Grade 10 Q&A: Chapter 4: Effects of Electric Current

Concept Questions

Q1: What is meant by the heating effect of electric current?

Answer: The heating effect of electric current refers to the phenomenon where a conductor becomes hot when electric current passes through it. This occurs due to the conversion of electrical energy into heat energy. When electric current flows through a conductor, the free electrons move through the conductor and collide with the atoms or ions of the conductor. These collisions transfer energy to the atoms, increasing their vibrational energy, which manifests as heat. This effect is the basis for many electrical appliances like electric irons, heaters, and incandescent bulbs.

Q2: State Joule's law of heating. What are the factors affecting the heat produced in a conductor?

Answer: Joule's law of heating states that the heat produced in a conductor is: 1. Directly proportional to the square of the current (I²) 2. Directly proportional to the resistance of the conductor (R) 3. Directly proportional to the time for which the current flows (t)

Mathematically, it is expressed as: H = I²Rt

Where: - H is the heat produced in joules (J) - I is the current in amperes (A) - R is the resistance in ohms (Ω) - t is the time in seconds (s)

The factors affecting the heat produced in a conductor are: 1. Current: Doubling the current increases the heat produced by four times 2. Resistance: Higher resistance materials produce more heat for the same current 3. Time: Longer duration of current flow produces more heat 4. Material of the conductor: Different materials have different resistances 5. Thickness of the conductor: Thinner conductors have higher resistance and produce more heat

Q3: What is an electric fuse? How does it work as a safety device?

Answer: An electric fuse is a safety device that protects electrical circuits and appliances from damage due to excessive current. It consists of a wire or strip of a low melting point metal (typically a tin-lead alloy) enclosed in a ceramic or glass holder.

Working principle: 1. The fuse is connected in series with the circuit it protects 2. Under normal conditions, the current flowing through the circuit is below the rated value of the fuse, and the fuse wire remains intact 3. When the current exceeds the rated value (due to overloading or short circuit), the fuse wire heats up according to Joule's heating effect (H = I²Rt) 4. The heat generated causes the fuse wire to melt and break the circuit 5. This interrupts the flow of current, preventing damage to the appliances and possible fire hazards

The fuse wire is designed to melt before the wiring in the circuit gets hot enough to cause damage, making it an essential safety device in electrical systems. Modern alternatives to traditional fuses include circuit breakers and MCBs (Miniature Circuit Breakers) that can be reset after tripping.

Q4: What is the magnetic effect of electric current? Who discovered it and how?

Answer: The magnetic effect of electric current refers to the phenomenon where a magnetic field is created around a conductor when electric current flows through it. This effect demonstrates the relationship between electricity and magnetism.

Discovery: The magnetic effect of electric current was discovered by Danish physicist Hans Christian Oersted in 1820. During a lecture demonstration, Oersted observed that a compass needle deflected when placed near a wire carrying electric current. This accidental discovery showed that: 1. Electric current produces a magnetic field around the conductor 2. The direction of the magnetic field depends on the direction of the current 3. The strength of the magnetic field depends on the magnitude of the current

This discovery was revolutionary as it established for the first time that electricity and magnetism are related phenomena, leading to the development of electromagnetism as a unified field of study. Oersted's discovery paved the way for further work by scientists like Ampere, Faraday, and Maxwell, ultimately leading to numerous practical applications including electromagnets, electric motors, generators, and transformers.

Q5: State the Right-Hand Thumb Rule. How is it used to determine the direction of the magnetic field?

Answer: The Right-Hand Thumb Rule (also known as Maxwell's corkscrew rule) is a method to determine the direction of the magnetic field around a current-carrying conductor.

Statement of the rule: If you hold a current-carrying straight conductor in your right hand such that your thumb points in the direction of the current, then your fingers will curl in the direction of the magnetic field around the conductor.

How to use it: 1. Imagine grasping the wire with your right hand 2. Position your thumb to point in the direction of conventional current (from positive to negative) 3. Your curled fingers will now indicate the direction of the magnetic field lines, which form concentric circles around the wire 4. The magnetic field lines are perpendicular to the conductor at all points

This rule can be extended to determine the magnetic field direction in solenoids and circular loops as well. For a solenoid, if your curled fingers follow the direction of current in the coils, your thumb will point toward the north pole of the electromagnetic field.

The Right-Hand Thumb Rule is an important tool in understanding and predicting the behavior of magnetic fields in various electromagnetic devices and experiments.

Q6: What is a solenoid? How does the magnetic field of a current-carrying solenoid compare to that of a bar magnet?

Answer: A solenoid is a coil of wire wound in the form of a helix (spiral). When electric current passes through this coil, it creates a magnetic field.

