CBSE Class 10 · Physics
CBSE Class 10 Physics Electricity Previous Year Questions
Help your child score full marks in CBSE Class 10 Physics Electricity with this carefully curated set of previous year board exam questions spanning 2016–2024. Each question comes with a detailed, step-by-step solution — covering Ohm’s Law, resistance, electric power, series and parallel circuits, and the heating effect of current — the most consistently tested topics in board exams.
CBSE Class 10 Physics Electricity — Questions with Solutions
A complete circuit is left on for several minutes, causing the connecting copper wire to become hot. As the temperature of the wire increases, the electrical resistance of the wire?
Solution
Answer: Option (C) is correct.
Explanation: When a complete circuit is left on for several minutes, the temperature of the wire increases. As the temperature of the wire increases, the electrical resistance of the wire also increases. This is due to the increased vibration of the metal atoms at higher temperatures, which obstructs the flow of electrons.
Explanation: When a complete circuit is left on for several minutes, the temperature of the wire increases. As the temperature of the wire increases, the electrical resistance of the wire also increases. This is due to the increased vibration of the metal atoms at higher temperatures, which obstructs the flow of electrons.
Plastic insulation surrounds a wire having diameter $d$ and length $l$ as shown. A decrease in the resistance of the wire would be produced by an increase in the:
Solution
Answer: Option (A) is correct.
The resistance $R$ of a wire is given by: $$R = \rho \times \frac{l}{A}$$ where $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area (proportional to $d^2$).
To decrease the resistance, we can increase the diameter of the wire, which increases the cross-sectional area $A$. Thus, the correct answer is (A) Diameter of the wire.
The resistance $R$ of a wire is given by: $$R = \rho \times \frac{l}{A}$$ where $\rho$ is the resistivity, $l$ is the length, and $A$ is the cross-sectional area (proportional to $d^2$).
To decrease the resistance, we can increase the diameter of the wire, which increases the cross-sectional area $A$. Thus, the correct answer is (A) Diameter of the wire.
A cylindrical conductor of length ‘$l$’ and uniform area of cross-section ‘$A$’ has resistance ‘$R$’. The area of cross-section of another conductor of the same material and same resistance but of length ‘$2l$’ is:
Solution
Answer: Option (C) is correct.
Using $R = \rho \dfrac{l}{A}$ for the first conductor and $R = \rho \dfrac{2l}{A’}$ for the second:
Since both resistances are equal: $$\frac{l}{A} = \frac{2l}{A’} \Rightarrow A’ = 2A$$ Thus, the area of cross-section of the second conductor is $\mathbf{2A}$.
Using $R = \rho \dfrac{l}{A}$ for the first conductor and $R = \rho \dfrac{2l}{A’}$ for the second:
Since both resistances are equal: $$\frac{l}{A} = \frac{2l}{A’} \Rightarrow A’ = 2A$$ Thus, the area of cross-section of the second conductor is $\mathbf{2A}$.
When a 4V battery is connected across an unknown resistor, there is a current of 100 mA in the circuit. The value of the resistance of the resistor is:
Solution
Answer: Option (B) is correct.
Using Ohm’s Law: $R = \dfrac{V}{I}$
Given: $V = 4 \, \text{V}$, $I = 100 \, \text{mA} = 0.1 \, \text{A}$ $$R = \frac{4}{0.1} = 40 \, \Omega$$ Thus, the value of the resistance is 40 Ω.
Using Ohm’s Law: $R = \dfrac{V}{I}$
Given: $V = 4 \, \text{V}$, $I = 100 \, \text{mA} = 0.1 \, \text{A}$ $$R = \frac{4}{0.1} = 40 \, \Omega$$ Thus, the value of the resistance is 40 Ω.
Two bulbs of 100 W and 40 W are connected in series. The current through the 100 W bulb is 1 A. The current through the 40 W bulb will be:
Solution
Answer: Option (D) is correct.
Explanation: In a series connection, the current through each device remains the same. Since the current through the 100 W bulb is 1 A, the current through the 40 W bulb is also 1 A.
Explanation: In a series connection, the current through each device remains the same. Since the current through the 100 W bulb is 1 A, the current through the 40 W bulb is also 1 A.
What is the maximum resistance which can be made using five resistors, each of 15 Ω?
Solution
Answer: Option (C) is correct.
