What is the potential difference across a 6 Ω resistor if a current of 3 A flows through it?

Correct Option: A. Solution: Using Ohm's law, $$V = IR = 3 \times 6 = 18$$ V.

What is the potential difference across a 6 V battery?

Correct Option: A. Solution: The potential difference across a 6 V battery is 6 V.

An electric heater has a resistance of 44 Ω and operates at a current of 5 A. What is the power consumed by the heater?

Correct Option: D. Solution: Power $P$ is given by $P = I^2R = 5^2 \times 44 = 1100 \text{ W}$.

Electricity

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CBSE Learning Objectives – Key Concepts & Skills You Must Know

Understand the concept of electricity and its significance in modern society.
Explain the flow of electric current in a circuit.
Identify the factors that regulate current in an electric circuit.
Describe the heating effect of electric current and its applications.

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter 11: Electricity

Electricity plays a crucial role in modern society, serving as a controllable and convenient energy source for various applications.

11.1 Electric Current and Circuit

Definition: Electric current is the flow of electric charge through a conductor, such as a metallic wire.
Electric Circuit: A continuous and closed path for electric current. If the circuit is broken, the current stops flowing.

Key Concepts

Flow of Charges: In metallic wires, electrons flow, but the conventional direction of current is opposite to the flow of electrons.
Measurement of Current: Electric current (I) is expressed as the amount of charge (Q) flowing through a cross-section per unit time (t).
- Formula: I = Q/t
- SI Unit: Ampere (A)
  - 1 A = 1 C/s
  - Smaller units: 1 mA = 10⁻³ A, 1 µA = 10⁻⁶ A

Example

A current of 0.5 A for 10 minutes results in:
- Q = I × t = 0.5 A × 600 s = 300 C

Resistors in Series and Parallel

Resistors in Series

Current: The same current flows through each resistor in series.
Total Resistance: The total resistance (Rₛ) is the sum of individual resistances (R₁, R₂, R₃).
- Formula: Rₛ = R₁ + R₂ + R₃

Resistors in Parallel

Current: The total current is the sum of the currents through each parallel branch.
Total Resistance: The equivalent resistance (Rₚ) can be calculated using:
- Formula: 1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃

Important Formulas

Formula	Description
I = Q/t	Current as charge per unit time
Rₛ = R₁ + R₂ + R₃	Total resistance in series
1/Rₚ = 1/R₁ + 1/R₂ + 1/R₃	Total resistance in parallel

Diagram Descriptions

Circuit Diagram: A typical electric circuit includes a battery, ammeter, resistors, and a switch. The current flows from the positive terminal of the battery through the circuit.
Graph: A linear graph showing the relationship between potential difference (V) and current (I) indicates Ohm's law.

Exercises

Calculate the equivalent resistance when a wire of resistance R is cut into five equal parts and connected in parallel.
Determine the current drawn from a 220 V supply by two lamps rated 100 W and 60 W connected in parallel.

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips in Electricity

Common Pitfalls

Misunderstanding Electric Current: Students often confuse the flow of electric current with the flow of positive charges. Remember, electric current is defined as the flow of electrons, which are negative charges, and its direction is opposite to the flow of electrons.
Incorrect Circuit Diagrams: When drawing circuit diagrams, students may forget to use the correct symbols for components like batteries, switches, and resistors. Always refer to the standard symbols to avoid confusion.
Confusing Series and Parallel Connections: Many students struggle with the differences between series and parallel circuits. In series, the current is the same through all components, while in parallel, the voltage across each component is the same. Make sure to understand these concepts clearly.
Forgetting Ohm's Law: Students often forget to apply Ohm's Law (V = IR) correctly when calculating voltage, current, or resistance in circuits. Always check if the values are correctly substituted into the formula.

