NCERT Science Notes - Class 10
Chapter 11 - Electricity

Welcome to AJs Chalo Seekhen. This webpage is dedicated to Class 10 | Science | Chapter - 11 | Electricity. In this chapter, introduces students to the fundamental concepts of electric current and circuits. It explains the relationship between voltage, current, and resistance through Ohm's Law, and dives into the intricacies of series and parallel circuits. The chapter also covers the heating effect of electric current, how electrical energy is converted into heat, and its applications. It lays the groundwork for understanding household circuits and electrical safety, making it a key chapter for grasping the basics of how electricity powers our world.

NCERT Science Notes - Class 8 Chapter 9 - Friction notes ajs, cbse notes class 10 ajslearning, cbse notes ajs, ajs notes class 10, ajslearning, ajs chalo seekhen

11.1 - Electric Current and Circuit

1. Definition of Current in Air and Water
   • Air Current: Movement of air particles, like wind.
   • Water Current: Flow of water, as observed in rivers.
     
     
2. Electric Current in Conductors
   • Electric Current: Flow of electric charge through a conductor, such as a metallic wire.
   • Example: In a torch, electric current flows from the cells (battery) through the bulb to make it glow.
     
     
3. Role of Switch in Electric Circuit
   • Switch: Device that connects or disconnects the electric circuit, controlling the flow of electric current.
   • Function: A switch provides a conducting link between the cell and the bulb. When turned off, it breaks the circuit, stopping the current and preventing the bulb from glowing.
     
     
4. Electric Circuit
   • Definition: A continuous and closed path of electric current is called an electric circuit.
   • Open Circuit: When the circuit is broken, current stops flowing, and the bulb does not glow.
     
     
5. Expression of Electric Current
   • Electric Current Measurement: Electric current is expressed as the amount of charge flowing through a specific area per unit time.
     • Formula: $I = \frac{Q}{t}$
       
     • Explanation: Here, $I$ is the current, $Q$ is the net charge, and $t$ is time.
   • SI Unit of Charge (Coulomb, C): Charge of 6 × 10¹⁸ electrons equals 1 coulomb (C).
   • Electron Charge: Each electron has a charge of -1.6 × 10⁻¹⁹ C.
     
     
6. Conventional Direction of Current
   • Conventional Current Flow: Initially thought to be the flow of positive charges, current is considered to flow in the direction opposite to electron movement.
   • Electron Flow: In metallic wires, the electrons constitute the flow of charges, moving from negative to positive terminals.
     
     
7. Units of Electric Current
   • Ampere (A): The unit of electric current, defined as the flow of one coulomb of charge per second.
     • 1 A = 1 C/s
   • Milliampere (mA): 1 mA = 10⁻³ A.
   • Microampere (µA): 1 µA = 10⁻⁶ A.
   • Named After: Andre-Marie Ampere, a French scientist who made significant contributions to the study of electromagnetism.
     
     
8. Instrument for Measuring Electric Current (Ammeter)
   • Ammeter: Instrument used to measure electric current in a circuit.
   • Connection in Circuit: Always connected in series in the circuit to measure current accurately.
     
     
9. Schematic Diagram of Electric Circuit
   • Components: Consists of a cell, an electric bulb, an ammeter, and a plug key.
   • Current Flow Direction: Flows from the positive terminal of the cell to the negative terminal through the bulb and ammeter.

Questions and Answers

1. Q: What is an electric current?
   A: Electric current is the flow of electric charge through a conductor.
2. Q: Define electric circuit.
   A: An electric circuit is a continuous and closed path that allows the flow of electric current.
3. Q: What is the function of a switch in an electric circuit?
   A: A switch provides a conducting link between the cell and the bulb. It controls the current flow, turning it on or off.
4. Q: Explain the conventional direction of current.
   A: Conventionally, the direction of current is opposite to the direction of electron flow, as current was first considered to be the flow of positive charges.
5. Q: What is the SI unit of electric charge?
   A: The SI unit of electric charge is the coulomb (C).
6. Q: How is electric current measured?
   A: Electric current is measured in amperes (A), where 1 ampere equals 1 coulomb per second (1 A = 1 C/s).
7. Q: How is an ammeter connected in a circuit?
   A: An ammeter is connected in series in the circuit to measure the electric current.
8. Q: What does 1 ampere of current represent?
   A: 1 ampere represents the flow of 1 coulomb of charge per second.