To obtain the maximum resistance, connect all resistors in series: $$R_{\text{total}} = 15 + 15 + 15 + 15 + 15 = 75 \, \Omega$$ Thus, the maximum resistance is 75 Ω.
To obtain the maximum resistance, connect all resistors in series: $$R_{\text{total}} = 15 + 15 + 15 + 15 + 15 = 75 \, \Omega$$ Thus, the maximum resistance is 75 Ω.
An electric bulb is connected to a 220V generator. The current is 0.50 A. What is the power of the bulb?
Solution
Answer: Option (B) is correct.
Using the power formula: $$P = V \times I = 220 \times 0.50 = 110 \, \text{W}$$ The power of the bulb is 110 W.
Using the power formula: $$P = V \times I = 220 \times 0.50 = 110 \, \text{W}$$ The power of the bulb is 110 W.
Which of the following represents voltage?
Solution
Answer: Option (D) is correct.
Voltage (potential difference) is defined as the work done per unit charge: $$V = \frac{W}{Q}$$ where $V$ is voltage, $W$ is work done, and $Q$ is charge.
Voltage (potential difference) is defined as the work done per unit charge: $$V = \frac{W}{Q}$$ where $V$ is voltage, $W$ is work done, and $Q$ is charge.
The maximum resistance which can be made using four resistors, each of resistance $\dfrac{1}{2} \, \Omega$, is:
Solution
Answer: Option (A) is correct.
To obtain the maximum resistance, connect all resistors in series: $$R_{\text{total}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 2 \, \Omega$$ Thus, the maximum resistance is 2 Ω.
To obtain the maximum resistance, connect all resistors in series: $$R_{\text{total}} = \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 2 \, \Omega$$ Thus, the maximum resistance is 2 Ω.
Unit of electric power may also be expressed as:
Solution
Answer: Option (A) is correct.
Electric power is given by $P = V \times I$, where $V$ is in volts and $I$ is in amperes. Therefore, the unit of electric power can also be expressed as volt-ampere.
Electric power is given by $P = V \times I$, where $V$ is in volts and $I$ is in amperes. Therefore, the unit of electric power can also be expressed as volt-ampere.
Which element will be used for electrical transmission lines?
Solution
Answer: Option (B) is correct.
Copper is used for electrical transmission lines because it is economical, less oxidative than other metals, and has low resistivity, which minimises energy loss during transmission.
Copper is used for electrical transmission lines because it is economical, less oxidative than other metals, and has low resistivity, which minimises energy loss during transmission.
At the time of short circuit, the electric current in the circuit:
Solution
Answer: Option (D) is correct.
At the time of a short circuit, the resistance in the circuit drops to near zero, causing the current to increase heavily. This excessive current can burn electrical devices. Hence, fuses and circuit breakers are used as protective devices.
At the time of a short circuit, the resistance in the circuit drops to near zero, causing the current to increase heavily. This excessive current can burn electrical devices. Hence, fuses and circuit breakers are used as protective devices.
Why is nichrome wire used in many electrical heating devices?
Solution
Answer: Option (D) is correct.
Nichrome wire is used as a heating element because: (i) It offers very large resistance, so a large amount of electric energy is converted into heat energy. (ii) It has a high melting point so it can be heated till red hot without melting.
Nichrome wire is used as a heating element because: (i) It offers very large resistance, so a large amount of electric energy is converted into heat energy. (ii) It has a high melting point so it can be heated till red hot without melting.
Assertion (A): A conductor has $+3.2 \times 10^{-19} \, \text{C}$ charge.
Reason (R): Conductor has gained two electrons.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of A.
(B) Both Assertion (A) and Reason (R) are true but R is NOT the correct explanation of Assertion (A).
(C) Assertion (A) is true but Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is true.
Reason (R): Conductor has gained two electrons.
(A) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of A.
(B) Both Assertion (A) and Reason (R) are true but R is NOT the correct explanation of Assertion (A).
(C) Assertion (A) is true but Reason (R) is false.
(D) Assertion (A) is false and Reason (R) is true.
Solution
Answer: Option (C) is correct.
If a conductor has a positive charge, it means it has lost electrons (not gained them). Charge of one electron $= 1.6 \times 10^{-19} \, \text{C}$. Number of electrons lost $= \dfrac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 2$.
Thus, Assertion (A) is true but Reason (R) is false because the conductor has lost 2 electrons, not gained them.
