Master Electricity: Solved Examples, Questions, and Expert Hints

Chapter 11: Electricity

Learn and practice with detailed explanations, solved examples, and extra questions.

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Introduction

Electricity plays a vital role in our daily lives. From lighting our homes to powering machines, its applications are endless. This chapter explores key concepts, solves practical problems, and provides additional resources for students to master this topic.

Topics

1. Electric Current and Circuit

Electric current is the flow of electric charge through a conductor. It is caused by the movement of electrons, which are negatively charged particles. An electric circuit is a closed path that allows current to flow. If the circuit is open, the current cannot flow.

• Formula: \( I = \frac{Q}{t} \), where \( I \) is current (amperes), \( Q \) is charge (coulombs), \( t \) is time (seconds).
• SI Unit: Ampere (A).
• Measurement: Current is measured using an ammeter, connected in series in the circuit.

Did You Know?

• The direction of conventional current is opposite to the flow of electrons.
• The SI unit of charge is the coulomb (C), and 1 coulomb is the charge carried by approximately \( 6.24 \times 10^{18} \) electrons.

2. Electric Potential and Potential Difference

Electric potential at a point is the amount of work needed to move a unit charge from infinity to that point. Potential difference between two points is the work done to move a unit charge from one point to another.

• Formula: \( V = \frac{W}{Q} \), where \( V \) is potential difference (volts), \( W \) is work done (joules), \( Q \) is charge (coulombs).
• SI Unit: Volt (V).
• Measurement: Measured using a voltmeter, connected in parallel across the points.

Quick Tips

• One volt is the potential difference when 1 joule of work is done to move 1 coulomb of charge.
• A battery provides a potential difference, which drives the current through the circuit.

Quick Quiz

What is the SI unit of electric current?

• Ohm
• Coulomb
• Ampere
• Joule

Key Formulas

• Current: \( I = \frac{Q}{t} \)
• Potential Difference: \( V = \frac{W}{Q} \)
• Ohm's Law: \( V = IR \)

Electric current is the flow of electric charge...

Solved Questions

Example: Electric Current

Problem: A current of 0.5 A flows through a circuit for 10 minutes. Find the charge flowing through the circuit.

Solution: Use the formula \( Q = I \times t \).

\( I = 0.5 \, \text{A}, \, t = 10 \, \text{min} = 600 \, \text{seconds} \).

\( Q = 0.5 \times 600 = 300 \, \text{C} \).

Answer: 300 coulombs.

Solved Examples

Example 11.1

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

Solution:

• Given:
  • Current (\( I \)) = 0.5 A
  • Time (\( t \)) = 10 minutes = 600 seconds
• Using the formula: \( Q = I \cdot t \)
• \( Q = 0.5 \times 600 = 300 \, C \)

Answer: The charge flowing through the circuit is \( 300 \, \text{C} \).

Example 11.2

Problem: How much work is done in moving a charge of 2 C across two points having a potential difference of 12 V?

Solution:

• Given:
  • Charge (\( Q \)) = 2 C
  • Potential difference (\( V \)) = 12 V
• Using the formula: \( W = V \cdot Q \)
• \( W = 12 \times 2 = 24 \, J \)

Answer: The work done is \( 24 \, \text{J} \).

End-of-Chapter Questions

Question 1

Problem: Define the unit of current.

Solution:

The unit of current is ampere (A). One ampere is defined as the flow of one coulomb of charge per second through a conductor. It is given by:

\( I = \frac{Q}{t} \)

Where:

• \( Q \): Electric charge in coulombs (C)
• \( t \): Time in seconds (s)

Answer: The unit of current is \( \text{A (ampere)} \).

Question 2

Problem: Calculate the number of electrons constituting one coulomb of charge.

Solution:

• Charge of one electron (\( e \)) = \( 1.6 \times 10^{-19} \, C \)
• Number of electrons (\( n \)) = \( \frac{1}{e} \)
• \( n = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18} \)

Answer: \( 6.25 \times 10^{18} \) electrons make up one coulomb of charge.

Example 11.3

Problem: A potential difference of 20 V is applied across a conductor of resistance 4 Ω. Calculate the current flowing through the conductor.

Solution:

• Given:
  • Potential difference (\( V \)) = 20 V
  • Resistance (\( R \)) = 4 Ω
• Using Ohm’s Law: \( V = IR \), rearrange to find \( I \):
• \( I = \frac{V}{R} = \frac{20}{4} = 5 \, \text{A} \)

Answer: The current flowing through the conductor is \( 5 \, \text{A} \).

Example 11.4

Problem: A wire of length 2 m and cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) has a resistance of 10 Ω. Calculate the resistivity of the material.

Solution:

• Given:
  • Length (\( l \)) = 2 m
  • Area (\( A \)) = \( 1 \times 10^{-6} \, \text{m}^2 \)
  • Resistance (\( R \)) = 10 Ω
• Using the formula for resistance: \( R = \rho \frac{l}{A} \), rearrange to find \( \rho \):
• \( \rho = R \cdot \frac{A}{l} = 10 \cdot \frac{1 \times 10^{-6}}{2} = 5 \times 10^{-6} \, \Omega \, \text{m} \)

Answer: The resistivity of the material is \( 5 \times 10^{-6} \, \Omega \, \text{m} \).

More End-of-Chapter Questions

Question 3

Problem: How much energy is consumed when a 500 W device runs for 3 hours?

Solution:

• Given:
  • Power (\( P \)) = 500 W = \( 0.5 \, \text{kW} \)
  • Time (\( t \)) = 3 hours
• Energy consumed (\( E \)) = \( P \times t = 0.5 \times 3 = 1.5 \, \text{kWh} \)

Answer: The energy consumed is \( 1.5 \, \text{kWh} \).