Characteristics of a current-carrying solenoid: 1. The magnetic field inside the solenoid is strong, uniform, and parallel to the axis of the solenoid 2. The magnetic field outside the solenoid is weaker and resembles that of a bar magnet 3. One end of the solenoid acts as a north pole and the other end as a south pole 4. The strength of the magnetic field depends on: - The current flowing through the solenoid - The number of turns per unit length - The presence of a core material (like soft iron)

Comparison with a bar magnet: 1. Similarity in field pattern: The magnetic field pattern outside both a solenoid and a bar magnet is similar, with field lines emerging from the north pole and entering at the south pole 2. Poles: Like a bar magnet, a solenoid has distinct north and south poles at its ends 3. Field strength control: Unlike a permanent bar magnet, the strength of a solenoid's magnetic field can be controlled by varying the current or number of turns 4. Reversibility: The polarity of a solenoid can be reversed by changing the direction of current, which is not possible with a permanent magnet 5. On/Off capability: The magnetic field of a solenoid can be turned on or off by controlling the current, while a bar magnet's field is permanent

When a soft iron core is placed inside a solenoid, it becomes an electromagnet, significantly increasing the strength of the magnetic field due to the magnetization of the iron core.

Q7: What is an electromagnet? List its properties and applications.

Answer: An electromagnet is a type of magnet in which the magnetic field is produced by an electric current. It typically consists of a solenoid (a coil of insulated wire) with a soft iron core.

Properties of electromagnets: 1. Temporary magnetism: The magnetic field exists only when current flows through the coil 2. Controllable strength: The strength of the magnetic field can be varied by changing: - The current in the coil - The number of turns in the coil - The core material 3. Reversible polarity: The magnetic poles can be reversed by changing the direction of current 4. Core material effect: Using a ferromagnetic core (like soft iron) greatly increases the magnetic field strength 5. Rapid response: Can be quickly magnetized and demagnetized by controlling the current

Applications of electromagnets: 1. Electric bells and buzzers: Use electromagnets to strike a metal gong when current flows 2. Relays: Use electromagnets to mechanically operate switches in circuits 3. Circuit breakers: Use electromagnets to trip the circuit when excess current flows 4. Magnetic cranes and lifts: Use powerful electromagnets to lift and move heavy magnetic materials in scrapyards and steel industries 5. Magnetic separation: Separate magnetic materials from non-magnetic materials in recycling and mining 6. MRI machines: Use very strong electromagnets for medical imaging 7. Particle accelerators: Use electromagnets to guide charged particles 8. Magnetic levitation trains (Maglev): Use electromagnets for propulsion and levitation 9. Loudspeakers: Convert electrical signals to sound using electromagnets 10. Hard disk drives: Use electromagnets to read and write data 11. Electric motors: Use electromagnets to convert electrical energy to mechanical energy 12. Generators: Use electromagnets to convert mechanical energy to electrical energy

Electromagnets have revolutionized many industries and technologies due to their controllable nature and versatility.

Q8: Explain the principle and working of an electric motor.

Answer: An electric motor is a device that converts electrical energy into mechanical energy. It works on the principle that a current-carrying conductor placed in a magnetic field experiences a force.

Principle: The working principle of an electric motor is based on the magnetic effect of electric current and specifically on the motor effect. When a current-carrying conductor is placed in a magnetic field, it experiences a force. The direction of this force is given by Fleming's Left-Hand Rule, which states that if the thumb, forefinger, and middle finger of the left hand are extended mutually perpendicular to each other with the forefinger pointing in the direction of the magnetic field and the middle finger pointing in the direction of the current, then the thumb points in the direction of the force acting on the conductor.

Construction of a simple DC motor: 1. Armature: A rectangular coil of insulated copper wire wound on a soft iron core 2. Field Magnet: Permanent magnets or electromagnets that provide a uniform magnetic field 3. Split-ring Commutator: Two half-rings of copper that reverse the current direction in the armature every half rotation 4. Carbon Brushes: Make sliding contact with the commutator to supply current to the armature 5. Axle: Supports the armature and allows it to rotate freely

Working: 1. When current flows through the armature coil, it creates its own magnetic field 2. This magnetic field interacts with the external magnetic field of the permanent magnets 3. According to Fleming's Left-Hand Rule, the forces acting on the two sides of the coil are in opposite directions, creating a torque that rotates the coil 4. After half a rotation, the commutator reverses the direction of current in the coil 5. This reversal ensures that the torque continues to act in the same direction, causing continuous rotation 6. The soft iron core enhances the magnetic effect and provides mechanical support

Factors affecting the rotation of the motor: 1. Strength of the current 2. Number of turns in the coil 3. Strength of the magnetic field 4. Area of the coil

Electric motors are fundamental to modern technology and are used in countless applications from small household appliances to large industrial machinery.

Q9: What is electrolysis? Explain the process with a suitable example.

Answer: Electrolysis is the process of chemical decomposition of an electrolyte by passing electric current through it. It is a technique that uses direct electric current (DC) to drive an otherwise non-spontaneous chemical reaction.

Components required for electrolysis: 1. Electrolyte: A solution or molten substance that conducts electricity due to the presence of ions 2. Electrodes: Conductors (usually metals) that connect the electrolyte to the external circuit - Anode: Positive electrode where oxidation occurs - Cathode: Negative electrode where reduction occurs 3. Battery or DC source: Provides the electrical energy for the process

Process of electrolysis: 1. When electrodes are connected to a battery and placed in an electrolyte: - Positive ions (cations) are attracted to and move towards the cathode - Negative ions (anions) are attracted to and move towards the anode 2. At the cathode, cations gain electrons (reduction) 3. At the anode, anions lose electrons (oxidation) 4. These redox reactions lead to chemical changes in the electrolyte

Example: Electrolysis of water (H₂O)

