Electricity

Introduction

• Electricity is a controllable, convenient energy form vital for modern society, powering homes, schools, hospitals, and industries.
• This chapter explores electric current, circuits, factors regulating current, and the heating effect of current with its applications.
• Understanding electricity’s principles is key to its safe and effective use in daily life.

11.1 Electric Current and Circuit

• Electric Current: Flow of electric charge through a conductor (e.g., metallic wire), analogous to water current in rivers.
• Electric Circuit: Continuous, closed path for current flow, including components like cells, switches, and bulbs (e.g., torch circuit).
• Switch: Connects or breaks circuit, controlling current flow (e.g., torch bulb glows only when switch is on).
• Charge Flow:
  • Electrons (negative charges) constitute current in metallic wires.
  • Conventionally, current direction is opposite to electron flow (positive charge flow direction).
• Current Definition:
  • Rate of charge flow: \( I = \frac{Q}{t} \), where \( I \) is current, \( Q \) is charge, \( t \) is time.
  • SI unit: Ampere (A); 1 A = 1 C/s (coulomb per second).
  • 1 C ≈ charge of \( 6 \times 10^{18} \) electrons (electron charge = \( 1.6 \times 10^{-19} \) C).
  • Small currents: milliampere (mA, \( 10^{-3} \) A), microampere (µA, \( 10^{-6} \) A).
• Measurement: Ammeter measures current, connected in series (Fig. 11.1).
• Example 11.1:
  • I = 0.5 A, t = 10 min = 600 s; \( Q = I t = 0.5 \times 600 = 300 \, \text{C} \).
• Current is the backbone of electrical systems, driven by charge movement in circuits.

11.2 Electric Potential and Potential Difference

• Potential Difference (V): Drives charge flow, analogous to pressure difference driving water flow.
• Source: Battery/cell creates potential difference via chemical reactions, even without current.
• Definition:
  • Work done per unit charge: \( V = \frac{W}{Q} \).
  • SI unit: Volt (V); 1 V = 1 J/C (1 joule per coulomb).
• Measurement: Voltmeter measures potential difference, connected in parallel.
• Example 11.2:
  • V = 12 V, Q = 2 C; \( W = V Q = 12 \times 2 = 24 \, \text{J} \).
• Potential difference is the energy push behind current flow.

11.3 Circuit Diagram

• Purpose: Schematic diagrams represent circuits using standard symbols for components (Table 11.1).
• Common Symbols:

  Component	Symbol
  Electric cell	Long/short line pair
  Battery	Multiple cell symbols
  Plug key (open)	Break in line
  Plug key (closed)	Connected line
  Wire joint	Dot at intersection
  Wires crossing	Lines without dot
  Electric bulb	Circle with filament
  Resistor	Zigzag or rectangle
  Variable resistor	Resistor with arrow
  Ammeter	Circle with A
  Voltmeter	Circle with V

• Circuit diagrams simplify complex electrical systems for analysis.

11.4 Ohm’s Law

• Definition: Potential difference (V) across a conductor is directly proportional to current (I) at constant temperature: \( V \propto I \), or \( V = I R \), where \( R \) is resistance.
• Resistance (R): Property opposing charge flow; SI unit: Ohm (Ω); 1 Ω = 1 V/A.
• Activity 11.1:
  • Circuit with nichrome wire, ammeter, voltmeter, and 1–4 cells (1.5 V each) (Fig. 11.2).
  • Measure I and V; calculate \( V/I \); plot V vs. I (straight line through origin, Fig. 11.3).
  • Constant \( V/I = R \), confirming Ohm’s law.
• Variable Resistance: Rheostat adjusts resistance to control current without changing voltage.
• Activity 11.2:
  • Test current through nichrome wire, torch bulb, 10 W bulb with 4×1.5 V cells.
  • Different components yield different currents, indicating varying resistances.
• Conductors vs. Resistors:
  • Good conductors (low R): Easy electron flow.
  • Resistors: Appreciable resistance.
  • Poor conductors/insulators (high R): High resistance to electron flow.
• Ohm’s law governs current-voltage relationships, critical for circuit design.