Question 4

Problem: Derive the formula for resistors in series.

Solution:

• When resistors \( R_1, R_2, \) and \( R_3 \) are connected in series, the total resistance (\( R_s \)) is the sum of individual resistances:
• Total voltage (\( V \)) = \( V_1 + V_2 + V_3 \)
• Using Ohm’s Law: \( V = IR, \, V_1 = IR_1, \, V_2 = IR_2, \, V_3 = IR_3 \)
• Substituting into the total voltage equation:
• \( IR_s = IR_1 + IR_2 + IR_3 \)
• Cancel \( I \) (common factor):
• \( R_s = R_1 + R_2 + R_3 \)

Answer: For resistors in series, \( R_s = R_1 + R_2 + R_3 \).

Example 11.5

Problem: Two resistors of 6 Ω and 12 Ω are connected in parallel. Calculate the total resistance of the combination.

Solution:

• For parallel resistors, the total resistance is given by:
• \( \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} \)
• \( \frac{1}{R_{\text{total}}} = \frac{1}{6} + \frac{1}{12} \)
• Finding the LCM of 6 and 12 gives 12:
• \( \frac{1}{R_{\text{total}}} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \)
• \( R_{\text{total}} = 4 \, \Omega \)

Answer: The total resistance is \( 4 \, \Omega \).

Example 11.6

Problem: A 100 W, 220 V bulb is connected to a 110 V supply. Calculate the power consumed by the bulb.

Solution:

• First, calculate the resistance of the bulb:
• \( P = \frac{V^2}{R} \), rearranging gives \( R = \frac{V^2}{P} \)
• \( R = \frac{220^2}{100} = \frac{48400}{100} = 484 \, \Omega \)
• Now, calculate power consumed with the new voltage:
• \( P = \frac{V^2}{R} \)
• \( P = \frac{110^2}{484} = \frac{12100}{484} \approx 25 \, \text{W} \)

Answer: The power consumed by the bulb is \( 25 \, \text{W} \).

Next Set of End-of-Chapter Questions

Question 5

Problem: Derive the formula for resistors in parallel.

Solution:

• For resistors \( R_1, R_2, \) and \( R_3 \) connected in parallel, the total current is:
• \( I = I_1 + I_2 + I_3 \)
• Using Ohm’s Law: \( I = \frac{V}{R_{\text{total}}}, I_1 = \frac{V}{R_1}, I_2 = \frac{V}{R_2}, I_3 = \frac{V}{R_3} \)
• Substituting into the total current equation:
• \( \frac{V}{R_{\text{total}}} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} \)
• Cancel \( V \) (common factor):
• \( \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \)

Answer: For resistors in parallel, \( \frac{1}{R_{\text{total}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \).

Question 6

Problem: A heater rated at 1000 W operates on a 250 V supply. Calculate the current drawn and its resistance.

Solution:

• Given:
  • Power (\( P \)) = 1000 W
  • Voltage (\( V \)) = 250 V
• Current (\( I \)) is given by \( I = \frac{P}{V} \):
• \( I = \frac{1000}{250} = 4 \, \text{A} \)
• Resistance (\( R \)) is given by \( R = \frac{V}{I} \):
• \( R = \frac{250}{4} = 62.5 \, \Omega \)

Answer: The current drawn is \( 4 \, \text{A} \) and the resistance is \( 62.5 \, \Omega \).

Example 11.7

Problem: A wire of resistance 5 Ω is stretched to double its length. What will be its new resistance?

Solution:

• When a wire is stretched, its length \( l \) doubles, and its cross-sectional area \( A \) is halved (constant volume).
• Resistance \( R \) is given by \( R = \rho \frac{l}{A} \).
• If \( l \to 2l \) and \( A \to \frac{A}{2} \), the new resistance \( R' \) becomes:
• \( R' = \rho \frac{2l}{A/2} = 4 \rho \frac{l}{A} = 4R \).
• \( R' = 4 \times 5 = 20 \, \Omega \).

Answer: The new resistance is \( 20 \, \Omega \).

Example 11.8

Problem: Calculate the heat produced in a resistor of 10 Ω when a current of 2 A flows through it for 5 minutes.

Solution:

• Given:
  • Resistance (\( R \)) = 10 Ω
  • Current (\( I \)) = 2 A
  • Time (\( t \)) = 5 minutes = \( 5 \times 60 = 300 \, \text{seconds} \)
• Using Joule’s law of heating: \( H = I^2 R t \)
• \( H = (2)^2 \cdot 10 \cdot 300 = 4 \cdot 10 \cdot 300 = 12000 \, \text{J} \).

Answer: The heat produced is \( 12000 \, \text{J} \) or \( 12 \, \text{kJ} \).

Next Set of End-of-Chapter Questions

Question 7

Problem: Why are electric appliances connected in parallel in household circuits?

Solution:

• In parallel connections:
  • Each appliance operates independently.
  • The failure of one appliance does not affect others.
  • The voltage across each appliance is the same as the supply voltage.
• It allows individual appliances to draw their required current without affecting others.

Answer: Parallel connections ensure independent operation, constant voltage, and safety in household circuits.

Question 8

Problem: A 500 W heater is used for 2 hours daily. Calculate the energy consumed in 10 days.

Solution:

• Given:
  • Power (\( P \)) = 500 W = \( 0.5 \, \text{kW} \)
  • Time (\( t \)) = \( 2 \, \text{hours/day} \times 10 \, \text{days} = 20 \, \text{hours} \)
• Energy consumed (\( E \)) = \( P \cdot t = 0.5 \cdot 20 = 10 \, \text{kWh} \).

Answer: The energy consumed is \( 10 \, \text{kWh} \).

Example 11.9

Problem: A 100 W bulb is used for 10 hours. Calculate the total energy consumed in kilowatt-hours (kWh).