Setup: - Electrolyte: Water with a small amount of sulfuric acid or sodium hydroxide to increase conductivity - Electrodes: Platinum or carbon electrodes - Power source: Battery or DC power supply

Reactions: 1. At the cathode (reduction): 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq) (Hydrogen gas is produced)

1. At the anode (oxidation): 2H₂O(l) → O₂(g) + 4H⁺(aq) + 4e⁻ (Oxygen gas is produced)

2. Overall reaction: 2H₂O(l) → 2H₂(g) + O₂(g)

Observations: - Hydrogen gas bubbles form at the cathode - Oxygen gas bubbles form at the anode - The volume of hydrogen produced is twice the volume of oxygen, confirming the 2:1 ratio in the water molecule (H₂O)

Applications of electrolysis include electroplating, electrorefining of metals, production of chemicals like chlorine and sodium hydroxide, and extraction of reactive metals from their ores.

Q10: What is electroplating? Describe the process with an example.

Answer: Electroplating is a process that uses electrical current to reduce dissolved metal ions and deposit them as a thin layer onto a conductive object. It is used to coat one metal with another for decorative purposes, corrosion resistance, or to improve other material characteristics like conductivity or durability.

Process of electroplating: 1. The object to be plated (workpiece) is made the cathode (negative electrode) 2. A piece of the plating metal is used as the anode (positive electrode) 3. Both electrodes are immersed in an electrolyte solution containing ions of the plating metal 4. When direct current is passed through the circuit: - The anode (plating metal) oxidizes, releasing metal ions into the solution - The cathode (workpiece) reduces these metal ions, which are deposited as a thin layer on its surface 5. The thickness of the coating can be controlled by adjusting the current and duration of the process

Example: Silver plating of copper

Setup: - Cathode: Copper object to be silver-plated - Anode: Pure silver metal - Electrolyte: Silver nitrate (AgNO₃) solution

Reactions: 1. At the anode (oxidation): Ag(s) → Ag⁺(aq) + e⁻ (Silver dissolves into the solution as silver ions)

1. At the cathode (reduction): Ag⁺(aq) + e⁻ → Ag(s) (Silver ions are reduced and deposited on the copper object)

2. Overall reaction: Ag(anode) → Ag(cathode) (Silver transfers from the anode to the cathode)

Applications of electroplating: 1. Decorative purposes: Gold or silver plating of jewelry and cutlery 2. Corrosion protection: Zinc plating (galvanizing) of iron to prevent rusting 3. Wear resistance: Chromium plating of car parts and tools 4. Electrical conductivity: Gold plating of electrical connectors 5. Radiation reflection: Silver plating of mirrors and reflective surfaces 6. Friction reduction: Nickel plating of engine parts 7. Building up worn parts: Rebuilding worn machine parts with metal plating

Factors affecting the quality of electroplating include current density, temperature, concentration of the electrolyte, pH of the solution, and cleanliness of the workpiece surface.

Application-Based Questions

Q11: Why are heating elements in electrical appliances like toasters and electric irons made of alloys like nichrome rather than pure metals?

Answer: Heating elements in electrical appliances like toasters and electric irons are made of alloys like nichrome (an alloy of nickel, chromium, and sometimes iron) rather than pure metals for several important reasons:

1. High resistivity: Nichrome has a high electrical resistivity (about 100 times that of copper), which means it generates more heat for the same current according to Joule's heating effect (H = I²Rt).

2. High melting point: Nichrome has a melting point of about 1400°C, allowing it to operate at high temperatures without melting. Pure metals like copper or aluminum would melt at much lower temperatures (copper melts at 1085°C).

3. Oxidation resistance: The chromium in nichrome forms a protective oxide layer that prevents further oxidation, allowing the element to last longer even at high temperatures. Pure metals would oxidize rapidly and deteriorate.

4. Low temperature coefficient of resistance: Nichrome's resistance changes very little with temperature, ensuring consistent heating performance. Pure metals' resistance increases significantly with temperature, which would cause fluctuations in heating.

5. Mechanical strength: Nichrome maintains its structural integrity at high temperatures, allowing it to be formed into coils or ribbons that don't sag or deform during operation.

6. Cost-effectiveness: While more expensive than some pure metals, nichrome's durability and performance characteristics make it cost-effective over the lifetime of the appliance.

These properties make nichrome and similar alloys ideal for heating elements, as they provide reliable, consistent heating while maintaining structural integrity and longevity, even under repeated heating and cooling cycles.

Q12: A household circuit has a refrigerator (800 W), two fans (60 W each), and three light bulbs (40 W each). Calculate the total current drawn from a 220 V supply. If the circuit has a 5 A fuse, will it blow?

Answer: To solve this problem, we need to calculate the total power consumption and then determine the current drawn.

Step 1: Calculate the total power consumption. Total power = Power of refrigerator + Power of fans + Power of light bulbs Total power = 800 W + (2 × 60 W) + (3 × 40 W) Total power = 800 W + 120 W + 120 W Total power = 1040 W

Step 2: Calculate the current using the power formula. P = VI Where: - P is power in watts (W) - V is voltage in volts (V) - I is current in amperes (A)

Rearranging to solve for current: I = P/V I = 1040 W / 220 V I = 4.73 A

Step 3: Determine if the fuse will blow. The circuit has a 5 A fuse, and the calculated current is 4.73 A. Since 4.73 A < 5 A, the fuse will not blow under normal operating conditions.