Tips for Success

Practice Drawing Circuit Diagrams: Regularly practice drawing circuit diagrams using the correct symbols. This will help you visualize and understand the circuits better.
Understand the Concepts: Focus on understanding the underlying concepts of electricity, such as current, voltage, resistance, and how they interact in different circuit configurations.
Work on Example Problems: Solve various example problems related to series and parallel circuits, as well as Ohm's Law applications. This will reinforce your understanding and help you avoid common mistakes.
Use Visual Aids: Utilize diagrams and charts to help memorize the relationships between different components in a circuit. Visual aids can make complex concepts easier to grasp.

CBSE Quiz & Practice Test – MCQs, True/False Questions with Solutions

Multiple Choice Questions

Copper

Aluminium

Tungsten

Iron

Correct Answer: C

Solution:

Tungsten is used for the filament of electric lamps because it has a high melting point, which allows it to glow brightly without melting.

2 Ω

3 Ω

4 Ω

5 Ω

Correct Answer: C

Solution:

According to Ohm's law, the resistance

R

can be calculated using the formula

R = \frac{V}{I}

, where

V

is the voltage and

I

is the current. Substituting the given values,

R = \frac{12 \text{ V}}{3 \text{ A}} = 4 \Omega

( )

(·)

Correct Answer: B

Solution:

The symbol (·) represents a closed plug key or switch in a circuit diagram.

Copper

Aluminium

Nichrome

Manganese

Correct Answer: C

Solution:

According to the resistivity values provided, Nichrome has a resistivity of

100 \times 10^{-6} \Omega \text{m}

, which is higher than that of Copper, Aluminium, and Manganese.

18 V

2 V

12 V

6 V

Correct Answer: A

Solution:

Using Ohm's law,

V = IR = 3 \times 6 = 18

2R

4R

R/2

R/4

Correct Answer: B

Solution:

When the wire is stretched to double its original length, its cross-sectional area becomes half (since volume is constant). The new resistance

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

0.5 A

1 A

1.5 A

2 A

Correct Answer: A

Solution:

The total resistance

R_{total}

in series is

R_1 + R_2 + R_3 = 4 + 6 + 12 = 22 \, \Omega

. The current

I = \frac{V}{R_{total}} = \frac{12}{22} \approx 0.545 \, A

. Rounding to one significant figure, the current is approximately 0.5 A.

The current doubles

The current halves

The current remains the same

The current becomes zero

Correct Answer: A

Solution:

According to Ohm's law, if the potential difference is doubled and resistance is constant, the current also doubles.

2 V

3 V

4 V

6 V

Correct Answer: C

Solution:

Potential difference (V) is calculated using the formula V = W/Q. Here, W = 12 J and Q = 3 C, so V = 12/3 = 4 V.

The resistance is halved

The resistance remains the same

The resistance is doubled

The resistance is quadrupled

Correct Answer: A

Solution:

Resistance is inversely proportional to the cross-sectional area, so doubling the area halves the resistance.

120 Ω

240 Ω

60 Ω

30 Ω

Correct Answer: B

Solution:

The power

P

is given by

P = \frac{V^2}{R}

. Solving for

R

, we get

R = \frac{V^2}{P} = \frac{120^2}{60} = 240 \, \Omega

220 W

440 W

1100 W

2200 W

Correct Answer: D

Solution:

The power

P

is given by

P = I^2 R = (5 \text{ A})^2 \times 44 \Omega = 1100 \text{ W}

To increase the voltage

To decrease the current

To open or close the circuit

To measure the resistance

Correct Answer: C

Solution:

A switch is used to open or close the circuit, allowing or stopping the flow of electric current.

To measure current

To measure resistance

To measure potential difference

To measure power

Correct Answer: C

Solution:

A voltmeter is used to measure the potential difference across two points in a circuit.