Solution:

• Given:
  • Power (\( P \)) = 100 W = \( 0.1 \, \text{kW} \)
  • Time (\( t \)) = 10 hours
• Energy consumed (\( E \)) = \( P \times t \):
• \( E = 0.1 \times 10 = 1 \, \text{kWh} \).

Answer: The total energy consumed is \( 1 \, \text{kWh} \).

Example 11.10

Problem: Calculate the total resistance of three resistors \( 10 \, \Omega \), \( 20 \, \Omega \), and \( 30 \, \Omega \) connected in series.

Solution:

• For resistors in series, total resistance \( R_s \) is:
• \( R_s = R_1 + R_2 + R_3 \)
• \( R_s = 10 + 20 + 30 = 60 \, \Omega \)

Answer: The total resistance is \( 60 \, \Omega \).

Next Set of End-of-Chapter Questions

Question 9

Problem: Two resistors of \( 4 \, \Omega \) and \( 6 \, \Omega \) are connected in parallel. This combination is connected in series with a \( 5 \, \Omega \) resistor. Find the total resistance.

Solution:

• First, calculate the equivalent resistance of the parallel combination:
• \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{4} + \frac{1}{6} \)
• Finding the LCM of 4 and 6 gives 12:
• \( \frac{1}{R_p} = \frac{3}{12} + \frac{2}{12} = \frac{5}{12} \)
• \( R_p = \frac{12}{5} = 2.4 \, \Omega \)
• Add the series resistor:
• \( R_{\text{total}} = R_p + R_3 = 2.4 + 5 = 7.4 \, \Omega \)

Answer: The total resistance is \( 7.4 \, \Omega \).

Question 10

Problem: Explain the factors affecting the resistance of a conductor.

Solution:

• Resistance (\( R \)) of a conductor depends on:
  • Length (\( l \)): Resistance is directly proportional to length (\( R \propto l \)).
  • Cross-sectional area (\( A \)): Resistance is inversely proportional to the area (\( R \propto \frac{1}{A} \)).
  • Material: Conductors with low resistivity (e.g., copper) have lower resistance, while insulators have high resistance.
  • Temperature: For most metals, resistance increases with temperature.

Answer: Resistance depends on length, cross-sectional area, material, and temperature of the conductor.

Extra Questions

1. Explain the significance of the SI unit of current.
2. Why is the direction of current taken as opposite to the flow of electrons?
3. A circuit carries 3 C of charge in 2 seconds. What is the current?

Extra Questions with Answers and Hints

Extra Question 1

Problem: A 60 W bulb and a 100 W bulb are connected in series to a 220 V supply. Which bulb will glow brighter?

Hint: The brightness of a bulb depends on the power it dissipates. For series connections, both bulbs have the same current flowing through them.

Solution:

• Given:
  • Power of bulb 1 (\( P_1 \)) = 60 W
  • Power of bulb 2 (\( P_2 \)) = 100 W
• Resistance of a bulb (\( R \)) is given by \( R = \frac{V^2}{P} \):
• \( R_1 = \frac{220^2}{60} \approx 806.67 \, \Omega \), \( R_2 = \frac{220^2}{100} \approx 484 \, \Omega \)
• In a series circuit, the power dissipated is \( P = I^2 R \). Since \( I \) is the same, the bulb with higher resistance (\( R_1 \)) will glow brighter.

Answer: The 60 W bulb will glow brighter.

Extra Question 2

Problem: Calculate the drift velocity of electrons in a copper wire of cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) carrying a current of 3 A. Assume the number density of electrons (\( n \)) is \( 8.5 \times 10^{28} \, \text{m}^{-3} \) and charge of an electron (\( e \)) is \( 1.6 \times 10^{-19} \, \text{C} \).

Hint: Drift velocity (\( v_d \)) is given by \( v_d = \frac{I}{n \cdot A \cdot e} \).

Solution:

• Given:
  • Current (\( I \)) = 3 A
  • Area (\( A \)) = \( 1 \times 10^{-6} \, \text{m}^2 \)
  • Number density (\( n \)) = \( 8.5 \times 10^{28} \, \text{m}^{-3} \)
  • Charge of an electron (\( e \)) = \( 1.6 \times 10^{-19} \, \text{C} \)
• Using the formula \( v_d = \frac{I}{n \cdot A \cdot e} \):
• \( v_d = \frac{3}{8.5 \times 10^{28} \cdot 1 \times 10^{-6} \cdot 1.6 \times 10^{-19}} \)
• \( v_d = \frac{3}{1.36 \times 10^4} \approx 2.2 \times 10^{-4} \, \text{m/s} \)

Answer: The drift velocity is \( 2.2 \times 10^{-4} \, \text{m/s} \).

Extra Question 3

Problem: Why are thick copper wires used in household wiring?

Hint: Consider the relationship between resistance and cross-sectional area.

Solution:

• Copper has low resistivity, which minimizes energy loss during power transmission.
• Thicker wires have a larger cross-sectional area, reducing resistance (\( R \propto \frac{1}{A} \)).
• Lower resistance ensures safe and efficient power delivery without overheating.

Answer: Thick copper wires are used because they have low resistance, ensuring efficient and safe power transmission.

Extra Question 4

Problem: Explain why nichrome is used as a heating element in appliances like toasters.

Hint: Consider the resistivity, melting point, and heating effect of nichrome.

Solution:

• Nichrome has high resistivity, which produces more heat for the same current as compared to other metals.
• It has a high melting point, allowing it to operate at high temperatures without melting.
• It does not oxidize easily, ensuring long-term durability as a heating element.

Answer: Nichrome is used for its high resistivity, high melting point, and resistance to oxidation.