However, it's important to note that: 1. This calculation assumes all appliances are running simultaneously 2. The refrigerator may draw a higher current momentarily when its compressor starts (starting current) 3. The circuit is operating very close to its maximum capacity (4.73 A is 94.6% of the 5 A limit)

For safety and to allow for momentary current surges, it would be advisable to either distribute some of these appliances to another circuit or upgrade to a higher-rated fuse (after ensuring the wiring can handle the higher current).

Q13: Explain how an electric bell works using the principle of electromagnets.

Answer: An electric bell works on the principle of electromagnets, converting electrical energy into mechanical energy to produce sound. Here's a detailed explanation of its working:

Components of an electric bell: 1. Electromagnet: A coil of insulated copper wire wound around an iron core 2. Armature: A soft iron bar mounted near the electromagnet, with a hammer attached to one end 3. Contact screw: An adjustable screw that makes contact with the armature 4. Gong: A metal bell that produces sound when struck by the hammer 5. Spring: Keeps the armature in contact with the contact screw when no current flows 6. Battery: Provides the electrical energy

Working principle: 1. Initial state: When the bell switch is off, the spring holds the armature against the contact screw, completing the circuit.

1. When the switch is turned on:
2. Current flows through the circuit: battery → switch → electromagnet → armature → contact screw → battery
3. The electromagnet gets magnetized and attracts the armature

4. The armature moves toward the electromagnet, causing the hammer to strike the gong, producing sound

5. Breaking the circuit:

6. As the armature moves toward the electromagnet, it breaks contact with the contact screw
7. This interrupts the current flow, and the electromagnet loses its magnetism

8. The spring pulls the armature back to its original position, making contact with the contact screw again

9. Repetition:

10. When the armature returns to its original position, it completes the circuit again
11. The electromagnet gets magnetized again, and the cycle repeats

12. This rapid making and breaking of the circuit causes the hammer to strike the gong repeatedly, producing a continuous ringing sound

13. When the switch is turned off:

14. The circuit is permanently broken
15. The electromagnet remains demagnetized
16. The armature stays in its rest position against the contact screw
17. The bell stops ringing

The frequency of the hammer strikes depends on the spring tension, the distance between the armature and electromagnet, and the strength of the electromagnet. Modern electric bells may use electronic circuits instead of mechanical contacts to control the current, but the basic principle of using an electromagnet to move a hammer remains the same.

Q14: A copper wire and a nichrome wire of the same length and diameter are connected in series to a battery. Which wire will get hotter and why?

Answer: When a copper wire and a nichrome wire of the same length and diameter are connected in series to a battery, the nichrome wire will get hotter. Here's the detailed explanation:

Key concept: In a series circuit, the same current flows through all components, but the heat generated depends on the resistance according to Joule's law of heating (H = I²Rt).

Analysis: 1. Resistivity comparison: - Resistivity of nichrome: approximately 1.0 × 10⁻⁶ Ω·m - Resistivity of copper: approximately 1.68 × 10⁻⁸ Ω·m - Nichrome has about 60 times higher resistivity than copper

1. Resistance calculation: For a wire, resistance R = ρL/A, where:
2. ρ is the resistivity
3. L is the length
4. A is the cross-sectional area

Since both wires have the same length and diameter (thus same cross-sectional area), their resistances are directly proportional to their resistivities: - Rnichrome = ρnichrome × L/A - Rcopper = ρcopper × L/A - Therefore, Rnichrome ≈ 60 × Rcopper

1. Heat generation: According to Joule's law, heat generated H = I²Rt
2. Since the wires are in series, the current (I) is the same through both
3. Time (t) is also the same for both wires
4. Therefore, the heat generated is directly proportional to the resistance

Hnichrome/Hcopper = Rnichrome/Rcopper ≈ 60

This means the nichrome wire will generate approximately 60 times more heat than the copper wire and will get much hotter.

Additional factors: - Thermal conductivity: Copper has higher thermal conductivity than nichrome, so it dissipates heat more efficiently, further increasing the temperature difference - Specific heat capacity: The materials have different heat capacities, affecting how quickly their temperatures rise for a given heat input

This property of nichrome (high resistance, high heat generation) is why it's commonly used in heating elements, while copper is preferred for electrical wiring where minimal heat generation is desired.

Q15: How does the strength of the magnetic field inside a current-carrying solenoid depend on the current, number of turns, and core material?