If a conductor has a positive charge, it means it has lost electrons (not gained them). Charge of one electron $= 1.6 \times 10^{-19} \, \text{C}$. Number of electrons lost $= \dfrac{3.2 \times 10^{-19}}{1.6 \times 10^{-19}} = 2$.
Thus, Assertion (A) is true but Reason (R) is false because the conductor has lost 2 electrons, not gained them.
Assertion: The resistance of a conductor is directly proportional to its length and inversely proportional to its area of cross-section.
Reason: Longer conductors have more resistance because there is a longer path for the current to travel, while larger cross-sectional areas allow for easier current flow.
Reason: Longer conductors have more resistance because there is a longer path for the current to travel, while larger cross-sectional areas allow for easier current flow.
Solution
Answer: Option (A) is correct.
The assertion correctly describes the factors affecting resistance ($R = \rho \dfrac{l}{A}$), where length and cross-sectional area are key determinants. The reason accurately explains why: a longer path increases collisions for electrons, while a wider cross-section allows more electron pathways, reducing resistance.
The assertion correctly describes the factors affecting resistance ($R = \rho \dfrac{l}{A}$), where length and cross-sectional area are key determinants. The reason accurately explains why: a longer path increases collisions for electrons, while a wider cross-section allows more electron pathways, reducing resistance.
A V-I graph for a nichrome wire is given below. What do you infer from this graph? Draw a labelled circuit diagram to obtain such a graph.
Answer
Inference from the V-I Graph:
The V-I graph for a nichrome wire shows a straight line passing through the origin. This indicates that the relationship between voltage (V) and current (I) is linear — the resistance of the wire is constant over the range of applied voltages. According to Ohm’s Law: $$V = I \times R$$ The slope of the graph represents the resistance (R) of the wire. We conclude that the nichrome wire obeys Ohm’s Law.
Circuit Diagram to Obtain V-I Graph:
The circuit consists of: (1) A DC power supply to vary the voltage. (2) A nichrome wire connected in series. (3) A voltmeter across the wire to measure potential difference (V). (4) An ammeter in series to measure current (I). (5) A key to complete or break the circuit.
The V-I graph for a nichrome wire shows a straight line passing through the origin. This indicates that the relationship between voltage (V) and current (I) is linear — the resistance of the wire is constant over the range of applied voltages. According to Ohm’s Law: $$V = I \times R$$ The slope of the graph represents the resistance (R) of the wire. We conclude that the nichrome wire obeys Ohm’s Law.
Circuit Diagram to Obtain V-I Graph:
The circuit consists of: (1) A DC power supply to vary the voltage. (2) A nichrome wire connected in series. (3) A voltmeter across the wire to measure potential difference (V). (4) An ammeter in series to measure current (I). (5) A key to complete or break the circuit.
What is electrical resistivity? Derive its S.I. unit.
Answer
Electrical resistivity of the material of a conductor is defined as the resistance offered by the conductor of length 1 metre and area of cross-section 1 m².
The formula for resistance is: $$R = \rho \times \frac{l}{A}$$ Rearranging for resistivity: $$\rho = \frac{R \times A}{l}$$ S.I. unit of Resistivity:
Unit of $R$ = ohm (Ω), unit of $A$ = m², unit of $l$ = m $$\rho = \frac{\text{Ω} \times \text{m}^2}{\text{m}} = \text{Ω m (ohm metre)}$$ Thus, the S.I. unit of electrical resistivity is ohm metre (Ω m).
The formula for resistance is: $$R = \rho \times \frac{l}{A}$$ Rearranging for resistivity: $$\rho = \frac{R \times A}{l}$$ S.I. unit of Resistivity:
Unit of $R$ = ohm (Ω), unit of $A$ = m², unit of $l$ = m $$\rho = \frac{\text{Ω} \times \text{m}^2}{\text{m}} = \text{Ω m (ohm metre)}$$ Thus, the S.I. unit of electrical resistivity is ohm metre (Ω m).
In a series electrical circuit comprising a resistor made up of a metallic wire, the ammeter reads 100 mA. If the length of the wire is doubled, how will the current in the circuit change? Justify your answer.
Answer
Given: Initial current $I = 100 \, \text{mA}$; length of wire is doubled.