1 Ω

2 Ω

4 Ω

8 Ω

Correct Answer: A

Solution:

The resistance of a wire is given by the formula

R = \rho \frac{l}{A}

, where

\rho

is the resistivity,

l

is the length, and

A

is the cross-sectional area. If the length is halved and the area is doubled, the new resistance

R' = \rho \frac{l/2}{2A} = \frac{1}{4} \rho \frac{l}{A} = \frac{1}{4} R = 1 Ω

Copper

Aluminium

Tungsten

Silver

Correct Answer: D

Solution:

Silver has the lowest resistivity at

1.60 \times 10^{-8} \Omega \text{m}

12 W

24 W

6 W

18 W

Correct Answer: B

Solution:

The power consumed by the resistor can be calculated using the formula

P = \frac{V^2}{R}

. Substituting the given values,

P = \frac{12^2}{6} = 24 \text{ W}

1 Ω

3 Ω

12 Ω

18 Ω

Correct Answer: B

Solution:

Using Ohm's Law,

V = IR

, where

V = 6 \text{ V}

and

I = 2 \text{ A}

. Thus,

R = \frac{V}{I} = \frac{6}{2} = 3 \text{ Ω}

Copper

Manganese

Aluminium

Iron

Correct Answer: B

Solution:

The resistivity of manganese is given as

1.84 \times 10^{-6} \, \Omega \, m

. Therefore, the material is manganese.

8 A

6 A

10 A

4 A

Correct Answer: A

Solution:

Using Ohm's Law, the resistance of the heater is

R = \frac{V}{I} = \frac{60}{4} = 15 \, \Omega

. When the potential difference is increased to 120 V, the current is

I = \frac{V}{R} = \frac{120}{15} = 8 \, A

12 V

24 V

48 V

6 V

Correct Answer: A

Solution:

The potential difference

V

is given by

V = \frac{W}{Q} = \frac{24 \text{ J}}{2 \text{ C}} = 12 \text{ V}

3 W

5 W

9 W

15 W

Correct Answer: A

Solution:

The total resistance in series is 15 Ω. The current through the circuit is I = V/R = 15 V / 15 Ω = 1 A. The power dissipated in the 5 Ω resistor is P = I²R = (1 A)² * 5 Ω = 5 W.

Voltmeter

Ammeter

Resistor

Battery

Correct Answer: B

Solution:

An ammeter is used to measure the current flowing through a circuit.

3 A

4 A

5 A

6 A

Correct Answer: C

Solution:

The total resistance in a parallel circuit is given by

\frac{1}{R_{total}} = \frac{1}{8 \, \Omega} + \frac{1}{12 \, \Omega}

. Solving,

R_{total} = 4.8 \, \Omega

. Using Ohm's Law,

I = \frac{V}{R} = \frac{24 \, V}{4.8 \, \Omega} = 5 \, A

2 Ω

3 Ω

4 Ω

6 Ω

Correct Answer: D

Solution:

Using Ohm's Law, V = IR, we can find the resistance: R = V/I = 12 V / 3 A = 4 Ω.

2 V

3 V

4 V

6 V

Correct Answer: C

Solution:

Total resistance

R_{total} = 2 + 3 + 5 = 10 \Omega

. Current

I = \frac{12}{10} = 1.2 \text{ A}

. Voltage drop across 3 Ω resistor is

V = IR = 1.2 \times 3 = 3.6 \text{ V}

160 W

220 W

100 W

60 W

Correct Answer: A

Solution:

In parallel, the total power consumed is the sum of the individual powers: 60 W + 100 W = 160 W.

6 V

12 V

3 V

0 V

Correct Answer: A

Solution:

The potential difference across a 6 V battery is 6 V.

0.5 A

1.0 A

2.0 A

0.25 A

Correct Answer: A

Solution:

The current flowing through the bulb can be calculated using the formula

P = VI

, rearranged to

I = \frac{P}{V} = \frac{60 \, W}{120 \, V} = 0.5 \, A

Copper

Aluminium

Manganese

Silver

Correct Answer: C

Solution:

Manganese has a resistivity of 1.84 x 10⁻⁶ Ω m, which is higher than the other options listed.