Extra Questions

Question 1

Problem: A 60 W bulb and a 100 W bulb are connected in series to a 220 V supply. Which bulb will glow brighter?

Hint: The brightness of a bulb depends on the power it dissipates. For series connections, the same current flows through both bulbs.

Question 2

Problem: Calculate the drift velocity of electrons in a copper wire of cross-sectional area \( 1 \times 10^{-6} \, \text{m}^2 \) carrying a current of 3 A. Assume the number density of electrons (\( n \)) is \( 8.5 \times 10^{28} \, \text{m}^{-3} \) and charge of an electron (\( e \)) is \( 1.6 \times 10^{-19} \, \text{C} \).

Hint: Use the formula \( v_d = \frac{I}{n \cdot A \cdot e} \).

Question 3

Problem: Why are thick copper wires used in household wiring?

Hint: Consider the relationship between resistance and cross-sectional area.

Question 4

Problem: Explain why nichrome is used as a heating element in appliances like toasters.

Hint: Think about resistivity, melting point, and oxidation resistance of nichrome.

Question 5

Problem: Two resistors of 10 Ω and 15 Ω are connected in parallel. The combination is connected in series with a 5 Ω resistor. Find the total resistance.

Hint: First, calculate the equivalent resistance of the parallel resistors using \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \), then add the series resistor.

Question 6

Problem: A wire is stretched to twice its original length. How does its resistance change?

Hint: Resistance is proportional to the square of the length when the volume remains constant.

Question 7

Problem: Why are household appliances connected in parallel instead of series?

Hint: Think about voltage distribution and the independence of devices in parallel connections.

Question 8

Problem: A 100 W bulb is connected to a 220 V supply. Calculate the current drawn and the resistance of the bulb.

Hint: Use \( P = VI \) to calculate current, then \( R = \frac{V}{I} \) to find resistance.

Question 9

Problem: A heater rated at 1500 W operates on a 220 V supply. How much energy does it consume in 5 hours?

Hint: Energy consumed is calculated as \( E = P \cdot t \). Ensure units are consistent.

Question 10

Problem: Explain why resistance of a conductor increases with temperature.

Hint: Consider how temperature affects the motion of electrons and the frequency of collisions in a conductor.

Question 11

Problem: A copper wire has a resistance of 10 Ω. What will happen to its resistance if the wire is cut into four equal parts and then connected in parallel?

Hint: The resistance of each part is proportional to its length. Use the formula for parallel resistance: \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \dots \).

Question 12

Problem: A 220 V electric iron draws a current of 5 A. Calculate the power consumed and the heat produced in 30 minutes.

Hint: Use \( P = VI \) to calculate power. Heat produced is \( H = P \cdot t \), where \( t \) is in seconds.

Question 13

Problem: Derive the formula for the resistivity of a material in terms of resistance, length, and cross-sectional area.

Hint: Start from the formula \( R = \rho \frac{l}{A} \) and rearrange to solve for \( \rho \).

Question 14

Problem: A 2 kW electric heater is used for 3 hours daily for a month. Calculate the total energy consumed and its cost if electricity costs \( 5 \, \text{Rs.}/\text{kWh} \).

Hint: Energy consumed is \( P \cdot t \), where \( t \) is the total time in hours. Multiply by the cost per unit to find the total cost.

Question 15

Problem: Why are alloys like nichrome used in heating elements instead of pure metals like copper?

Hint: Consider the properties of nichrome, such as high resistivity, durability, and resistance to oxidation.

Question 16

Problem: Two bulbs rated 60 W and 100 W are connected in parallel to a 220 V supply. Which bulb will consume more power?

Hint: In a parallel connection, voltage across each bulb is the same. Use \( P = \frac{V^2}{R} \) to compare power dissipation.

Question 17

Problem: A 12 V battery is connected across a resistor, and a current of 2.5 mA flows through it. Calculate the resistance of the resistor.

Hint: Use Ohm’s law: \( V = IR \). Rearrange to find \( R \).

Question 18

Problem: Derive the expression for heat produced in a resistor using Joule’s law of heating.

Hint: Start with \( V = IR \), then substitute \( P = VI \) into \( H = P \cdot t \).

Question 19

Problem: Explain why metals like copper and aluminum are used for electrical wiring, but not alloys.

Hint: Consider the resistivity, conductivity, and cost-effectiveness of metals versus alloys.

Question 20

Problem: A resistor of \( 5 \, \Omega \) is connected in series with another resistor of \( 10 \, \Omega \). This combination is connected to a 24 V battery. Calculate the current through the circuit.

Hint: First calculate the total resistance (\( R_s = R_1 + R_2 \)), then use \( I = \frac{V}{R} \).

Question 21

Problem: A lamp is marked 220 V, 60 W. Calculate its resistance and the current flowing through it when connected to a 220 V supply.

Hint: Use \( P = \frac{V^2}{R} \) to find resistance, then \( I = \frac{V}{R} \) for current.

Question 22

Problem: Why do we use a fuse in an electric circuit? Explain the working principle of a fuse.

Hint: Think about the role of a fuse in protecting a circuit from excessive current. Consider its low melting point.

More Solved Questions

1. What does an electric circuit mean?

Answer:

An electric circuit is a continuous and closed path through which an electric current flows. It typically consists of components such as a cell or battery, connecting wires, and other elements like resistors, switches, or bulbs.

Example: In a torch, when the switch is turned on, the circuit becomes complete, allowing current to flow and light the bulb.

2. Define the unit of current.

Answer:

The unit of current is the ampere (A). One ampere is defined as the flow of one coulomb of charge per second through a conductor.

Formula: \( I = \frac{Q}{t} \)

Here, \( Q \) is the charge in coulombs, and \( t \) is the time in seconds.