The resistance $R$ of a wire is directly proportional to its length $l$. If the length doubles, the resistance also doubles: $$R_{\text{new}} = 2 \times R$$ Since the voltage $V$ remains constant, by Ohm’s Law $V = I \times R$, if resistance doubles, the current is halved: $$I_{\text{new}} = \frac{I}{2} = \frac{100 \, \text{mA}}{2} = 50 \, \text{mA}$$ Conclusion: The ammeter will now read 50 mA.
The resistance $R$ of a wire is directly proportional to its length $l$. If the length doubles, the resistance also doubles: $$R_{\text{new}} = 2 \times R$$ Since the voltage $V$ remains constant, by Ohm’s Law $V = I \times R$, if resistance doubles, the current is halved: $$I_{\text{new}} = \frac{I}{2} = \frac{100 \, \text{mA}}{2} = 50 \, \text{mA}$$ Conclusion: The ammeter will now read 50 mA.
Calculate the resistance of a 1 km long copper wire of area of cross-section $2 \times 10^{-2} \, \text{cm}^2$. The resistivity of copper is $1.623 \times 10^{-8} \, \Omega\text{-metre}$.
Answer
Using the formula $R = \rho \times \dfrac{l}{A}$
Given:
$\rho = 1.623 \times 10^{-8} \, \Omega \, \text{m}$
$l = 1 \, \text{km} = 1000 \, \text{m}$
$A = 2 \times 10^{-2} \, \text{cm}^2 = 2 \times 10^{-6} \, \text{m}^2$
Substituting: $$R = \frac{1.623 \times 10^{-8} \times 1000}{2 \times 10^{-6}} = \frac{1.623 \times 10^{-5}}{2 \times 10^{-6}} = 0.81 \, \Omega$$ The resistance of the copper wire is 0.81 Ω.
Given:
$\rho = 1.623 \times 10^{-8} \, \Omega \, \text{m}$
$l = 1 \, \text{km} = 1000 \, \text{m}$
$A = 2 \times 10^{-2} \, \text{cm}^2 = 2 \times 10^{-6} \, \text{m}^2$
Substituting: $$R = \frac{1.623 \times 10^{-8} \times 1000}{2 \times 10^{-6}} = \frac{1.623 \times 10^{-5}}{2 \times 10^{-6}} = 0.81 \, \Omega$$ The resistance of the copper wire is 0.81 Ω.
(a) List the factors on which the resistance of a conductor in the shape of a wire depends.
(b) Why are metals good conductors of electricity, whereas glass is a bad conductor of electricity? Give reason.
(c) Why are alloys commonly used in electrical heating devices? Give reason.
(b) Why are metals good conductors of electricity, whereas glass is a bad conductor of electricity? Give reason.
(c) Why are alloys commonly used in electrical heating devices? Give reason.
Answer
(a) Factors affecting resistance:
1. Length of the conductor (l): Resistance is directly proportional to length.
2. Area of cross-section (A): Resistance is inversely proportional to the area of cross-section.
3. Resistivity of the material (ρ): Resistance depends on the material’s resistivity.
4. Temperature (T): Resistance is directly proportional to temperature for most materials.
(b) Metals vs. Glass:
Metals are good conductors because they have free electrons that move easily when a potential difference is applied. Glass is a bad conductor (insulator) because it has no free electrons — its electrons are tightly bound to atoms and cannot carry current.
(c) Alloys in heating devices:
Alloys such as nichrome are used because they have higher resistivity than pure metals (generating more heat per unit current) and a high melting point, making them durable at elevated temperatures without oxidising or melting.
1. Length of the conductor (l): Resistance is directly proportional to length.
2. Area of cross-section (A): Resistance is inversely proportional to the area of cross-section.
3. Resistivity of the material (ρ): Resistance depends on the material’s resistivity.
4. Temperature (T): Resistance is directly proportional to temperature for most materials.
(b) Metals vs. Glass:
Metals are good conductors because they have free electrons that move easily when a potential difference is applied. Glass is a bad conductor (insulator) because it has no free electrons — its electrons are tightly bound to atoms and cannot carry current.
(c) Alloys in heating devices:
Alloys such as nichrome are used because they have higher resistivity than pure metals (generating more heat per unit current) and a high melting point, making them durable at elevated temperatures without oxidising or melting.
(a) An electric iron consumes energy at a rate of 880 W when heating is at the maximum rate and 330 W when heating is at the minimum. If the source voltage is 220 V, calculate the current and resistance in each case.