Ohm meter (Ω m)

Volt (V)

Ampere (A)

Coulomb (C)

Correct Answer: A

Solution:

Resistivity is measured in ohm meters (Ω m), which is a characteristic property of the material.

220 W

440 W

880 W

1100 W

Correct Answer: D

Solution:

Power

P

is given by

P = I^2R = 5^2 \times 44 = 1100 \text{ W}

2 A

3 A

5 A

7 A

Correct Answer: C

Solution:

The total resistance

R_t

in a parallel circuit is given by

\frac{1}{R_t} = \frac{1}{4} + \frac{1}{6}

. Solving,

R_t = 2.4 \text{ Ω}

. The total current

I = \frac{V}{R_t} = \frac{12}{2.4} = 5 \text{ A}

The current is halved.

The current remains the same.

The current doubles.

The current quadruples.

Correct Answer: C

Solution:

According to Ohm's law,

V = IR

. If the potential difference

V

is doubled and resistance

R

remains constant, the current

I

must also double.

1 Ω

2 Ω

4 Ω

8 Ω

Correct Answer: A

Solution:

The resistance of the new wire is calculated as

R = \frac{1}{2} \times \frac{1}{2} \times 4 = 1

Ω.

0.45 A

0.5 A

0.55 A

0.6 A

Correct Answer: B

Solution:

The power

P

is given by

P = VI

, where

V

is the voltage and

I

is the current. Rearranging gives

I = \frac{P}{V} = \frac{100 \text{ W}}{220 \text{ V}} \approx 0.45 \text{ A}

4 A

8 A

16 A

2 A

Correct Answer: B

Solution:

Using Ohm's law, the current is directly proportional to the voltage. Doubling the voltage doubles the current, so the current will be 8 A.

10 Ω

20 Ω

40 Ω

80 Ω

Correct Answer: C

Solution:

Resistance is proportional to length and inversely proportional to cross-sectional area. When stretched to twice its length, the area is halved, so

R_{new} = 4 \times R_{original} = 40 \Omega

All components stop working.

The rest of the components continue to work.

The circuit becomes open.

The voltage across all components becomes zero.

Correct Answer: B

Solution:

In a parallel circuit, each component is independently connected to the power source. If one component fails, the rest continue to work as they still have a complete path to the power source.

6 V

12 V

24 V

48 V

Correct Answer: C

Solution:

Using Ohm's law, the potential difference V = I × R = 2 A × 12 Ω = 24 V.

It remains the same

It doubles

It quadruples

It halves

Correct Answer: C

Solution:

Resistance is proportional to the length and inversely proportional to the area. Doubling the length and halving the area quadruples the resistance.

0.73 A

1.45 A

0.27 A

2.00 A

Correct Answer: B

Solution:

The total power is 160 W. Using the formula

I = \frac{P}{V}

, the total current is

\frac{160}{220} = 0.73 + 0.45 = 1.45

Voltmeter

Ammeter

Ohmmeter

Galvanometer

Correct Answer: B

Solution:

An ammeter is used to measure the current flowing through a circuit.

The resistance is halved

The resistance remains the same

The resistance is doubled

The resistance is quadrupled

Correct Answer: C

Solution:

Resistance is directly proportional to the length of the wire, so doubling the length doubles the resistance.

It flows from negative to positive terminal.

It flows from positive to negative terminal.

It flows in the same direction as electron flow.

It flows in a random direction.

Correct Answer: B

Solution:

Conventional current is considered to flow from the positive terminal to the negative terminal, which is opposite to the direction of electron flow.

5 W

10 W

15 W

20 W

Correct Answer: C

Solution:

The equivalent resistance

R_{eq}

for resistors in parallel is given by

\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} = \frac{1}{10} + \frac{1}{20} = \frac{3}{20}

. Thus,

R_{eq} = \frac{20}{3} \, \Omega

. The total power consumed is

P = \frac{V^2}{R_{eq}} = \frac{10^2}{20/3} = 15 \, W

1100 W

880 W

440 W

2200 W

Correct Answer: A

Solution:

Power consumed by the heater is given by

P = \frac{V^2}{R} = \frac{220^2}{44} = 1100 \, W

60 W bulb

100 W bulb

Both will glow equally

Neither will glow

Correct Answer: A

Solution:

In a series circuit, the current is the same through both bulbs. The bulb with the higher resistance (60 W bulb) will have a higher voltage drop across it and will dissipate more power, thus glowing brighter.

