Cheat Sheet: Electricity - Class 10 PDF Download

Introduction

• Electricity: A controllable, convenient form of energy powering homes, schools, hospitals, and industries, resulting from the movement of charged particles (electrons or ions).

• Historical Context: Originated from Greek words “Electrica” and “Elektron.” Thales observed materials’ attracting capacity when rubbed together.

• Conductors: Materials (e.g., copper, aluminum) allowing free flow of electric charge due to free electrons.

• Insulators: Materials (e.g., rubber, glass, plastic) resisting charge flow.

Electric Current and Circuit

• Current Electricity: Deals with moving charges.

• Electric Circuit: Continuous, closed path for electric current flow, including components like cell, switch, and conductor.

• Electric Current: Rate of flow of electric charge through a conductor’s cross-section, I = Q/t, where I is current (amperes, A), Q is charge (coulombs, C), t is time (seconds, s). 1 A = 1 C s⁻¹.

• Charge: 1 A corresponds to 6.25 × 10¹⁸ electrons per second (electron charge = -1.6 × 10⁻¹⁹ C).

• Direction: Conventional current is opposite to electron flow (positive charge flow direction).

• Example: For I = 0.75 A, t = 8 min = 480 s, Q = It = 0.75 × 480 = 360 C.

Electric Potential & Potential Difference

• Analogy: Charge flow in a circuit is like water flow in a pipe; current flows from higher to lower electric potential, similar to water from higher to lower pressure.

• Potential Difference: Work done to move a unit positive charge between two points, V = W/Q, where V is potential difference (volts, V), W is work (joules, J), Q is charge (coulombs, C). 1 V = 1 J C⁻¹.

• Source: Cell or battery maintains potential difference via chemical action between electrodes.

• Example: For Q = 3 C, V = 9 V, W = VQ = 9 × 3 = 27 J.

Circuit Diagram

• Purpose: Simplified representation of circuit components using standard symbols, showing their relative positions and connections for electricians.

• Components: Include cell, battery, switch (open/closed), wires, bulb, resistor, ammeter, voltmeter, rheostat.

Ohm’s Law

• Definition: At constant temperature, current (I) through a conductor is directly proportional to potential difference (V) across its ends: V ∝ I, or V = IR, where R is resistance (ohms, Ω). 1 Ω = 1 V/1 A.

• Resistance: Measure of opposition to charge flow; constant for a conductor at fixed temperature.

• Activity:

  • Objective: Study V vs. I relationship using a 0.5 m nichrome wire, ammeter, voltmeter, four 1.5 V cells, and circuit board.

  • Procedure: Connect circuit with one cell, measure I and V, repeat with 2, 3, 4 cells. Tabulate results. V/I is constant (R), and V-I graph is a straight line through origin.

  • Observation: V/I = R (constant resistance); doubling R halves I (I ∝ 1/R).

• Instruments:

  • Ammeter: Measures current in series, in amperes.

  • Voltmeter: Measures potential difference in parallel, in volts.

  • Rheostat: Variable resistor to regulate current without changing voltage source.

Factors Affecting Resistance of a Conductor

• Dependence: Resistance (R) depends on:

  • Length (L): R ∝ L.

  • Cross-sectional area (A): R ∝ 1/A.

  • Material’s nature (resistivity, ρ).

• Formula: R = ρL/A, where ρ is resistivity (Ω m), specific to material and temperature.

• Resistivity:

  • Defined as resistance of a 1 m³ cube with current perpendicular to opposite faces.

  • Metals/alloys: 10⁻⁸ to 10⁻⁶ Ω m (good conductors, e.g., copper, aluminum for transmission lines).

  • Insulators: 10¹² to 10¹⁷ Ω m (e.g., rubber, glass).

  • Alloys (e.g., nichrome): Higher ρ, resist oxidation, minimal ρ change with temperature, used in heaters, irons.

  • Tungsten: High melting point, used in bulb filaments.

• Temperature Effect: For metals, R and ρ increase with temperature.

• Examples:

  • Wire (R = 30 Ω, L = 1.5 m, d = 0.4 mm): ρ = RA/L = 30 × π(0.2 × 10⁻³)² / 1.5 ≈ 2.51 × 10⁻⁶ Ω m.

  • Wire (R = 6 Ω, L, A): New wire (L/3, 3A), R₂ = ρ(L/3)/(3A) = (1/9)R = 1 Ω.

Resistance of System of Resistors

Series Arrangement

• Characteristics:

  • Same current through all resistors.

  • Total voltage = sum of individual voltage drops (V = V₁ + V₂ + V₃ + ...).

  • Equivalent resistance: Rₛ = R₁ + R₂ + R₃ + ...

• Disadvantages: If one component fails, the circuit breaks; all components get the same current, unsuitable for devices with different current needs.