(b) What is the heating effect of electric current?
(b) What is the heating effect of electric current?
Answer
(a) At maximum heating ($P = 880 \, \text{W}$, $V = 220 \, \text{V}$): $$R = \frac{V^2}{P} = \frac{220 \times 220}{880} = 55 \, \Omega$$
$$I = \frac{P}{V} = \frac{880}{220} = 4 \, \text{A}$$ At minimum heating ($P = 330 \, \text{W}$, $V = 220 \, \text{V}$): $$R = \frac{V^2}{P} = \frac{220 \times 220}{330} \approx 146.66 \, \Omega$$
$$I = \frac{P}{V} = \frac{330}{220} = 1.5 \, \text{A}$$
(b) Heating effect of electric current:
When an electric current flows through a conductor, the moving electrons collide with atoms of the conductor, causing them to vibrate and release energy as heat. The amount of heat produced $H$ is given by Joule’s Law: $$H = I^2 R t$$ where $I$ is current, $R$ is resistance, and $t$ is time.
(b) Heating effect of electric current:
When an electric current flows through a conductor, the moving electrons collide with atoms of the conductor, causing them to vibrate and release energy as heat. The amount of heat produced $H$ is given by Joule’s Law: $$H = I^2 R t$$ where $I$ is current, $R$ is resistance, and $t$ is time.
Fill in the blanks:
1. The SI unit of current is ……… .
2. According to ……… Law, the potential difference across the ends of a resistor is directly proportional to the ……… through it, provided its ……… remains constant.
3. The resistance of a conductor depends directly on its ……… , inversely on its ……… and also on the ……… of the conductor.
4. The SI unit of resistivity is ……… .
5. If the potential difference across the ends of a conductor is doubled, the current flowing through it gets ……… .
1. The SI unit of current is ……… .
2. According to ……… Law, the potential difference across the ends of a resistor is directly proportional to the ……… through it, provided its ……… remains constant.
3. The resistance of a conductor depends directly on its ……… , inversely on its ……… and also on the ……… of the conductor.
4. The SI unit of resistivity is ……… .
5. If the potential difference across the ends of a conductor is doubled, the current flowing through it gets ……… .
Answer
1. The SI unit of current is ampere.
2. According to Ohm’s Law, the potential difference across the ends of a resistor is directly proportional to the current through it, provided its temperature remains constant.
3. The resistance of a conductor depends directly on its length, inversely on its area of cross-section, and also on the material of the conductor.
4. The SI unit of resistivity is ohm-metre (Ω m).
5. If the potential difference across the ends of a conductor is doubled, the current flowing through it gets doubled.
2. According to Ohm’s Law, the potential difference across the ends of a resistor is directly proportional to the current through it, provided its temperature remains constant.
3. The resistance of a conductor depends directly on its length, inversely on its area of cross-section, and also on the material of the conductor.
4. The SI unit of resistivity is ohm-metre (Ω m).
5. If the potential difference across the ends of a conductor is doubled, the current flowing through it gets doubled.
An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery as shown in the circuit.
(a) The total resistance of the circuit.
(b) The current through the circuit.
(c) The potential difference across: (i) The electric lamp (ii) The conductor
(d) The power of the lamp.
(a) The total resistance of the circuit.
(b) The current through the circuit.
(c) The potential difference across: (i) The electric lamp (ii) The conductor
(d) The power of the lamp.
Answer
(a) Total resistance: $$R_{\text{total}} = 20 + 4 = 24 \, \Omega$$ (b) Current through the circuit: $$I = \frac{V}{R} = \frac{6}{24} = 0.25 \, \text{A}$$ (c) Potential difference:
(i) Across electric lamp: $V = 0.25 \times 20 = 5 \, \text{V}$
(ii) Across conductor: $V = 0.25 \times 4 = 1 \, \text{V}$
(d) Power of the lamp: $$P = V \times I = 5 \times 0.25 = 1.25 \, \text{W}$$
(i) Across electric lamp: $V = 0.25 \times 20 = 5 \, \text{V}$
(ii) Across conductor: $V = 0.25 \times 4 = 1 \, \text{V}$
(d) Power of the lamp: $$P = V \times I = 5 \times 0.25 = 1.25 \, \text{W}$$
100 J of heat is produced each second in a 4 Ω resistor. The potential difference across the resistor will be:
Answer
The heat produced per second is the power ($P = 100 \, \text{W}$). Using:
$$P = \frac{V^2}{R}$$
$$100 = \frac{V^2}{4} \Rightarrow V^2 = 400 \Rightarrow V = \sqrt{400} = 20 \, \text{V}$$ The potential difference across the resistor is 20 V.