5 Ω

10 Ω

15 Ω

20 Ω

Correct Answer: B

Solution:

According to Ohm's Law,

V = IR

. Here,

R = \frac{V}{I} = \frac{5}{0.5} = 10 \, \Omega

4 Ω

8 Ω

16 Ω

32 Ω

Correct Answer: C

Solution:

Resistance is directly proportional to the length of the wire. Doubling the length while keeping the cross-sectional area constant will double the resistance. Hence, the new resistance is 16 Ω.

2 A

3 A

4 A

5 A

Correct Answer: D

Solution:

The total resistance

R_t

in parallel is given by

\frac{1}{R_t} = \frac{1}{R_1} + \frac{1}{R_2}

. Thus,

\frac{1}{R_t} = \frac{1}{10} + \frac{1}{15} = \frac{5}{30} = \frac{1}{6}

. Therefore,

R_t = 6 \Omega

. The total current

I

is given by

I = \frac{V}{R_t} = \frac{30 \text{ V}}{6 \Omega} = 5 \text{ A}

8 Ω

16 Ω

2 Ω

12 Ω

Correct Answer: B

Solution:

When a wire is stretched to double its length, its resistance becomes four times the original resistance because resistance is directly proportional to the length of the wire. Therefore, the new resistance is

R = 4 \times 4 \, \Omega = 16 \, \Omega

0.73 A

1.36 A

1.73 A

2.36 A

Correct Answer: C

Solution:

The power

P

is given by

P = VI

. For the 60 W bulb,

I_1 = \frac{60}{220} = 0.27

A. For the 100 W bulb,

I_2 = \frac{100}{220} = 0.45

A. The total current

I = I_1 + I_2 = 0.27 + 0.45 = 0.72

25 Ω

10 Ω

15 Ω

30 Ω

Correct Answer: A

Solution:

In a series circuit, the total resistance is the sum of the individual resistances. Therefore, the total resistance is

R = 5 \, \Omega + 8 \, \Omega + 12 \, \Omega = 25 \, \Omega

The total resistance increases

The total resistance decreases

The total resistance remains the same

The total resistance becomes infinite

Correct Answer: B

Solution:

Adding a resistor in parallel decreases the total resistance because the overall conductance increases.

TV set

Toaster

Both use the same energy

Cannot be determined

Correct Answer: A

Solution:

The energy used by the TV set is P * t = 250 W * 2 h = 500 Wh. The energy used by the toaster is P * t = 1200 W * (10/60) h = 200 Wh. Therefore, the TV set uses more energy.

15 W

30 W

45 W

60 W

Correct Answer: A

Solution:

Power is given by

P = \frac{V^2}{R}

. The resistance of the bulb is

R = \frac{120^2}{60} = 240 \Omega

. At 60 V,

P = \frac{60^2}{240} = 15 \text{ W}

1 Ω

2 Ω

4 Ω

8 Ω

Correct Answer: B

Solution:

The resistance

R

is given by

R = \rho \frac{l}{A}

, where

\rho

is the resistivity,

l

is the length, and

A

is the area. If the length is halved and the area is doubled, the new resistance

R' = \rho \frac{l/2}{2A} = \frac{1}{4} \rho \frac{l}{A} = \frac{1}{4} \times 8 \Omega = 2 \Omega

The TV consumes more energy

The toaster consumes more energy

Both consume the same energy

Cannot be determined

Correct Answer: A

Solution:

Energy consumed by TV = 250 W × 1 hr = 250 Wh. Energy consumed by toaster = 1200 W × (10/60) hr = 200 Wh. The TV consumes more energy.