Answer: The strength of the magnetic field inside a current-carrying solenoid depends on several factors:

1. Current (I):
2. The magnetic field strength is directly proportional to the current flowing through the solenoid
3. Doubling the current doubles the magnetic field strength
4. Mathematical relationship: B ∝ I

5. This occurs because more current means more electrons flowing per second, creating a stronger magnetic effect

6. Number of turns per unit length (n):

7. The magnetic field strength is directly proportional to the number of turns per unit length
8. Doubling the number of turns per unit length doubles the magnetic field strength
9. Mathematical relationship: B ∝ n

10. This happens because more turns create more overlapping magnetic fields that add up

11. Core material (μ):

12. The magnetic field strength is directly proportional to the permeability (μ) of the core material
13. Ferromagnetic materials like iron, nickel, and cobalt significantly increase the magnetic field strength
14. Mathematical relationship: B ∝ μ
15. This occurs because ferromagnetic materials become magnetized in the same direction as the solenoid's field, amplifying it

The complete mathematical relationship is given by: B = μ₀μᵣnI

Where: - B is the magnetic field strength in teslas (T) - μ₀ is the permeability of free space (4π × 10⁻⁷ H/m) - μᵣ is the relative permeability of the core material (dimensionless) - n is the number of turns per unit length (turns/m) - I is the current in amperes (A)

Practical implications: - Air core solenoid: μᵣ ≈ 1, providing the baseline magnetic field strength - Iron core solenoid: μᵣ can be 100-5000, dramatically increasing field strength - Length effect: For a finite solenoid, the field is strongest at the center and weaker near the ends - Temperature effect: High temperatures can reduce the permeability of ferromagnetic materials

This relationship explains why electromagnets typically use: 1. High currents (within safe limits) 2. Many turns of wire 3. Ferromagnetic cores to achieve the strongest possible magnetic fields for applications like lifting heavy objects, MRI machines, and particle accelerators.

Q16: A student sets up an electroplating experiment to coat a copper key with silver. The key is connected to the negative terminal of the battery, but no silver gets deposited on the key. Identify the possible errors and explain how to correct them.

Answer: In an electroplating experiment to coat a copper key with silver, several possible errors could prevent silver deposition even when the key is correctly connected to the negative terminal (cathode) of the battery. Here are the potential issues and their corrections:

Possible Error 1: Incorrect electrolyte solution - Problem: The electrolyte might not contain silver ions (Ag⁺) - Correction: Use a proper silver salt solution such as silver nitrate (AgNO₃) or silver cyanide complex (K[Ag(CN)₂]) - Explanation: Without silver ions in the solution, there are no ions to be reduced and deposited on the cathode

Possible Error 2: Improper anode material - Problem: The anode might not be made of silver - Correction: Use a pure silver plate or rod as the anode - Explanation: The anode should be made of the plating metal (silver) to replenish the solution with silver ions as the process continues

Possible Error 3: Poor electrical connections - Problem: There might be loose connections or breaks in the circuit - Correction: Check and secure all electrical connections; ensure the key is making good electrical contact with the circuit - Explanation: Without a complete circuit, current cannot flow, and no electroplating will occur

Possible Error 4: Insufficient voltage/current - Problem: The battery voltage might be too low to overcome the resistance in the circuit - Correction: Use a power supply with adequate voltage (typically 3-6V for silver plating) - Explanation: Electroplating requires sufficient electrical potential to drive the redox reactions

Possible Error 5: Contaminated or oxidized key surface - Problem: The copper key might have oils, oxides, or other contaminants preventing adhesion - Correction: Clean the key thoroughly with a mild acid solution, then rinse with distilled water before placing in the electrolyte - Explanation: Contaminants can prevent proper bonding of silver to the copper surface

Possible Error 6: Incorrect pH or temperature - Problem: The solution conditions might not be optimal for silver plating - Correction: Adjust the pH to the recommended range (typically slightly acidic for silver plating) and ensure room temperature operation - Explanation: Extreme pH or temperature can affect the efficiency of the electroplating process

Possible Error 7: Current density too low - Problem: The current might be spread over too large an area - Correction: Adjust the current or the distance between electrodes to achieve the proper current density - Explanation: Proper silver plating requires an appropriate current density (typically 1-5 A/dm²)

A properly set up silver electroplating experiment should have: 1. A clean copper key connected to the negative terminal (cathode) 2. A silver anode connected to the positive terminal 3. A silver salt solution as the electrolyte 4. Secure electrical connections 5. Appropriate voltage/current 6. Proper solution conditions (pH, temperature)

When correctly set up, silver ions (Ag⁺) from the solution will be reduced to silver metal (Ag) at the cathode (key), creating a silver coating on the copper key.

Q17: Explain how the earth wire in domestic circuits acts as a safety measure. What could happen if an appliance with a metal body is not properly earthed?

Answer: The earth wire (also called the ground wire) in domestic circuits is a crucial safety feature that protects users from electric shocks and prevents damage to appliances. It is typically a green or yellow-green wire connected to the metal body of electrical appliances.

How the earth wire works as a safety measure:

1. Normal operation: Under normal conditions, current flows from the live wire through the appliance and back through the neutral wire, completing the circuit. The earth wire carries no current.

2. Fault condition: If there's an insulation failure or internal wiring fault that causes the live wire to touch the metal body of an appliance:

3. Without earthing: The metal body becomes "live" at the supply voltage (typically 220-240V), creating an electric shock hazard for anyone who touches it

4. With proper earthing: The earth wire provides a low-resistance path for the fault current to flow directly to the ground

5. Circuit protection activation: When a fault occurs in a properly earthed system:

6. The fault current flowing through the earth wire is very high (due to low resistance)
7. This high current quickly triggers the fuse or circuit breaker to blow or trip

8. The circuit is disconnected before a person touching the appliance can receive a serious shock

9. ELCB/RCD protection: Modern installations also include Earth Leakage Circuit Breakers (ELCB) or Residual Current Devices (RCD) that detect even small leakage currents to earth (typically 30mA) and disconnect the supply within milliseconds.