11.5 Factors on Which the Resistance of a Conductor Depends

• Activity 11.3:
  • Circuit with nichrome wire, ammeter, cell (Fig. 11.5).
  • Test: (1) Nichrome wire length \( l \); (2) Length 2\( l \); (3) Thicker wire, length \( l \); (4) Copper wire, same length/area as (1).
  • Observations: Current halves with doubled length; increases with thicker wire; changes with material.
• Factors:
  • Resistance \( R \propto l \) (length).
  • Resistance \( R \propto \frac{1}{A} \) (area of cross-section).
  • Material-dependent via resistivity (\( \rho \)).
• Formula: \( R = \rho \frac{l}{A} \), where \( \rho \) is resistivity (Ω·m), characteristic of material.
• Resistivity:
  • Conductors (e.g., metals): \( 10^{-8} \) to \( 10^{-6} \) Ω·m.
  • Alloys: Higher than metals (e.g., nichrome: \( 100 \times 10^{-6} \) Ω·m).
  • Insulators (e.g., glass): \( 10^{10} \) to \( 10^{17} \) Ω·m.
  • Varies with temperature.
• Table 11.2: Resistivity at 20°C:

  Material	Resistivity (Ω·m)
  Silver	\( 1.60 \times 10^{-8} \)
  Copper	\( 1.62 \times 10^{-8} \)
  Aluminium	\( 2.63 \times 10^{-8} \)
  Tungsten	\( 5.20 \times 10^{-8} \)
  Nickel	\( 6.84 \times 10^{-8} \)
  Iron	\( 10.0 \times 10^{-8} \)
  Chromium	\( 12.9 \times 10^{-8} \)
  Mercury	\( 94.0 \times 10^{-8} \)
  Manganese	\( 1.84 \times 10^{-6} \)
  Constantan	\( 49 \times 10^{-6} \)
  Manganin	\( 44 \times 10^{-6} \)
  Nichrome	\( 100 \times 10^{-6} \)
  Glass	\( 10^{10} - 10^{14} \)
  Hard rubber	\( 10^{13} - 10^{16} \)
  Ebonite	\( 10^{15} - 10^{17} \)
  Diamond	\( 10^{12} - 10^{13} \)
  Paper (dry)	\( 10^{12} \)

• Applications:
  • Alloys (e.g., nichrome): High resistivity, oxidation resistance for heaters, toasters.
  • Tungsten: High melting point for bulb filaments.
  • Copper/Aluminium: Low resistivity for transmission lines.
• Example 11.3:
  • (a) Bulb: V = 220 V, R = 1200 Ω; \( I = \frac{V}{R} = \frac{220}{1200} = 0.18 \, \text{A} \).
  • (b) Heater: V = 220 V, R = 100 Ω; \( I = \frac{220}{100} = 2.2 \, \text{A} \).
• Example 11.4:
  • V = 60 V, I = 4 A; \( R = \frac{V}{I} = \frac{60}{4} = 15 \, \Omega \).
  • New V = 120 V; \( I = \frac{V}{R} = \frac{120}{15} = 8 \, \text{A} \).
• Example 11.5:
  • R = 26 Ω, l = 1 m, d = 0.3 mm = \( 3 \times 10^{-4} \) m; \( A = \frac{\pi d^2}{4} \).
  • \( \rho = \frac{R A}{l} = \frac{26 \cdot \pi (3 \times 10^{-4})^2}{4 \cdot 1} = 1.84 \times 10^{-6} \, \Omega \cdot \text{m} \).
  • Material: Manganese (Table 11.2).
• Example 11.6:
  • First wire: \( R_1 = \rho \frac{l}{A} = 4 \, \Omega \).
  • Second wire: \( R_2 = \rho \frac{l/2}{2A} = \frac{1}{4} \cdot 4 = 1 \, \Omega \).
• Resistance tailoring via length, area, and material is crucial for circuit functionality.