0.0324 Ω

0.324 Ω

3.24 Ω

32.4 Ω

Correct Answer: B

Solution:

Using the formula R = ρl/A, where ρ = 1.62 x 10⁻⁸ Ω m, l = 2 m, and A = 1 x 10⁻6 m², the resistance R = (1.62 x 10⁻⁸ Ω m * 2 m) / 1 x 10⁻6 m² = 0.324 Ω.

8 Ω

16 Ω

32 Ω

64 Ω

Correct Answer: C

Solution:

When the length of a wire is doubled, its resistance becomes four times the original resistance due to

R \propto l

. Therefore, the new resistance is

8 \times 4 = 32 \text{ Ω}

Copper

Aluminium

Tungsten

Silver

Correct Answer: C

Solution:

Tungsten is chosen for the filament of incandescent bulbs due to its high melting point and moderate resistivity, which allows it to withstand high temperatures without melting.

0.4 A

0.6 A

0.8 A

1.0 A

Correct Answer: A

Solution:

The total resistance

R_t

in series is

10 Ω + 20 Ω = 30 Ω

. Using Ohm's law,

I = \frac{V}{R_t} = \frac{12}{30} = 0.4

20 Ω

40 Ω

10 Ω

30 Ω

Correct Answer: B

Solution:

When a wire is stretched to double its length, its resistance becomes four times the original resistance because resistance is directly proportional to the square of the length. Thus, the new resistance is

4 \times 10 = 40 \text{ Ω}

1.5 A

3 A

4.5 A

6 A

Correct Answer: C

Solution:

The total resistance in parallel is given by

\frac{1}{R_{total}} = \frac{1}{5} + \frac{1}{10} = \frac{3}{10}

, so

R_{total} = \frac{10}{3} \text{ Ω}

. The total current is

I = \frac{V}{R_{total}} = \frac{15}{\frac{10}{3}} = 4.5 \text{ A}

1.2 A

2.0 A

0.8 A

1.0 A

Correct Answer: D

Solution:

The total resistance in the series circuit is the sum of the resistances:

R_{total} = 4 \, \Omega + 6 \, \Omega = 10 \, \Omega

. Using Ohm's Law,

I = \frac{V}{R} = \frac{12 \, V}{10 \, \Omega} = 1.2 \, A

, which is higher than silver's resistivity of

1.60 \times 10^{-8} \Omega \text{m}

Correct Answer: False

Solution:

Electric current is conventionally considered the flow of positive charges, but in reality, it is the flow of electrons, which are negative charges.

Correct Answer: False

Solution:

The resistance of a conductor is inversely proportional to its area of cross-section, as given by the formula

R \propto \frac{1}{A}

Correct Answer: True

Solution:

Conventionally, the direction of electric current is taken as opposite to the direction of the flow of electrons, which are negative charges. Therefore, the flow of electric current is considered to be the flow of positive charges.

Correct Answer: False

Solution:

The resistivity of a material generally depends on its temperature, as it is a characteristic property of the material.

Correct Answer: True

Solution:

The resistance of a uniform metallic conductor is directly proportional to its length and inversely proportional to its area of cross-section, as given by the formula

R = \rho \frac{l}{A}

Electricity

AI Learning Assistant

CBSE Learning Objectives – Key Concepts & Skills You Must Know

CBSE Revision Notes & Quick Summary for Last-Minute Study

Chapter 11: Electricity

11.1 Electric Current and Circuit

Key Concepts

Example

Resistors in Series and Parallel

Resistors in Series

Resistors in Parallel

Important Formulas

Diagram Descriptions

Exercises

CBSE Exam Tips, Important Questions & Common Mistakes to Avoid

Common Mistakes and Exam Tips in Electricity

Common Pitfalls

Tips for Success

Multiple Choice Questions

Q1.Which material is typically used for the filament of electric lamps due to its high melting point?