Consequences of improper earthing:

If an appliance with a metal body is not properly earthed, several dangerous situations can arise:

1. Electric shock hazard: If a live wire contacts the metal casing due to insulation failure:
2. The entire metal body becomes energized at mains voltage
3. A person touching the appliance becomes part of the circuit to ground

4. Current flows through the person's body, causing an electric shock that can be fatal

5. Delayed or no circuit breaking: Without the low-resistance earth path:

6. Fault currents may be too small to trip circuit breakers or blow fuses

7. The appliance can remain "live" indefinitely, creating an ongoing hazard

8. Fire risk: Electrical faults without proper earthing can cause:

9. Arcing and sparking inside the appliance
10. Overheating of components

11. Potential ignition of surrounding materials

12. Damage to sensitive electronics: Without proper earthing:

13. Voltage surges have no path to ground
14. Electronic components may be damaged by voltage spikes

15. Electromagnetic interference can affect performance

16. Electrocution risk increases in wet areas: In bathrooms, kitchens, or outdoor areas:

17. Water reduces body resistance
18. The risk of severe or fatal shocks is much higher
19. Proper earthing is especially critical in these locations

Proper earthing is not optional but a critical safety requirement for all electrical installations. It should be regularly tested and maintained to ensure it provides effective protection against electric shocks and fire hazards.

Q18: A student wants to electroplate a steel spoon with copper. Draw a labeled diagram of the experimental setup and explain the process.

Answer: Experimental Setup for Copper Electroplating a Steel Spoon

[Note: In an actual diagram, the following components would be illustrated with appropriate labels]

Components: 1. Power Supply: A DC power source (battery or DC adapter) with positive and negative terminals 2. Anode: Pure copper plate or rod connected to the positive terminal 3. Cathode: Steel spoon connected to the negative terminal 4. Electrolyte: Copper sulfate (CuSO₄) solution in a non-metallic container 5. Connecting Wires: Insulated copper wires with alligator clips for connections

Process Explanation:

1. Preparation of the steel spoon (cathode):
2. Clean the spoon thoroughly to remove dirt, grease, and oxides
3. This can be done using mild detergent, followed by dilute acid (like vinegar)
4. Rinse with distilled water to remove all cleaning agents

5. The clean surface ensures proper adhesion of the copper coating

6. Preparation of the electrolyte:

7. Prepare a solution of copper sulfate (CuSO₄) in distilled water
8. Add a small amount of sulfuric acid (H₂SO₄) to increase conductivity and prevent formation of copper hydroxide

9. The typical concentration is about 200-250 g/L of copper sulfate with 50-75 g/L of sulfuric acid

10. Setting up the circuit:

11. Connect the copper plate (anode) to the positive terminal of the power supply
12. Connect the steel spoon (cathode) to the negative terminal
13. Immerse both the anode and cathode in the copper sulfate solution
14. Ensure they don't touch each other to prevent short-circuiting

15. Position them parallel to each other for uniform deposition

16. Electroplating process:

17. Turn on the power supply (typically 1-6 volts DC)
18. Adjust the current to the appropriate density (approximately 2-5 A/dm² of cathode surface)
19. The electrochemical reactions begin:
    • At the anode (copper plate): Cu(s) → Cu²⁺(aq) + 2e⁻ (oxidation)
    • At the cathode (steel spoon): Cu²⁺(aq) + 2e⁻ → Cu(s) (reduction)
20. Copper ions from the solution are reduced and deposited on the spoon

21. The copper anode dissolves, replenishing the copper ions in the solution

22. Monitoring and completion:

23. Maintain the process for the required time (depends on desired thickness)
24. A typical plating rate is about 1 micrometer per minute at standard current density
25. The solution may need occasional stirring to maintain uniform concentration
26. Once the desired thickness is achieved, turn off the power supply

27. Remove the spoon, rinse it with distilled water, and dry it

28. Finishing:

29. The plated spoon may be buffed or polished to achieve a bright finish
30. A clear lacquer can be applied to prevent tarnishing

Key factors affecting the quality of copper plating: - Cleanliness of the steel spoon - Purity of the copper anode - Composition and pH of the electrolyte - Current density - Temperature of the solution - Duration of the process - Distance between electrodes

This process creates a durable and attractive copper coating on the steel spoon, improving its appearance and corrosion resistance.

Numerical Problems

Q19: A heating coil has a resistance of 100 Ω. Calculate the heat produced when a current of 2 A flows through it for 5 minutes.

Answer: To calculate the heat produced by the heating coil, we'll use Joule's law of heating: H = I²Rt

Where: - H is the heat produced in joules (J) - I is the current in amperes (A) - R is the resistance in ohms (Ω) - t is the time in seconds (s)

Given: - Current (I) = 2 A - Resistance (R) = 100 Ω - Time (t) = 5 minutes = 5 × 60 = 300 seconds

Substituting these values into the formula: H = I²Rt H = (2 A)² × 100 Ω × 300 s H = 4 × 100 × 300 J H = 120,000 J = 120 kJ

Therefore, the heat produced by the heating coil is 120 kilojoules (120 kJ).

