ = (30 × 3.1416 × (4 × 10⁻⁴)²) / (4 × 1.5) = 2.51 × 10⁻⁶ Ω·m
📌 Solved Example 4: Resized Wire Resistance
Question: A wire of length l and cross-sectional area A has resistance 6 Ω. What is the resistance of another wire of the same material with length l/3 and cross-sectional area 3A?
For wire 1: R₁ = ρ(l/A) = 6 Ω
For wire 2: R₂ = ρ((l/3)/(3A)) = ρ(l/9A) = (1/9) × R₁
R₂ = (1/9) × 6 Ω = 0.67 Ω (approximately).
6. Resistance of System of Resistors
Resistors in an electric circuit can be connected in series or parallel arrangement.
Series Combination
- Current flowing through all resistors is the same.
- Total voltage equals the sum of voltage drops: V = V₁ + V₂ + V₃.
- Total equivalent resistance:
Parallel Combination
- Voltage across each resistor is the same and equals the total applied voltage.
- Current divides among resistors inversely proportional to resistance. I = I₁ + I₂ + I₃.
- Reciprocal of equivalent resistance:
For two resistors:
Why Parallel Connection is Preferred in Household Circuits
- In series, if one component fails, the circuit breaks and all components stop working.
- In series, the same current flows through all appliances regardless of their requirement.
- Parallel circuits divide current so each appliance draws the required current with separate switches, and total circuit resistance decreases.
📌 Solved Example 5: Series Circuit and Power Analysis
Question: An electric lamp of resistance 20 Ω and a conductor of 4 Ω are connected in series to a 6 V battery. Calculate (a) total resistance, (b) circuit current, (c) potential difference across the lamp and conductor, and (d) power of the lamp.
Given: R₁ = 20 Ω; R₂ = 4 Ω; V = 6 V
- Total resistance: R = R₁ + R₂ = 20 + 4 = 24 Ω.
- Current: I = V/R = 6/24 = 0.25 A.
- Voltage across lamp: V₁ = I × R₁ = 0.25 × 20 = 5 V.
- Voltage across conductor: V₂ = I × R₂ = 0.25 × 4 = 1 V.
- Power of lamp: P = V₁ × I = 5 × 0.25 = 1.25 W.
7. Heating Effect of Electric Current
According to Joule’s law of heating, when current I flows through resistance R for time t under voltage V, the electrical energy consumed or heat produced is:
Unit: Joule (J).
- Formula H = I²Rt is preferred for series circuits (constant current I).
- Formula H = (V²t)/R is preferred for parallel circuits (constant voltage V).
📌 Solved Example 6: Electric Heater Power and Resistance
Question: An electric heater uses 900 W at maximum heat and 400 W at minimum heat, operating at 240 V. Calculate current and resistance in each case.
Formulae: I = P/V and R = V/I
(a) Maximum heating: P = 900 W, V = 240 V
- Current I = 900/240 = 3.75 A
- Resistance R = 240/3.75 = 64 Ω
(b) Minimum heating: P = 400 W, V = 240 V
- Current I = 400/240 = 1.67 A
- Resistance R = 240/1.67 ≈ 144 Ω
Practical Applications of Heating Effect
- Electric lighting – Electric bulb: The filament is made of high melting-point tungsten. It is filled with inert nitrogen and argon gas to prolong filament life.
- Electric fuse: A safety device placed in series. It consists of a wire with an appropriate low melting point. Excess current heats the wire beyond its melting point, breaking the circuit.
8. Electric Power
- SI unit: Watt (W), where 1 watt = 1 volt × 1 ampere = 1 VA.
- Commercial unit of electrical energy: Kilowatt-hour (kWh).
- Conversion: 1 kWh = 1000 W × 3600 s = 3.6 × 10⁶ J.
⚡ Quick Formula Revision
| Quantity / Concept | Formula | SI Unit |
|---|---|---|
| Electric current | I = Q/t | Ampere (A) |
| Electric charge | Q = I × t | Coulomb (C) |
| Potential difference | V = W/Q | Volt (V) |
| Ohm’s law | V = I × R | — |
| Resistance | R = V/I | Ohm (Ω) |
| Resistivity relation | R = ρL/A | Ω·m for ρ |
| Series resistance | Rs = R₁ + R₂ + ... | Ohm (Ω) |
| Parallel resistance | 1/Rp = 1/R₁ + 1/R₂ + ... | Ohm (Ω) |
| Heat produced | H = I²Rt | Joule (J) |
| Electric power | P = VI = I²R = V²/R | Watt (W) |
| Electrical energy | E = P × t | Joule (J) or kWh |
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