(a) Define Power and state its S.I. unit.
(b) A torch bulb is rated 5 V and 500 mA. Calculate:
(i) Power (ii) Resistance (iii) Energy consumed when it is lighted for 2½ hours.
(b) A torch bulb is rated 5 V and 500 mA. Calculate:
(i) Power (ii) Resistance (iii) Energy consumed when it is lighted for 2½ hours.
Answer
(a) Power: Power is the rate at which electrical energy is converted into other forms of energy. It is given by:
$$P = \frac{W}{t}$$
The S.I. unit of power is the watt (W), defined as one joule per second ($1 \, \text{W} = 1 \, \text{J/s}$).
(b) Given: $V = 5 \, \text{V}$, $I = 500 \, \text{mA} = 0.5 \, \text{A}$, $t = 2.5 \times 60 \times 60 = 9000 \, \text{s}$
(i) Power: $$P = V \times I = 5 \times 0.5 = 2.5 \, \text{W}$$ (ii) Resistance: $$R = \frac{V}{I} = \frac{5}{0.5} = 10 \, \Omega$$ (iii) Energy consumed: $$E = P \times t = 2.5 \times 9000 = 22{,}500 \, \text{J}$$
(b) Given: $V = 5 \, \text{V}$, $I = 500 \, \text{mA} = 0.5 \, \text{A}$, $t = 2.5 \times 60 \times 60 = 9000 \, \text{s}$
(i) Power: $$P = V \times I = 5 \times 0.5 = 2.5 \, \text{W}$$ (ii) Resistance: $$R = \frac{V}{I} = \frac{5}{0.5} = 10 \, \Omega$$ (iii) Energy consumed: $$E = P \times t = 2.5 \times 9000 = 22{,}500 \, \text{J}$$
The diagram given is a schematic diagram of a household circuit. The house has 5 usable spaces where electrical connections are made. The mains have a voltage of 220 V and the net current from the mains is 22 A.
(a) What is the mode of connection to all the spaces in the house from the mains?
(b) Spaces 5 and 4 have the same resistance, and spaces 3 and 2 have respective resistances of 20 Ω and 30 Ω. Space 1 has a resistance double that of space 5. What is the net resistance for space 5?
(c) What is the current in space 3?
(d) What should be placed between the main connection and the rest of the house’s electrical appliances to save them from accidental high electric current?
(a) What is the mode of connection to all the spaces in the house from the mains?
(b) Spaces 5 and 4 have the same resistance, and spaces 3 and 2 have respective resistances of 20 Ω and 30 Ω. Space 1 has a resistance double that of space 5. What is the net resistance for space 5?
(c) What is the current in space 3?
(d) What should be placed between the main connection and the rest of the house’s electrical appliances to save them from accidental high electric current?
Answer
(a) All spaces are connected in parallel — each receives the full mains voltage of 220 V.
(b) Let resistance of space 5 (and 4) = $R$. Space 1 = $2R$, Space 2 = 30 Ω, Space 3 = 20 Ω.
Equivalent resistance: $$R_{\text{eq}} = \frac{V}{I} = \frac{220}{22} = 10 \, \Omega$$ For parallel combination: $$\frac{1}{10} = \frac{1}{2R} + \frac{1}{R} + \frac{1}{30} + \frac{1}{20} + \frac{1}{R} = \frac{2}{R} + \frac{5}{60}$$ $$\frac{1}{10} – \frac{5}{60} = \frac{6-5}{60} = \frac{1}{60} = \frac{2}{R} \Rightarrow R = 120 \, \Omega$$ The net resistance for space 5 is 120 Ω.
(c) $$I_3 = \frac{V}{R_3} = \frac{220}{20} = 11 \, \text{A}$$ The current in space 3 is 11 A.
(d) An electric fuse (or circuit breaker) should be placed between the main connection and the rest of the house’s appliances. It protects the circuit by melting and breaking the circuit if the current exceeds a safe limit.