We can also express this in terms of energy in watt-hours: Energy = Power × Time Power = I²R = (2 A)² × 100 Ω = 400 W Time = 5 minutes = 5/60 = 0.0833 hours Energy = 400 W × 0.0833 h = 33.33 Wh = 0.0333 kWh

The heat produced is equivalent to 0.0333 kilowatt-hours of electrical energy.

Q20: An electric iron draws a current of 3.4 A when connected to a 220 V supply. Calculate: (a) the power consumed, (b) the energy consumed in 2 hours, and (c) the cost of operation if electricity costs ₹6 per kWh.

Answer:

Given: - Current (I) = 3.4 A - Voltage (V) = 220 V - Time (t) = 2 hours - Cost of electricity = ₹6 per kWh

(a) Power consumed: Power (P) = Voltage (V) × Current (I) P = 220 V × 3.4 A P = 748 W = 0.748 kW

(b) Energy consumed in 2 hours: Energy (E) = Power (P) × Time (t) E = 0.748 kW × 2 h E = 1.496 kWh

(c) Cost of operation: Cost = Energy consumed × Cost per unit Cost = 1.496 kWh × ₹6 per kWh Cost = ₹8.976 ≈ ₹8.98

Therefore: (a) The power consumed by the electric iron is 748 watts or 0.748 kilowatts. (b) The energy consumed in 2 hours is 1.496 kilowatt-hours. (c) The cost of operation for 2 hours is ₹8.98.

This calculation demonstrates how to determine the energy consumption and operating cost of electrical appliances, which is useful for household budgeting and energy conservation efforts.

Q21: A copper wire of length 2 m and cross-sectional area 1.7 × 10⁻⁶ m² has a resistance of 0.02 Ω. Calculate the resistivity of copper.

Answer:

To calculate the resistivity of copper, we'll use the formula for the resistance of a conductor: R = ρL/A

Where: - R is the resistance in ohms (Ω) - ρ (rho) is the resistivity in ohm-meters (Ω·m) - L is the length of the conductor in meters (m) - A is the cross-sectional area in square meters (m²)

Given: - Resistance (R) = 0.02 Ω - Length (L) = 2 m - Cross-sectional area (A) = 1.7 × 10⁻⁶ m²

Rearranging the formula to solve for resistivity: ρ = RA/L

Substituting the values: ρ = (0.02 Ω × 1.7 × 10⁻⁶ m²) / 2 m ρ = (0.02 × 1.7 × 10⁻⁶) / 2 Ω·m ρ = 0.034 × 10⁻⁶ / 2 Ω·m ρ = 0.017 × 10⁻⁶ Ω·m ρ = 1.7 × 10⁻⁸ Ω·m

Therefore, the resistivity of copper is 1.7 × 10⁻⁸ Ω·m.

This value is consistent with the standard resistivity of copper, which is approximately 1.68 × 10⁻⁸ Ω·m at 20°C. The slight difference could be due to factors such as temperature, purity of the copper, or rounding in the given values.

The low resistivity of copper explains why it's widely used in electrical wiring, as it allows current to flow with minimal resistance and heat generation.

Q22: A solenoid 20 cm long has 400 turns of wire. Calculate the magnetic field strength at the center of the solenoid when a current of 2 A flows through it. (μ₀ = 4π × 10⁻⁷ H/m)

Answer:

To calculate the magnetic field strength at the center of a solenoid, we use the formula: B = μ₀nI

Where: - B is the magnetic field strength in teslas (T) - μ₀ is the permeability of free space (4π × 10⁻⁷ H/m) - n is the number of turns per unit length (turns/m) - I is the current in amperes (A)

Given: - Length of solenoid (L) = 20 cm = 0.2 m - Number of turns (N) = 400 - Current (I) = 2 A - Permeability of free space (μ₀) = 4π × 10⁻⁷ H/m

Step 1: Calculate the number of turns per unit length (n). n = N/L n = 400 turns / 0.2 m n = 2000 turns/m

Step 2: Calculate the magnetic field strength (B). B = μ₀nI B = 4π × 10⁻⁷ H/m × 2000 turns/m × 2 A B = 4π × 10⁻⁷ × 4000 T B = 16π × 10⁻⁷ T B = 16 × 3.14159 × 10⁻⁷ T B = 50.27 × 10⁻⁷ T B = 5.027 × 10⁻⁶ T B ≈ 5.03 × 10⁻⁶ T (or 5.03 μT)

Therefore, the magnetic field strength at the center of the solenoid is approximately 5.03 × 10⁻⁶ teslas or 5.03 microteslas.

This calculation assumes an ideal solenoid that is much longer than its diameter. For a real solenoid with finite length, the field at the center would be slightly less than this calculated value. If the solenoid had an iron core instead of air, the magnetic field would be significantly stronger due to the higher permeability of iron.