Example 11.1: Calculation of Electric Charge Flow

Problem Statement:

• A current of 0.5 A is drawn by the filament of an electric bulb for 10 minutes.
• Objective: Find the amount of electric charge that flows through the circuit.

1. Given Data:
   • Current, $I = 0.5 \, \text{A}$
     
   • Time, $t = 10 \, \text{min} = 600 \, \text{s}$
     (since 1 minute = 60 seconds, so $10 \times 60 = 600$ seconds)
2. Formula Used:
   From the formula for electric charge, $Q = I \times t$
   
3. Calculation: $$Q=0.5\text{A}\times 600\text{s}=300\text{C}$$
4. Result:
   • Amount of Electric Charge, $Q$: 300 C

Q: If a current of 0.5 A flows for 10 minutes, what is the total electric charge passing through the circuit?
A: The total electric charge passing through the circuit is 300 C.

11.2 - Electric Potential and Potential Difference

Example 11.2: Work Done in Moving a Charge

11.3 - Circuit Diagram

Ohm's Law

1. Introduction to Ohm's Law:

• Question: Is there a relationship between the potential difference (V) across a conductor and the current (I) through it?
• Activity Objective: To explore the relationship between voltage and current.

Activity 11.1: Exploring Ohm's Law

Setup Instructions:

• Components Needed:
  • Nichrome wire (length: 0.5 m)
  • Ammeter
  • Voltmeter
  • Four cells (1.5 V each)
• Nichrome Definition: An alloy of nickel, chromium, manganese, and iron metals.

1. Single Cell Connection:
   • Connect only one cell as the source in the circuit.
   • Record the readings:
     • Ammeter (I): Current through the nichrome wire.
     • Voltmeter (V): Potential difference across the nichrome wire.
   • Tabulate Results: Record values in the table.
2. Two Cells Connection:
   • Connect two cells in the circuit.
   • Record the respective readings of the ammeter and voltmeter.
3. Three Cells Connection:
   • Repeat the process with three cells.
   • Record readings.
4. Four Cells Connection:
   • Repeat the process with four cells.
   • Record readings.
5. Calculate Ratio:
   • For each pair of voltage (V) and current (I), calculate the ratio $\frac{V}{I}$IV​.
6. Graph Plotting:
   • Plot a graph of V against I.
   • Observe the nature of the graph.

• Graph Nature: The graph will be a straight line passing through the origin, indicating a linear relationship between V and I.
• Constant Ratio: The value of $\frac{V}{I}$​ is approximately the same in each case.

Conclusion

• Ohm’s Law Statement: The potential difference (V) across a conductor is directly proportional to the current (I) flowing through it, expressed mathematically as: $$V = I \times R$$where $R$ is the resistance of the conductor.

Ohm’s Law and Resistance

Activity 11.2: Exploring Current through Different Components

Objective: To investigate how the current varies when different components are used in a circuit and to understand the concept of resistance in conductors.

Materials Needed

• Nichrome wire, Torch bulb, 10 W bulb, Ammeter (0 – 5 A range), Plug key, Connecting wires, Four dry cells (1.5 V each)

1. Circuit Setup:
   • Connect four dry cells in series (total voltage = 6 V) with the ammeter.
   • Leave a gap $XY$ in the circuit as shown in Fig. 11.4.
   • Complete the circuit by connecting the nichrome wire in the gap $XY$.
2. Measuring Current with Nichrome Wire:
   • Plug in the key and note the ammeter reading.
   • Important Note: Always remove the key from the plug after measuring the current to prevent damage.
3. Replace with Torch Bulb:
   • Disconnect the nichrome wire and replace it with the torch bulb.
   • Measure the current through the torch bulb using the ammeter.
4. Replace with 10 W Bulb:
   • Again, remove the previous component and replace it with the 10 W bulb.
   • Measure the current using the ammeter.
5. Analysis of Results:
   • Question: Are the ammeter readings different for different components connected in the gap $XY$?
   • Observation: The readings will vary for different components.
   • Discussion: This indicates that different components offer different levels of resistance to the flow of electric current.

• Variation in Current: The current differs for different components because some components provide an easier path for the flow of electric current while others resist it.
• Electron Movement: In an electric circuit, the flow of electrons constitutes electric current. However, electrons are not entirely free to move due to:
  • Attraction of Atoms: Electrons are attracted to the atoms within the conductor, causing resistance to their flow.