11.6 Resistance of a System of Resistors

• Combinations: Series and parallel resistor arrangements affect total resistance and current flow.

11.6.1 Resistors in Series

• Characteristics (Fig. 11.6):
  • Same current through all resistors.
  • Total potential difference is sum of individual drops: \( V = V_1 + V_2 + V_3 \).
• Activity 11.4:
  • Series circuit with resistors (e.g., 1 Ω, 2 Ω, 3 Ω), 6 V battery.
  • Ammeter shows constant current regardless of position.
• Activity 11.5:
  • Measure total V and individual \( V_1, V_2, V_3 \); confirm \( V = V_1 + V_2 + V_3 \).
• Equivalent Resistance:
  • \( V = I R_s \), \( V_1 = I R_1 \), etc.; thus \( R_s = R_1 + R_2 + R_3 \).
• Example 11.7:
  • Lamp (20 Ω), conductor (4 Ω), 6 V battery.
  • (a) \( R_s = 20 + 4 = 24 \, \Omega \).
  • (b) \( I = \frac{V}{R_s} = \frac{6}{24} = 0.25 \, \text{A} \).
  • (c) Lamp: \( V_1 = I R_1 = 0.25 \times 20 = 5 \, \text{V} \); Conductor: \( V_2 = 0.25 \times 4 = 1 \, \text{V} \).
• Series circuits have additive resistance, suitable for uniform current needs.

11.6.2 Resistors in Parallel

• Characteristics (Fig. 11.7):
  • Same potential difference across all resistors.
  • Total current is sum of branch currents: \( I = I_1 + I_2 + I_3 \).
• Activity 11.6:
  • Parallel circuit with resistors \( R_1, R_2, R_3 \), battery, ammeter, voltmeter (Fig. 11.10).
  • Confirm same V across each resistor; measure branch currents \( I_1, I_2, I_3 \).
  • Total current: \( I = I_1 + I_2 + I_3 \).
• Equivalent Resistance:
  • \( I = \frac{V}{R_p} \), \( I_1 = \frac{V}{R_1} \), etc.; thus \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \).
• Example 11.8:
  • \( R_1 = 5 \, \Omega, R_2 = 10 \, \Omega, R_3 = 30 \, \Omega \), V = 12 V.
  • (a) Currents: \( I_1 = \frac{12}{5} = 2.4 \, \text{A} \), \( I_2 = \frac{12}{10} = 1.2 \, \text{A} \), \( I_3 = \frac{12}{30} = 0.4 \, \text{A} \).
  • (b) Total current: \( I = 2.4 + 1.2 + 0.4 = 4 \, \text{A} \).
  • (c) \( \frac{1}{R_p} = \frac{1}{5} + \frac{1}{10} + \frac{1}{30} = \frac{1}{3} \); \( R_p = 3 \, \Omega \).
• Example 11.9:
  • \( R_1 = 10 \, \Omega, R_2 = 40 \, \Omega, R_3 = 30 \, \Omega, R_4 = 20 \, \Omega, R_5 = 60 \, \Omega \), V = 12 V.
  • \( \frac{1}{R'} = \frac{1}{10} + \frac{1}{40} = \frac{5}{40} \); \( R' = 8 \, \Omega \).
  • \( \frac{1}{R''} = \frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{6}{60} \); \( R'' = 10 \, \Omega \).
  • (a) Total resistance: \( R = R' + R'' = 8 + 10 = 18 \, \Omega \).
  • (b) Current: \( I = \frac{12}{18} = 0.67 \, \text{A} \).
• Advantages of Parallel:
  • Each device gets same voltage, suitable for different current needs.
  • Failure of one device doesn’t break circuit.
  • Lower total resistance increases current capacity.
• Parallel circuits are practical for household wiring due to flexibility and reliability.