• Example: Lamp (20 Ω) and conductor (4 Ω) with 6 V battery:

  • Total resistance: Rₛ = 20 + 4 = 24 Ω.

  • Current: I = V/R = 6/24 = 0.25 A.

  • Voltage across lamp: V₁ = 0.25 × 20 = 5 V.

  • Voltage across conductor: V₂ = 0.25 × 4 = 1 V.

  • Power of lamp: P = V₁I = 5 × 0.25 = 1.25 W.

Parallel Arrangement

• Characteristics:

  • Same voltage across all resistors.

  • Total current = sum of currents through each resistor (I = I₁ + I₂ + I₃ + ...).

  • Equivalent resistance: 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃ + ...

• Advantages: Each device draws required current; one failure doesn’t affect others; lower total resistance allows stronger current. Used in household circuits.

• Note: Household circuits use parallel connections for independent operation and reduced total resistance.

Heating Effect of Electric Current

• Joule’s Law: Heat produced in a conductor with resistance R, current I, time t: H = I²Rt (for series, same I) or H = V²t/R (for parallel, same V).

• Mechanism: Electrical energy dissipates as heat in resistive circuits.

• Example: Heater (P = 900 W, V = 240 V): I = P/V = 900/240 = 3.75 A, R = V/I = 240/3.75 = 64 Ω. At P = 400 W: I = 400/240 ≈ 1.67 A, R = 240/1.67 ≈ 143.71 Ω.

Practical Applications of Heating Effect of Electric Current

• Electric Bulbs: Tungsten filaments (high melting point) produce light at high temperatures; bulbs filled with inert gases (nitrogen, argon) to extend filament life.

• Electric Fuse: Series-connected wire of low-melting-point metal/alloy melts at high current, breaking circuit to protect appliances.

Electric Power

• Definition: Rate of doing electric work, P = VI = I²R = V²/R. Unit: watt (W), 1 W = 1 V × 1 A.

• Energy Units: Watt-hour (Wh); kilowatt-hour (kWh) = 1000 Wh = 3.6 × 10⁶ J (commercial unit).

Exercises

1. Charge Flow: I = 0.75 A, t = 8 min = 480 s, Q = 0.75 × 480 = 360 C.

2. Work Done: Q = 3 C, V = 9 V, W = 9 × 3 = 27 J.

3. Resistivity: Wire (R = 30 Ω, L = 1.5 m, d = 0.4 mm): ρ = 30 × π(0.2 × 10⁻³)² / 1.5 ≈ 2.51 × 10⁻⁶ Ω m.

4. Resistance Change: Wire (R = 6 Ω, L, A): New wire (L/3, 3A), R₂ = (1/9) × 6 = 1 Ω.

5. Series Circuit: Lamp (20 Ω), conductor (4 Ω), V = 6 V:

   • Rₛ = 20 + 4 = 24 Ω.

   • I = 6/24 = 0.25 A.

   • V₁ (lamp) = 0.25 × 20 = 5 V, V₂ (conductor) = 0.25 × 4 = 1 V.

   • P (lamp) = 5 × 0.25 = 1.25 W.

6. Heater: P = 900 W, V = 240 V: I = 900/240 = 3.75 A, R = 240/3.75 = 64 Ω. P = 400 W: I = 400/240 ≈ 1.67 A, R = 240/1.67 ≈ 143.71 Ω.

Summary Points

• Electricity is energy from moving charges (electrons/ions), generated from fossil fuels, nuclear, or renewables (FAQ 1).

• Current (I = Q/t) flows from negative to positive terminal through conductors (e.g., copper, aluminum) due to potential difference, following least resistance path (FAQ 2).

• Conductors (low resistance, e.g., metals) allow easy charge flow; insulators (high resistance, e.g., rubber, glass) resist flow (FAQ 3).

• Resistance (R = V/I) opposes current flow, depends on length (R ∝ L), area (R ∝ 1/A), and material (FAQ 4).

• Ohm’s law: V = IR; resistance constant at fixed temperature.

• Resistivity (ρ): Metals (10⁻⁸ to 10⁻⁶ Ω m), insulators (10¹² to 10¹⁷ Ω m); alloys (higher ρ) used in appliances, tungsten in bulbs.

• Series: Rₛ = R₁ + R₂ + ..., same current, higher resistance (FAQ 6).

• Parallel: 1/Rₚ = 1/R₁ + 1/R₂ + ..., same voltage, lower resistance, used in households.

• Joule’s heating: H = I²Rt or V²t/R; used in bulbs, fuses (FAQ 8).

• Power: P = VI, energy in kWh (1 kWh = 3.6 × 10⁶ J).

• Safety hazards: Shocks, fires from faulty wiring, wet hands (lower body resistance to ~