• Definition of Resistance: Resistance is the property of a material that opposes the flow of electric current.
• Classification of Conductors:
  • Good Conductor: A component of a given size that offers low resistance (e.g., copper wire).
  • Resistor: A conductor with appreciable resistance used to control current flow.
  • Poor Conductor: A component of identical size that offers higher resistance.
  • Insulator: A material with very high resistance that does not conduct electricity well (e.g., rubber).

• The activity demonstrates that different materials exhibit varying levels of resistance, influencing the amount of current that can flow through them. Components like nichrome wire, bulbs, and resistors affect the flow of electrons differently based on their inherent resistive properties.

Activity 11.3: Investigating Factors Affecting Resistance

Objective: To study how resistance varies with length, cross-sectional area, and material type, and its effect on electric current.

Materials

• Cell, Ammeter, Nichrome wire (length l, marked (1))
• Nichrome wire (length $2l$, marked (2))
• Thicker nichrome wire (length $l$, marked (3))
• Copper wire (length $l$, marked (4)), Plug key

1. With Nichrome Wire $l$ (1):
   • Current: $I_1$
     
2. With Nichrome Wire $2l$ (2):
   • Current: $I_2$
     
3. With Thicker Nichrome Wire $l$ (3):
   • Current: $I_3$
     
4. With Copper Wire $l$ (4):
   • Current: $I_4$
     

• $I_2 < I_1$: Current decreases with increased length.
• $I_3 > I_1$: Current increases with increased cross-sectional area.
• $I_4 > I_1$​: Current increases with a better conducting material (copper).

• Resistance increases with length and decreases with cross-sectional area. Different materials affect resistance, influencing the current flow.

Resistance of Conductors

Observation Summary:

• The ammeter reading decreases to half when the length of the wire is doubled.
• The ammeter reading increases with a thicker wire of the same material and length.
• A change in ammeter reading occurs when a wire of different material (same length and cross-section) is used.

1. Ohm’s Law Relation:
   • Resistance ( $R$) depends on:
     • Length ( $l$): Directly proportional to length.
       $R \propto l \quad (11.8)$
       
     • Area of Cross-Section ( $A$A): Inversely proportional to the area.
       $R \propto \frac{1}{A} \quad (11.9)$
       
2. Combined Relationship:
   • Combining both relationships gives:
     $R \propto \frac{l}{A} \quad (11.10)$
     
   • Thus,
     $R = \rho \cdot \frac{l}{A}$ 
     where ρ (rho) is the resistivity of the material.

• Definition: A constant that represents how strongly a material opposes the flow of electric current.
• SI Unit: Ohm meter ( $\Omega \cdot m$Ω⋅m).
• Characteristics:
  • Metals and alloys have low resistivity: $10^{-8} \, \Omega \cdot m$ to $10^{-6} \, \Omega \cdot m$ (good conductors).
  • Insulators (e.g., rubber, glass) have high resistivity: $10^{12} \, \Omega \cdot m$ to $10^{17} \, \Omega \cdot m$
    

• Both resistance and resistivity vary with temperature.
• Alloys generally have higher resistivity than their constituent metals.
• Alloys are preferred for heating devices due to their higher stability at high temperatures.

• Tungsten: Used for filaments in electric bulbs.
• Copper and Aluminum: Commonly used for electrical transmission lines.

Examples

Example 11.3

Problem: (a) How much current will an electric bulb draw from a 220 V source if the resistance of the bulb filament is 1200 Ω?
(b) How much current will an electric heater coil draw from a 220 V source if the resistance of the heater coil is 100 Ω?

Solution: (a)

• Given:
  • $V = 220 \, \text{V}$
    
  • $R = 1200 \, \Omega$
    
• Calculation: $$I=\frac{V}{R}=\frac{220\text{V}}{1200\mathrm{\Omega }}=0.18\text{A}$$

• Given:
  • $V = 220 \, \text{V}$
    
  • $R = 100 \, \Omega$
    
• Calculation: $$I = \frac{V}{R} = \frac{220 \, \text{V}}{100 \, \Omega} = 2.2 \, \text{A}$$
  

Problem:
The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A from the source. What current will the heater draw if the potential difference is increased to 120 V?

Solution:

• Given:
  • $V = 60 \, \text{V}$
    
  • $I=4\text{A}$
• Calculate resistance using Ohm's law: $$R=\frac{V}{I}=\frac{60\text{V}}{4\text{A}}=15\mathrm{\Omega }$$
• For $V = 120 \, \text{V}$V=120V: $$I=\frac{V}{R}=\frac{120\text{V}}{15\mathrm{\Omega }}=8\text{A}$$

Example 11.5

Problem:
Resistance of a metal wire of length 1 m is 26 Ω at 20°C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Using Table 11.2, predict the material of the wire.