3. Calculate the number of electrons constituting one coulomb of charge.

Solution:

The charge of one electron is \( 1.6 \times 10^{-19} \, \text{C} \).

Number of electrons in 1 coulomb of charge:

\[ n = \frac{1 \, \text{C}}{1.6 \times 10^{-19} \, \text{C/electron}} = 6.25 \times 10^{18} \, \text{electrons}. \]

Answer: Approximately \( 6.25 \times 10^{18} \) electrons make up one coulomb of charge.

Example: Calculate the current through a filament

Problem: A current of 0.5 A flows through a filament for 10 minutes. Calculate the total charge passing through the filament.

Solution:

We use the formula: \( Q = I \times t \).

Here, \( I = 0.5 \, \text{A}, \, t = 10 \, \text{minutes} = 600 \, \text{seconds} \).

\( Q = 0.5 \times 600 = 300 \, \text{C} \).

Answer: 300 coulombs of charge pass through the filament.

4. How much work is done in moving a charge of 2 C across two points having a potential difference of 12 V?

Solution:

We use the formula: \( V = \frac{W}{Q} \).

Rearranging, \( W = V \times Q \).

\( V = 12 \, \text{V}, \, Q = 2 \, \text{C} \).

\( W = 12 \times 2 = 24 \, \text{J} \).

Answer: The work done is 24 joules.

5. A 220 V electric heater draws a current of 5 A. Calculate its resistance.

Solution:

We use Ohm's Law: \( V = IR \).

Rearranging, \( R = \frac{V}{I} \).

\( V = 220 \, \text{V}, \, I = 5 \, \text{A} \).

\( R = \frac{220}{5} = 44 \, \Omega \).

Answer: The resistance of the heater is \( 44 \, \Omega \).

6. A current of 3 A flows through a conductor with a resistance of 15 Ω. Find the potential difference across the conductor.

Solution:

From Ohm's Law, \( V = IR \).

\( I = 3 \, A, \, R = 15 \, \Omega \).

\( V = 3 \times 15 = 45 \, V \).

Answer: The potential difference is 45 volts.

7. An electric bulb is rated 220 V and 100 W. What is its resistance?

Solution:

We use \( P = \frac{V^2}{R} \). Rearranging, \( R = \frac{V^2}{P} \).

\( V = 220 \, V, \, P = 100 \, W \).

\( R = \frac{220^2}{100} = \frac{48400}{100} = 484 \, \Omega \).

Answer: The resistance is \( 484 \, \Omega \).

8. A wire has a resistance of 10 Ω. If the length is doubled and the cross-sectional area is halved, what will be the new resistance?

Solution:

The resistance \( R \) is given by \( R = \rho \frac{l}{A} \), where \( l \) is the length and \( A \) is the cross-sectional area.

If \( l \) is doubled and \( A \) is halved, the new resistance \( R' \) becomes:

\( R' = \rho \frac{2l}{A/2} = 4 \rho \frac{l}{A} = 4R \).

\( R' = 4 \times 10 = 40 \, \Omega \).

Answer: The new resistance is \( 40 \, \Omega \).

9. A 100 W electric lamp is used for 6 hours. Calculate the energy consumed in kWh.

Solution:

Energy consumed \( E = P \times t \), where \( P \) is power (kW) and \( t \) is time (hours).

\( P = 100 \, W = 0.1 \, \text{kW}, \, t = 6 \, \text{hours} \).

\( E = 0.1 \times 6 = 0.6 \, \text{kWh} \).

Answer: The energy consumed is \( 0.6 \, \text{kWh} \).

10. Explain why the resistance of a conductor increases with temperature.

Answer:

As the temperature of a conductor increases, the atoms vibrate more vigorously. This increased vibration hinders the free flow of electrons, causing an increase in resistance. This effect is prominent in metals.

11. A 12 V battery is connected across a resistor. If the current through the circuit is 2.5 mA, find the resistance of the resistor.

Solution:

We use Ohm's Law: \( V = IR \). Rearranging, \( R = \frac{V}{I} \).

\( V = 12 \, \text{V}, \, I = 2.5 \, \text{mA} = 2.5 \times 10^{-3} \, \text{A} \).

\( R = \frac{12}{2.5 \times 10^{-3}} = \frac{12}{0.0025} = 4800 \, \Omega \).

Answer: The resistance of the resistor is \( 4800 \, \Omega \) or \( 4.8 \, \text{k}\Omega \).

12. Two resistors of 5 Ω and 10 Ω are connected in series. Calculate the total resistance.

Solution:

For resistors in series, \( R_s = R_1 + R_2 \).

\( R_s = 5 + 10 = 15 \, \Omega \).

Answer: The total resistance is \( 15 \, \Omega \).

13. Two resistors of 6 Ω and 12 Ω are connected in parallel. Calculate the equivalent resistance.

Solution:

For resistors in parallel, \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} \).

\( \frac{1}{R_p} = \frac{1}{6} + \frac{1}{12} = \frac{2}{12} + \frac{1}{12} = \frac{3}{12} = \frac{1}{4} \).

\( R_p = 4 \, \Omega \).

Answer: The equivalent resistance is \( 4 \, \Omega \).

14. How much energy is consumed by a 500 W heater used for 3 hours?

Solution:

Energy consumed \( E = P \times t \).

\( P = 500 \, \text{W} = 0.5 \, \text{kW}, \, t = 3 \, \text{hours} \).

\( E = 0.5 \times 3 = 1.5 \, \text{kWh} \).

Answer: The energy consumed is \( 1.5 \, \text{kWh} \).

15. A wire of resistance 20 Ω is stretched to double its original length. What is the new resistance?

Solution:

When the length of a wire is doubled, its resistance increases by a factor of 4 because resistance is proportional to the square of the length (for constant volume).