11.7 Heating Effect of Electric Current

• Principle: Energy from battery dissipates as heat in resistive circuits (Joule’s heating).
• Energy:
  • Work to move charge Q across V: \( W = V Q \).
  • Power: \( P = \frac{W}{t} = V I \).
  • Heat energy: \( H = V I t \).
  • Using Ohm’s law (\( V = I R \)): \( H = I^2 R t \) (Joule’s law).
• Joule’s Law:
  • Heat \( H \propto I^2 \) (given R).
  • Heat \( H \propto R \) (given I).
  • Heat \( H \propto t \).
• Example 11.10:
  • Electric iron: V = 220 V, P = 840 W (max), 360 W (min).
  • Max: \( I = \frac{P}{V} = \frac{840}{220} = 3.82 \, \text{A} \); \( R = \frac{V}{I} = \frac{220}{3.82} = 57.60 \, \Omega \).
  • Min: \( I = \frac{360}{220} = 1.64 \, \text{A} \); \( R = \frac{220}{1.64} = 134.15 \, \Omega \).
• Example 11.11:
  • H = 100 J, R = 4 Ω, t = 1 s; \( I = \sqrt{\frac{H}{R t}} = \sqrt{\frac{100}{4 \cdot 1}} = 5 \, \text{A} \).
  • \( V = I R = 5 \times 4 = 20 \, \text{V} \).
• Heating effect is both a challenge and a useful phenomenon in electrical devices.

11.7.1 Practical Applications of Heating Effect

• Challenges: Unwanted heating in circuits increases component temperature, altering properties.
• Applications:
  • Heaters: Electric irons, toasters, ovens, kettles use resistive heating.
  • Electric Bulbs: Tungsten filament (high melting point, 3380°C) heats to emit light; filled with inert gases (nitrogen, argon) to prolong life.
  • Fuses: Low-melting-point alloy wire in series melts at high current, breaking circuit to protect devices (e.g., 5 A fuse for 1 kW iron at 220 V, drawing 4.54 A).
• Joule’s heating powers essential household devices and safety mechanisms.

11.8 Electric Power

• Definition: Rate of energy consumption: \( P = V I \).
• Using Ohm’s Law: \( P = I^2 R = \frac{V^2}{R} \).
• Unit: Watt (W); 1 W = 1 V × 1 A.
• Larger Unit: Kilowatt (kW) = 1000 W.
• Energy Unit:
  • Watt-hour (Wh): Energy for 1 W used for 1 hour.
  • Kilowatt-hour (kWh): \( 1 \, \text{kWh} = 1000 \, \text{W} \times 3600 \, \text{s} = 3.6 \times 10^6 \, \text{J} \).
• Note: Electrons are not consumed; energy is paid for to move electrons through devices.
• Example 11.12:
  • Bulb: V = 220 V, I = 0.50 A; \( P = V I = 220 \times 0.50 = 110 \, \text{W} \).
• Example 11.13:
  • Refrigerator: 400 W, 8 h/day, 30 days.
  • Energy: \( 400 \times 8 \times 30 = 96000 \, \text{Wh} = 96 \, \text{kWh} \).
  • Cost: \( 96 \times 3.00 = \text{Rs } 288 \).
• Power and energy calculations are essential for electrical billing and efficiency.