Correct Answer: C

Q2.In a circuit, a 12 V battery is connected to a resistor of unknown resistance. If the current flowing through the circuit is 3 A, what is the resistance of the resistor?

Correct Answer: C

Q3.Which of the following is a symbol for a closed plug key or switch in a circuit diagram?

Correct Answer: B

Q4.Which of the following materials has the highest resistivity?

Correct Answer: C

Q5.What is the potential difference across a 6 Ω resistor if a current of 3 A flows through it?

Correct Answer: A

Q6.A wire of length lll and cross-sectional area AAA has a resistance RRR. If the wire is stretched to double its original length without changing its volume, what will be the new resistance?

Correct Answer: B

Q7.A circuit consists of a 12 V battery and three resistors of 4 Ω, 6 Ω, and 12 Ω connected in series. What is the total current flowing through the circuit?

Correct Answer: A

Q8.What happens to the current through a heater if the potential difference is increased from 60 V to 120 V, assuming the resistance remains constant?

Correct Answer: A

Q9.What is the potential difference across a conductor if 12 J of energy is used to move 3 C of charge?

Correct Answer: C

Q10.How does the resistance of a wire change if its cross-sectional area is doubled?

Correct Answer: A

Q11.A 60 W light bulb is connected to a 120 V source. What is the resistance of the bulb?

Correct Answer: B

Q12.An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. What is the rate at which heat is developed in the heater?

Correct Answer: D

Q13.What is the role of a switch in an electric circuit?

Correct Answer: C

Q14.What is the function of a voltmeter in an electric circuit?

Correct Answer: C

Q15.A wire of a certain material has a resistance of 4 Ω. If the length of the wire is halved and the cross-sectional area is doubled, what will be the new resistance of the wire?

Correct Answer: A

Q16.Which material has the lowest resistivity among the following?

Correct Answer: D

Q17.Consider a circuit consisting of a 12 V battery, a switch, and a resistor of resistance 6 Ω. If the switch is closed, what is the power consumed by the resistor?

Correct Answer: B

Q18.A circuit consists of a 6 V battery connected in series with a resistor of unknown resistance. If the current flowing through the circuit is 2 A, what is the resistance of the resistor?

Correct Answer: B

Q19.If the resistivity of a material is given by 1.84×10−6 Ω m1.84 \times 10^{-6} \, \Omega \, m1.84×10−6Ωm, which material is it likely to be?

Correct Answer: B

Q20.An electric heater is connected to a 60 V source and draws a current of 4 A. If the potential difference is increased to 120 V, what will be the new current drawn by the heater?

Correct Answer: A

Q21.What is the potential difference across a conductor if 24 J of energy is used to move 2 C of charge through it?

Correct Answer: A

Q22.Two resistors, 5 Ω and 10 Ω, are connected in series across a 15 V battery. What is the power dissipated in the 5 Ω resistor?

Correct Answer: A

Q23.Which component is used to measure the current in a circuit?

Correct Answer: B

Q24.In a parallel circuit, two resistors of 8 Ω and 12 Ω are connected across a 24 V battery. What is the total current supplied by the battery?

Correct Answer: C

Q25.A circuit consists of a 12 V battery, a switch, and a resistor of unknown resistance. If the switch is closed and the current measured by the ammeter is 3 A, what is the resistance of the resistor?

Correct Answer: D

Q26.In a series circuit with a 12 V battery and three resistors of 2 Ω, 3 Ω, and 5 Ω, what is the voltage drop across the 3 Ω resistor?

Correct Answer: C

Q27.If a 60 W lamp and a 100 W lamp are connected in parallel to a 220 V supply, what is the total power consumed?

Correct Answer: A

Q28.What is the potential difference across a 6 V battery?

Correct Answer: A

Q29.A 60 W light bulb is connected to a 120 V power source. What is the current flowing through the bulb?

Correct Answer: A

Q30.Which of the following materials has the highest resistivity?

Correct Answer: C

Q31.What is the unit of resistivity?