\( R_{\text{new}} = 4 \times R_{\text{original}} = 4 \times 20 = 80 \, \Omega \).

Answer: The new resistance is \( 80 \, \Omega \).

16. Three resistors of 4 Ω, 6 Ω, and 8 Ω are connected in parallel. Calculate the total resistance.

Solution:

For resistors in parallel, \( \frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \).

\( \frac{1}{R_p} = \frac{1}{4} + \frac{1}{6} + \frac{1}{8} \).

Finding the LCM of 4, 6, and 8 gives 24:

\( \frac{1}{R_p} = \frac{6}{24} + \frac{4}{24} + \frac{3}{24} = \frac{13}{24} \).

\( R_p = \frac{24}{13} = 1.85 \, \Omega \) (approximately).

Answer: The total resistance is approximately \( 1.85 \, \Omega \).

17. An electric motor operates on 220 V and draws a current of 2 A. Calculate its power and energy consumed in 5 hours.

Solution:

Power \( P = VI \).

\( V = 220 \, \text{V}, \, I = 2 \, \text{A} \).

\( P = 220 \times 2 = 440 \, \text{W} = 0.44 \, \text{kW} \).

Energy consumed \( E = P \times t \).

\( t = 5 \, \text{hours} \).

\( E = 0.44 \times 5 = 2.2 \, \text{kWh} \).

Answer: The motor’s power is \( 440 \, \text{W} \), and the energy consumed is \( 2.2 \, \text{kWh} \).

18. Why is tungsten used as the filament in electric bulbs?

Answer:

Tungsten is used as the filament in electric bulbs because:

• It has a high melting point (3380°C), allowing it to withstand high temperatures.
• It emits bright light when heated.
• It has high resistivity, producing the required heating effect.

19. Why are conductors of electric heating devices, such as toasters and irons, made of alloys?

Answer:

Alloys are used because:

• They have higher resistivity than pure metals, producing more heat.
• They do not oxidize or burn easily at high temperatures.
• They have better durability and stability at elevated temperatures.

20. A 220 V, 60 W bulb is connected to a 110 V supply. Calculate the power consumed by the bulb.

Solution:

Power consumed is calculated using \( P = \frac{V^2}{R} \). First, find \( R \):

\( R = \frac{V^2}{P} = \frac{220^2}{60} = \frac{48400}{60} \approx 806.67 \, \Omega \).

Now calculate power with \( V = 110 \, \text{V} \):

\( P = \frac{V^2}{R} = \frac{110^2}{806.67} = \frac{12100}{806.67} \approx 15 \, \text{W} \).

Answer: The power consumed by the bulb is approximately \( 15 \, \text{W} \).

21. An electric iron has a rating of 1000 W, 220 V. Calculate the current drawn and its resistance.

Solution:

Power \( P = VI \), rearranging: \( I = \frac{P}{V} \).

\( P = 1000 \, \text{W}, \, V = 220 \, \text{V} \).

\( I = \frac{1000}{220} \approx 4.55 \, \text{A} \).

Resistance \( R = \frac{V}{I} = \frac{220}{4.55} \approx 48.35 \, \Omega \).

Answer: The current is \( 4.55 \, \text{A} \) and the resistance is \( 48.35 \, \Omega \).

22. Calculate the total resistance of a combination of resistors: 3 Ω and 6 Ω in series, and this combination in parallel with 2 Ω.

Solution:

First, find the resistance in series:

\( R_s = 3 + 6 = 9 \, \Omega \).

Now, find the equivalent resistance in parallel with \( R_p = 2 \, \Omega \):

\( \frac{1}{R_{\text{total}}} = \frac{1}{R_s} + \frac{1}{R_p} = \frac{1}{9} + \frac{1}{2} \).

\( \frac{1}{R_{\text{total}}} = \frac{2}{18} + \frac{9}{18} = \frac{11}{18} \).

\( R_{\text{total}} = \frac{18}{11} \approx 1.64 \, \Omega \).

Answer: The total resistance is approximately \( 1.64 \, \Omega \).

23. An electric motor takes 8 A from a 220 V line. Determine its power and energy consumed in 2 hours.

Solution:

Power \( P = VI \):

\( V = 220 \, \text{V}, \, I = 8 \, \text{A} \).

\( P = 220 \times 8 = 1760 \, \text{W} = 1.76 \, \text{kW} \).

Energy consumed \( E = P \times t \):

\( t = 2 \, \text{hours} \), \( E = 1.76 \times 2 = 3.52 \, \text{kWh} \).

Answer: The motor’s power is \( 1.76 \, \text{kW} \), and the energy consumed is \( 3.52 \, \text{kWh} \).

24. A 1.5 V battery is connected to a 10 Ω resistor. Find the current and power delivered to the resistor.

Solution:

Current \( I = \frac{V}{R} \):

\( V = 1.5 \, \text{V}, \, R = 10 \, \Omega \).

\( I = \frac{1.5}{10} = 0.15 \, \text{A} \).

Power \( P = VI \):

\( P = 1.5 \times 0.15 = 0.225 \, \text{W} \).

Answer: The current is \( 0.15 \, \text{A} \) and the power is \( 0.225 \, \text{W} \).

25. Why are domestic circuits connected in parallel?

Answer:

• In parallel circuits, each device operates independently. If one device fails, others continue to work.
• The voltage across each device remains the same, ensuring proper operation.
• Parallel connections allow each device to draw the current it requires, improving efficiency.

26. Two wires of the same material have lengths \( l_1 = l_2 \) and cross-sectional areas \( A_1 = 2A_2 \). Compare their resistances.

Solution:

Resistance \( R \) is given by \( R = \rho \frac{l}{A} \).