Key Questions and Answers

1. Electric Circuit:
   • Continuous, closed path for current flow, including cell, switch, and components.
2. Unit of Current:
   • Ampere (A); 1 A = 1 C/s.
3. Electrons per Coulomb:
   • Charge of electron = \( 1.6 \times 10^{-19} \, \text{C} \); Electrons per C = \( \frac{1}{1.6 \times 10^{-19}} \approx 6.25 \times 10^{18} \).
4. Device for Potential Difference:
   • Battery or cell.
5. Potential Difference of 1 V:
   • 1 J of work done to move 1 C of charge.
6. Energy per Coulomb (6 V Battery):
   • \( V = \frac{W}{Q} \); \( W = V Q = 6 \, \text{J} \) for 1 C.
7. Resistance Factors:
   • Length (\( R \propto l \)), area (\( R \propto \frac{1}{A} \)), material (via \( \rho \)), temperature.
8. Thick vs. Thin Wire:
   • Thick wire (larger A) has lower R, allowing more current.
9. Halved Potential Difference:
   • \( I = \frac{V}{R} \); if V halves, I halves (R constant).
10. Alloy vs. Pure Metal:
    • Alloys have higher resistivity, resist oxidation at high temperatures, suitable for heating elements.
11. Table 11.2 Analysis:
    • (a) Iron (\( 10.0 \times 10^{-8} \)) vs. Mercury (\( 94.0 \times 10^{-8} \)): Iron is better conductor (lower \( \rho \)).
    • (b) Silver (\( 1.60 \times 10^{-8} \)) is best conductor (lowest \( \rho \)).
12. Heater Cord vs. Element:
    • Cord (low R, e.g., copper) produces less heat (\( H \propto R \)); element (high R, e.g., nichrome) glows due to high heat.
13. Heat Generated:
    • Q = 96000 C, V = 50 V, t = 1 h = 3600 s.
    • \( H = V Q = 50 \times 96000 = 4.8 \times 10^6 \, \text{J} \).
14. Heat in Iron:
    • R = 20 Ω, I = 5 A, t = 30 s; \( H = I^2 R t = 5^2 \times 20 \times 30 = 15000 \, \text{J} \).
15. Rate of Energy Delivery:
    • Power (\( P = V I \)) determines energy delivery rate.
16. Motor Power and Energy:
    • I = 5 A, V = 220 V; \( P = V I = 220 \times 5 = 1100 \, \text{W} \).
    • t = 2 h; Energy = \( P \times t = 1100 \times 2 \times 3600 = 7.92 \times 10^6 \, \text{J} \).