Solution:

• Given:
  • $R = 26 \, \Omega$
    
  • Diameter $d = 0.3 \, \text{mm} = 0.3 \times 10^{-3} \, \text{m}$
    
  • Length $l=1\text{m}$
• Area of cross-section $A$: $$A = \frac{\pi d^2}{4} = \frac{\pi (0.3 \times 10^{-3})^2}{4} \approx 7.07 \times 10^{-8} \, \text{m}^2$$
  
• Calculate resistivity: $$\rho = \frac{RA}{l} = \frac{26 \, \Omega \cdot 7.07 \times 10^{-8} \, \text{m}^2}{1 \, \text{m}} \approx 1.84 \times 10^{-6} \, \Omega \cdot \text{m}$$
  
• Material prediction: From Table 11.2, this resistivity corresponds to manganese.

Example 11.6

Problem:
A wire of given material having length $l$l and area of cross-section $A$ has a resistance of 4 Ω. What would be the resistance of another wire of the same material having length $\frac{l}{2}$​ and area of cross-section $2A$?Solution:

• Given:
  • Resistance of first wire $R_1 = 4 \, \Omega$
    
  • New wire has:
    • Length $l_2 = \frac{l}{2}$
      
    • Area of cross-section $A_2 = 2A$
      
• Calculate resistance of the new wire $R_2$​:
• Using $R = \rho \cdot \frac{l}{A}$: $$R_2=\rho \cdot \frac{l\mathrm{/}2}{2A}=\frac{\rho l}{4A}=\frac{R_1}{4}=\frac{4\mathrm{\Omega }}{4}=1\mathrm{\Omega }$$

Activity 11.5: Measuring Potential Difference Across Resistors in Series

11.6.2 - Resistors in Parallel

Example 11.9

In Fig. 11.12, resistors R₁​=10Ω, R₂​=40Ω, R₃​=30Ω, R₄​=20Ω, and R₅​=60Ω are connected to a 12 V battery.

Solution

1. Equivalent Resistance of $R_1$​ and $R_2$​: Since $R_1$​ and $R_2$​ are in parallel, we can replace them with an equivalent resistance $R'$. $$\frac{1}{R'} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{10} + \frac{1}{40} = \frac{5}{40}$$
   $$R' = \frac{40}{5} = 8 \, \Omega$$
   
2. Equivalent Resistance of $R_3$​, $R_4$​, and $R_5$​: Since $R_3$​, $R_4$​, and $R_5$​ are also in parallel, we can replace them with an equivalent resistance $R''$. $$\frac{1}{R''} = \frac{1}{R_3} + \frac{1}{R_4} + \frac{1}{R_5} = \frac{1}{30} + \frac{1}{20} + \frac{1}{60} = \frac{6}{60}$$
   $$R'' = \frac{60}{6} = 10 \, \Omega$$
   
3. Total Resistance in the Circuit: Since $R'$ and $R''$are in series, the total resistance $R$R of the circuit is the sum of $R'$ and $R''$: $$R=R^{\mathrm{\prime }}+R^{\mathrm{\prime }\mathrm{\prime }}=8\mathrm{\Omega }+10\mathrm{\Omega }=18\mathrm{\Omega \; <br>\; <br>}$$
4. Total Current in the Circuit: Using Ohm’s law to calculate the total current $I$: $$I=\frac{V}{R}=\frac{12\text{V}}{18\mathrm{\Omega }}=0.67\text{A}$$

• (a) The total resistance in the circuit is 18 $\Omega$.
• (b) The total current flowing in the circuit is 0.67 A.

• Series Circuit Limitation: In a series circuit, the current is the same throughout, which is impractical for devices needing different currents, like a bulb and heater. If one component fails, the entire circuit breaks, as often seen in decorative "fairy lights."
• Parallel Circuit Advantage: A parallel circuit divides the current among devices, reducing total resistance and allowing each device to receive the required current independently.

11.7 - Heating Effect of Electric Current

11.7.1 - Practical Applications of Heating Effect of Electric Current

11.8 - Electric Power

More to Know!

NCERT Science Notes - Class 10 | Science | Chapter - 11 | Electricity

NCERT Science Notes - Class 10 | Science | Chapter - 11 | Electricity