For the two wires:

\( R_1 = \rho \frac{l_1}{A_1} \) and \( R_2 = \rho \frac{l_2}{A_2} \).

Given \( l_1 = l_2 \) and \( A_1 = 2A_2 \), we have:

\( \frac{R_1}{R_2} = \frac{\rho \frac{l_1}{A_1}}{\rho \frac{l_2}{A_2}} = \frac{A_2}{A_1} = \frac{1}{2} \).

Answer: \( R_1 : R_2 = 1 : 2 \).

27. A toaster with a power rating of 1000 W operates at 250 V. Calculate its resistance and the current drawn.

Solution:

From \( P = \frac{V^2}{R} \), rearrange to find \( R \):

\( R = \frac{V^2}{P} = \frac{250^2}{1000} = \frac{62500}{1000} = 62.5 \, \Omega \).

Current \( I = \frac{V}{R} \):

\( I = \frac{250}{62.5} = 4 \, \text{A} \).

Answer: The resistance is \( 62.5 \, \Omega \), and the current drawn is \( 4 \, \text{A} \).

28. Why is the cord of an electric heater not hot while its heating element glows?

Answer:

The resistance of the cord is much lower than the resistance of the heating element. According to Joule’s law of heating (\( H = I^2 R t \)), heat produced depends on resistance. Since the cord has a low resistance, it produces negligible heat, while the heating element with high resistance glows due to significant heat production.

29. A 10 Ω resistor and a 15 Ω resistor are connected in parallel. This combination is connected in series with a 5 Ω resistor. Find the total resistance.

Solution:

First, calculate the equivalent resistance of the parallel combination:

\( \frac{1}{R_p} = \frac{1}{10} + \frac{1}{15} = \frac{3}{30} + \frac{2}{30} = \frac{5}{30} = \frac{1}{6} \).

\( R_p = 6 \, \Omega \).

Now, add the series resistor:

\( R_{\text{total}} = R_p + 5 = 6 + 5 = 11 \, \Omega \).

Answer: The total resistance is \( 11 \, \Omega \).

30. Calculate the energy consumed by a 60 W bulb operating for 4 hours daily over 30 days.

Solution:

Energy consumed \( E = P \times t \), where \( P = 60 \, \text{W} = 0.06 \, \text{kW} \), and \( t \) is total hours.

Daily usage: \( 4 \, \text{hours} \). Total usage: \( 4 \times 30 = 120 \, \text{hours} \).

\( E = 0.06 \times 120 = 7.2 \, \text{kWh} \).

Answer: The energy consumed is \( 7.2 \, \text{kWh} \).

31. Explain why tungsten is not used in electric transmission lines.

Answer:

Tungsten is not used in electric transmission lines because:

• It has very high resistance, leading to significant energy loss as heat.
• Its weight makes it unsuitable for long-distance overhead lines.
• Conductors like copper and aluminum, with lower resistances, are preferred for efficient power transmission.

32. Derive the expression for heat produced in a conductor using Joule’s law of heating.

Solution:

Joule’s law states that heat produced in a conductor is:

\( H = I^2 R t \), where \( H \) is heat, \( I \) is current, \( R \) is resistance, and \( t \) is time.

• The work done by the current (\( W \)) is equal to \( VIt \), where \( V \) is voltage.
• From Ohm's law, \( V = IR \).
• Substituting \( V = IR \) into \( W = VIt \), we get \( W = I(IR)t = I^2Rt \).
• Thus, \( H = I^2 R t \).

Answer: The heat produced in a conductor is \( H = I^2 R t \).

33. A 5 Ω resistor is connected to a 12 V battery. Find the current and the heat produced in 2 minutes.

Solution:

Current \( I = \frac{V}{R} \):

\( V = 12 \, \text{V}, \, R = 5 \, \Omega \).

\( I = \frac{12}{5} = 2.4 \, \text{A} \).

Heat produced \( H = I^2 R t \):

\( t = 2 \, \text{minutes} = 120 \, \text{seconds} \).

\( H = (2.4)^2 \times 5 \times 120 = 13.824 \times 600 = 8294.4 \, \text{J} \).

Answer: The current is \( 2.4 \, \text{A} \), and the heat produced is \( 8294.4 \, \text{J} \).

34. Derive the formula for the equivalent resistance of three resistors connected in parallel.

Solution:

For three resistors \( R_1 \), \( R_2 \), and \( R_3 \) connected in parallel, the total current \( I \) is the sum of currents through each resistor:

\( I = I_1 + I_2 + I_3 \).

Using Ohm’s law, \( I = \frac{V}{R_{\text{eq}}}, \, I_1 = \frac{V}{R_1}, \, I_2 = \frac{V}{R_2}, \, I_3 = \frac{V}{R_3} \):

\( \frac{V}{R_{\text{eq}}} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3} \).

Dividing through by \( V \):

\( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \).

Answer: The equivalent resistance for resistors in parallel is:

\( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \).

35. Calculate the power dissipated in a resistor when 3 A of current flows through it, and its resistance is 5 Ω.

Solution:

Power \( P = I^2 R \):

\( I = 3 \, \text{A}, \, R = 5 \, \Omega \).

\( P = (3)^2 \times 5 = 9 \times 5 = 45 \, \text{W} \).

Answer: The power dissipated is \( 45 \, \text{W} \).

36. Explain why appliances like electric bulbs and fans are connected in parallel in domestic wiring.

Answer:

• In parallel connections, each appliance gets the same voltage as the supply voltage.
• If one appliance fails, others can still operate because they have independent circuits.
• Appliances can be operated independently without affecting the performance of others.

37. A heater of resistance 50 Ω is connected to a 220 V supply. Calculate the current and heat produced in 10 minutes.

Solution:

Current \( I = \frac{V}{R} \):

\( V = 220 \, \text{V}, \, R = 50 \, \Omega \).