Exercises

1. Wire Cut into Five Parts:
   • R per part = \( \frac{R}{5} \). Parallel: \( \frac{1}{R'} = 5 \cdot \frac{5}{R} = \frac{25}{R} \); \( R' = \frac{R}{25} \).
   • Ratio: \( \frac{R}{R'} = 25 \). Answer: (d).
2. Not Electrical Power:
   • Power: \( P = V I, I^2 R, \frac{V^2}{R} \). \( I R^2 \) is not power. Answer: (b).
3. Bulb Power at 110 V:
   • 220 V, 100 W; \( R = \frac{V^2}{P} = \frac{220^2}{100} = 484 \, \Omega \).
   • At 110 V: \( P = \frac{V^2}{R} = \frac{110^2}{484} = 25 \, \text{W} \). Answer: (d).
4. Heat in Series vs. Parallel:
   • Series: \( R_s = 2R \), \( H_s = \frac{V^2}{2R} t \). Parallel: \( R_p = \frac{R}{2} \), \( H_p = \frac{V^2}{\frac{R}{2}} t = \frac{2V^2}{R} t \).
   • Ratio: \( \frac{H_s}{H_p} = \frac{1}{4} \). Answer: (c).
5. Voltmeter Connection:
   • Connected in parallel across the two points.
6. Copper Wire Resistance:
   • \( \rho = 1.6 \times 10^{-8} \, \Omega \cdot \text{m} \), d = 0.5 mm = \( 5 \times 10^{-4} \, \text{m} \), R = 10 Ω.
   • \( A = \frac{\pi (5 \times 10^{-4})^2}{4} \approx 1.963 \times 10^{-7} \, \text{m}^2 \).
   • \( l = \frac{R A}{\rho} = \frac{10 \cdot 1.963 \times 10^{-7}}{1.6 \times 10^{-8}} \approx 122.7 \, \text{m} \).
   • Doubled d: \( A' = 4A \), \( R' = \frac{\rho l}{4A} = \frac{R}{4} = 2.5 \, \Omega \).
7. V-I Graph and Resistance:
   • Plot V vs. I; slope = R. Using points (1.0 A, 3.4 V), (4.0 A, 13.2 V):
   • Slope = \( \frac{13.2 - 3.4}{4.0 - 1.0} = \frac{9.8}{3} \approx 3.27 \, \Omega \).
8. Unknown Resistor:
   • V = 12 V, I = 2.5 mA = \( 2.5 \times 10^{-3} \, \text{A} \).
   • \( R = \frac{V}{I} = \frac{12}{2.5 \times 10^{-3}} = 4800 \, \Omega = 4.8 \, \text{k}\Omega \).
9. Series Circuit Current:
   • R = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω, V = 9 V.
   • \( I = \frac{V}{R} = \frac{9}{13.4} \approx 0.67 \, \text{A} \) (same through 12 Ω).
10. Parallel 176 Ω Resistors:
    • V = 220 V, I = 5 A; \( R_p = \frac{V}{I} = \frac{220}{5} = 44 \, \Omega \).
    • Each R = 176 Ω; \( \frac{1}{R_p} = \frac{n}{176} \); \( n = \frac{176}{44} = 4 \).
11. 6 Ω Resistors:
    • (i) 9 Ω: Two in parallel (\( \frac{6 \cdot 6}{6 + 6} = 3 \, \Omega \)), third in series (3 + 6 = 9 Ω).
    • (ii) 4 Ω: Two in series (6 + 6 = 12 Ω), parallel with third (\( \frac{12 \cdot 6}{12 + 6} = 4 \, \Omega \)).
12. Bulbs in Parallel:
    • Bulb: 10 W, 220 V; \( I = \frac{P}{V} = \frac{10}{220} \approx 0.0455 \, \text{A} \).
    • Total I = 5 A; \( n = \frac{5}{0.0455} \approx 110 \).
13. Hot Plate Currents:
    • Each coil: 24 Ω, V = 220 V.
    • Separate: \( I = \frac{220}{24} \approx 9.17 \, \text{A} \).
    • Series: \( R = 24 + 24 = 48 \, \Omega \); \( I = \frac{220}{48} \approx 4.58 \, \text{A} \).
    • Parallel: \( \frac{1}{R} = \frac{2}{24} \); \( R = 12 \, \Omega \); \( I = \frac{220}{12} \approx 18.33 \, \text{A} \).
14. Power in 2 Ω Resistor:
    • (i) Series: 1 Ω + 2 Ω = 3 Ω, V = 6 V; \( I = \frac{6}{3} = 2 \, \text{A} \); \( P = I^2 R = 2^2 \times 2 = 8 \, \text{W} \).
    • (ii) Parallel: V = 4 V across 2 Ω; \( P = \frac{V^2}{R} = \frac{4^2}{2} = 8 \, \text{W} \).
15. Lamps in Parallel:
    • 100 W: \( I = \frac{100}{220} \approx 0.455 \, \text{A} \); 60 W: \( I = \frac{60}{220} \approx 0.273 \, \text{A} \).
    • Total: \( 0.455 + 0.273 \approx 0.727 \, \text{A} \).
16. Energy Comparison:
    • TV: 250 W, 1 h = \( 250 \times 3600 = 9 \times 10^5 \, \text{J} \).
    • Toaster: 1200 W, 10 min = \( \frac{10}{60} \, \text{h} \); \( 1200 \times \frac{10}{60} \times 3600 = 7.2 \times 10^5 \, \text{J} \).
    • TV uses more energy.
17. Heater Heat Rate:
    • R = 44 Ω, I = 5 A; \( P = I^2 R = 5^2 \times 44 = 1100 \, \text{W} \).
18. Explanations:
    • (a) Tungsten: High melting point (3380°C), retains heat for light emission.
    • (b) Alloys: Higher resistivity, oxidation resistance for heating elements.
    • (c) Series: Single failure breaks circuit, unequal current unsuitable for devices.
    • (d) Resistance: \( R \propto \frac{1}{A} \); larger area reduces R.
    • (e) Copper/Aluminium: Low resistivity, efficient for transmission.