Q23: In an electroplating process, a current of 2 A is passed through a copper sulfate solution for 30 minutes. Calculate the mass of copper deposited on the cathode. (Electrochemical equivalent of copper = 3.3 × 10⁻⁷ kg/C)

Answer:

To calculate the mass of copper deposited during electroplating, we'll use Faraday's First Law of Electrolysis: m = Z × I × t

Where: - m is the mass of the substance deposited in kilograms (kg) - Z is the electrochemical equivalent of the substance in kg/C - I is the current in amperes (A) - t is the time in seconds (s)

Given: - Current (I) = 2 A - Time (t) = 30 minutes = 30 × 60 = 1800 seconds - Electrochemical equivalent of copper (Z) = 3.3 × 10⁻⁷ kg/C

Substituting these values into the formula: m = Z × I × t m = 3.3 × 10⁻⁷ kg/C × 2 A × 1800 s m = 3.3 × 10⁻⁷ × 2 × 1800 kg m = 3.3 × 10⁻⁷ × 3600 kg m = 11.88 × 10⁻⁴ kg m = 1.188 × 10⁻³ kg m = 1.188 g

Therefore, the mass of copper deposited on the cathode is 1.188 grams.

Note: We can verify this result using Faraday's constant and the molar mass of copper: - Faraday's constant (F) = 96,485 C/mol - Molar mass of copper = 63.55 g/mol - In copper sulfate, copper has a valency of 2 (Cu²⁺)

The mass deposited can be calculated as: m = (M × I × t) / (n × F) Where M is the molar mass, n is the valency, and F is Faraday's constant.

m = (63.55 g/mol × 2 A × 1800 s) / (2 × 96,485 C/mol) m = (63.55 × 2 × 1800) / (2 × 96,485) g m = 228,780 / 192,970 g m = 1.186 g

The slight difference (1.188 g vs. 1.186 g) is due to rounding in the electrochemical equivalent value.

Q24: A household uses the following electrical appliances: two 60 W fans for 8 hours each, three 100 W bulbs for 5 hours each, a 1000 W iron for 30 minutes, and a 1500 W geyser for 1 hour daily. Calculate the monthly electricity bill if the cost is ₹5 per kWh.

Answer:

To calculate the monthly electricity bill, we need to find the total energy consumed by all appliances in a month and multiply it by the cost per unit.

Given: - Two 60 W fans used for 8 hours each daily - Three 100 W bulbs used for 5 hours each daily - A 1000 W iron used for 30 minutes (0.5 hours) daily - A 1500 W geyser used for 1 hour daily - Cost of electricity = ₹5 per kWh - Assuming a 30-day month

Step 1: Calculate the daily energy consumption for each appliance.

Fans: Energy = Power × Time × Number of appliances Energy = 60 W × 8 h × 2 Energy = 960 Wh = 0.96 kWh

Bulbs: Energy = Power × Time × Number of appliances Energy = 100 W × 5 h × 3 Energy = 1500 Wh = 1.5 kWh

Iron: Energy = Power × Time Energy = 1000 W × 0.5 h Energy = 500 Wh = 0.5 kWh

Geyser: Energy = Power × Time Energy = 1500 W × 1 h Energy = 1500 Wh = 1.5 kWh

Step 2: Calculate the total daily energy consumption. Total daily energy = 0.96 + 1.5 + 0.5 + 1.5 = 4.46 kWh

Step 3: Calculate the monthly energy consumption. Monthly energy = Daily energy × Number of days Monthly energy = 4.46 kWh × 30 days Monthly energy = 133.8 kWh

Step 4: Calculate the monthly electricity bill. Monthly bill = Monthly energy × Cost per unit Monthly bill = 133.8 kWh × ₹5 per kWh Monthly bill = ₹669

Therefore, the monthly electricity bill for the household is ₹669.

This calculation helps households understand their energy consumption patterns and identify opportunities for energy conservation. For example, replacing the 100 W bulbs with energy-efficient LED bulbs (15-20 W) could significantly reduce the electricity bill.

Q25: A copper wire of resistance 10 Ω at 20°C is heated to 100°C. If the temperature coefficient of resistance of copper is 0.004 per °C, calculate the resistance of the wire at 100°C.

Answer:

To calculate the resistance of a conductor at a different temperature, we use the formula: Rt = R₀[1 + α(t - t₀)]

Where: - Rt is the resistance at temperature t - R₀ is the resistance at reference temperature t₀ - α is the temperature coefficient of resistance - t is the final temperature - t₀ is the reference temperature

Given: - Resistance at 20°C (R₀) = 10 Ω - Final temperature (t) = 100°C - Reference temperature (t₀) = 20°C - Temperature coefficient of resistance of copper (α) = 0.004 per °C

Substituting these values into the formula: R₁₀₀ = R₂₀[1 + α(100 - 20)] R₁₀₀ = 10[1 + 0.004(80)] R₁₀₀ = 10[1 + 0.32] R₁₀₀ = 10 × 1.32 R₁₀₀ = 13.2 Ω

Therefore, the resistance of the copper wire at 100°C is 13.2 ohms.

This increase in resistance with temperature is a characteristic property of most metals and is due to increased vibration of atoms at higher temperatures, which impedes the flow of electrons. This property is important in the design of electrical systems, as it affects the performance of conductors under varying temperature conditions.

For example, power transmission lines can experience significant increases in resistance during hot weather, leading to increased power losses and reduced efficiency. Similarly, the heating elements in appliances like toasters and electric heaters rely on this property to generate heat from electrical energy.

References

1. Maharashtra State Board 10th Standard Science Syllabus 2025-26
2. NCERT Science Textbook for Class 10
3. Electricity and Magnetism - Concepts and Applications
4. Practical Applications of Electrical Effects - Modern Perspective
5. Journal of Electrochemistry - Principles and Applications