\( I = \frac{220}{50} = 4.4 \, \text{A} \).

Heat \( H = I^2 R t \):

\( t = 10 \, \text{minutes} = 600 \, \text{seconds} \).

\( H = (4.4)^2 \times 50 \times 600 = 19.36 \times 50 \times 600 = 580800 \, \text{J} \).

Answer: The current is \( 4.4 \, \text{A} \), and the heat produced is \( 580800 \, \text{J} \) or \( 580.8 \, \text{kJ} \).

38. A 40 W lamp is used for 5 hours daily. Calculate its energy consumption in 15 days.

Solution:

Energy consumed \( E = P \times t \), where \( P = 40 \, \text{W} = 0.04 \, \text{kW} \), and \( t \) is total time in hours.

Daily usage: \( 5 \, \text{hours} \). Total usage: \( 5 \times 15 = 75 \, \text{hours} \).

\( E = 0.04 \times 75 = 3.0 \, \text{kWh} \).

Answer: The energy consumption is \( 3.0 \, \text{kWh} \).

39. Explain why the resistance of alloys is higher than that of pure metals.

Answer:

• In alloys, the arrangement of atoms is irregular, causing more collisions between electrons and atoms.
• This increases resistivity, as the free movement of electrons is hindered.
• Alloys like nichrome and manganin are specifically used in heating devices due to their higher resistance and stability at high temperatures.

40. Two 12 Ω resistors are connected in parallel, and this combination is connected in series with a 6 Ω resistor. Calculate the total resistance.

Solution:

First, calculate the equivalent resistance of the parallel combination:

\( \frac{1}{R_p} = \frac{1}{12} + \frac{1}{12} = \frac{2}{12} = \frac{1}{6} \).

\( R_p = 6 \, \Omega \).

Now, add the series resistor:

\( R_{\text{total}} = R_p + 6 = 6 + 6 = 12 \, \Omega \).

Answer: The total resistance is \( 12 \, \Omega \).

41. Derive the relationship between power, current, and resistance.

Solution:

We know that:

\( P = VI \) and \( V = IR \) (from Ohm’s Law).

Substituting \( V = IR \) into \( P = VI \):

\( P = (IR)I = I^2R \).

Thus, \( P = I^2R \).

Alternatively, using \( I = \frac{V}{R} \):

\( P = V \left( \frac{V}{R} \right) = \frac{V^2}{R} \).

Answer: Power can be expressed as \( P = I^2R \) or \( P = \frac{V^2}{R} \).

42. A 60 W electric bulb is connected to a 120 V power supply. Calculate the current and resistance of the bulb.

Solution:

Power \( P = VI \). Rearranging, \( I = \frac{P}{V} \):

\( P = 60 \, \text{W}, \, V = 120 \, \text{V} \).

\( I = \frac{60}{120} = 0.5 \, \text{A} \).

Resistance \( R = \frac{V}{I} = \frac{120}{0.5} = 240 \, \Omega \).

Answer: The current is \( 0.5 \, \text{A} \) and the resistance is \( 240 \, \Omega \).

43. Why are copper and aluminum preferred for overhead power lines?

Answer:

• Copper and aluminum have low resistivity, reducing energy loss during transmission.
• They are ductile, making them easy to draw into wires.
• Aluminum is lightweight and economical, while copper has better conductivity.

44. A 100 Ω resistor is connected across a 220 V power supply. Calculate the power dissipated.

Solution:

Power \( P = \frac{V^2}{R} \):

\( V = 220 \, \text{V}, \, R = 100 \, \Omega \).

\( P = \frac{220^2}{100} = \frac{48400}{100} = 484 \, \text{W} \).

Answer: The power dissipated is \( 484 \, \text{W} \).

45. A 5 Ω and a 10 Ω resistor are connected in series. The combination is connected across a 30 V battery. Calculate the current in the circuit.

Solution:

For resistors in series, total resistance:

\( R_s = R_1 + R_2 = 5 + 10 = 15 \, \Omega \).

Current \( I = \frac{V}{R_s} \):

\( V = 30 \, \text{V}, \, R_s = 15 \, \Omega \).

\( I = \frac{30}{15} = 2 \, \text{A} \).

Answer: The current in the circuit is \( 2 \, \text{A} \).

46. A 2 kW heater is used for 3 hours daily. Calculate the total energy consumed in 10 days.

Solution:

Energy consumed \( E = P \times t \), where \( P = 2 \, \text{kW} \), \( t = \text{total hours} \).

Daily usage: \( 3 \, \text{hours} \). Total usage: \( 3 \times 10 = 30 \, \text{hours} \).

\( E = 2 \times 30 = 60 \, \text{kWh} \).

Answer: The total energy consumed is \( 60 \, \text{kWh} \).

47. Compare the advantages of parallel and series circuits.

Answer:

• Parallel Circuits:
• Each component operates independently.
• Same voltage is supplied to all components.
• Failure of one component does not affect others.
• Series Circuits:
• Easy to set up and requires fewer wires.
• Total resistance increases, limiting current flow.

48. Calculate the equivalent resistance of three resistors (4 Ω, 6 Ω, and 12 Ω) connected in parallel.

Solution:

For parallel resistors:

\( \frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} \).

\( \frac{1}{R_{\text{eq}}} = \frac{1}{4} + \frac{1}{6} + \frac{1}{12} \).

Finding the LCM of 4, 6, and 12 gives 12:

\( \frac{1}{R_{\text{eq}}} = \frac{3}{12} + \frac{2}{12} + \frac{1}{12} = \frac{6}{12} = \frac{1}{2} \).

\( R_{\text{eq}} = 2 \, \Omega \).

Answer: The equivalent resistance is \( 2 \, \Omega \).

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