(b) Let resistance of space 5 (and 4) = $R$. Space 1 = $2R$, Space 2 = 30 Ω, Space 3 = 20 Ω.
Equivalent resistance: $$R_{\text{eq}} = \frac{V}{I} = \frac{220}{22} = 10 \, \Omega$$ For parallel combination: $$\frac{1}{10} = \frac{1}{2R} + \frac{1}{R} + \frac{1}{30} + \frac{1}{20} + \frac{1}{R} = \frac{2}{R} + \frac{5}{60}$$ $$\frac{1}{10} – \frac{5}{60} = \frac{6-5}{60} = \frac{1}{60} = \frac{2}{R} \Rightarrow R = 120 \, \Omega$$ The net resistance for space 5 is 120 Ω.
(c) $$I_3 = \frac{V}{R_3} = \frac{220}{20} = 11 \, \text{A}$$ The current in space 3 is 11 A.
(d) An electric fuse (or circuit breaker) should be placed between the main connection and the rest of the house’s appliances. It protects the circuit by melting and breaking the circuit if the current exceeds a safe limit.
(a) State Ohm’s law. Represent it mathematically.
(b) Define 1 ohm.
(c) What is the resistance of a conductor through which a current of 0.5 A flows when a potential difference of 2 V is applied across its ends?
(b) Define 1 ohm.
(c) What is the resistance of a conductor through which a current of 0.5 A flows when a potential difference of 2 V is applied across its ends?
Answer
(a) Ohm’s Law: The current flowing through a conductor is directly proportional to the potential difference across its ends, provided temperature and other physical conditions remain constant. Mathematically:
$$V = I \times R$$ (b) Definition of 1 Ohm: 1 ohm is defined as the resistance of a conductor when a current of 1 ampere flows through it under a potential difference of 1 volt:
$$1 \, \text{ohm} = \frac{1 \, \text{volt}}{1 \, \text{ampere}}$$ (c) Resistance Calculation: $$R = \frac{V}{I} = \frac{2}{0.5} = 4 \, \text{ohms}$$ The resistance of the conductor is 4 Ω.
The current flowing through a resistor connected in a circuit and the potential difference developed across its ends are as shown in the diagram by milliammeter and voltmeter readings respectively.
(a) What are the least counts of these meters?
(b) What is the resistance of the resistor?
(a) What are the least counts of these meters?
(b) What is the resistance of the resistor?
Answer
(a) Least Counts:
Least count of milliammeter: $$\frac{100}{10} = 10 \, \text{mA}$$ Least count of voltmeter: $$\frac{1}{10} = 0.1 \, \text{V}$$ (b) Resistance Calculation:
$I = 250 \, \text{mA} = 0.25 \, \text{A}$, $V = 2.4 \, \text{V}$ $$R = \frac{V}{I} = \frac{2.4}{0.25} = 9.6 \, \Omega$$ Final Answers: Least count of milliammeter = 10 mA; Least count of voltmeter = 0.1 V; Resistance = 9.6 Ω.
Least count of milliammeter: $$\frac{100}{10} = 10 \, \text{mA}$$ Least count of voltmeter: $$\frac{1}{10} = 0.1 \, \text{V}$$ (b) Resistance Calculation:
$I = 250 \, \text{mA} = 0.25 \, \text{A}$, $V = 2.4 \, \text{V}$ $$R = \frac{V}{I} = \frac{2.4}{0.25} = 9.6 \, \Omega$$ Final Answers: Least count of milliammeter = 10 mA; Least count of voltmeter = 0.1 V; Resistance = 9.6 Ω.
Why are the heating elements of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer
The heating elements are made of alloys rather than pure metals because:
1. Higher resistivity: Alloys generally have higher resistivity than their constituent pure metals. This higher resistivity means more heat is generated for a given current, which is ideal for heating applications.
2. Durability at high temperatures: Alloys like nichrome are more resistant to oxidation and corrosion at high temperatures, making them longer-lasting and more reliable in heating devices.
1. Higher resistivity: Alloys generally have higher resistivity than their constituent pure metals. This higher resistivity means more heat is generated for a given current, which is ideal for heating applications.
2. Durability at high temperatures: Alloys like nichrome are more resistant to oxidation and corrosion at high temperatures, making them longer-lasting and more reliable in heating devices.
Some work is done to move a charge $Q$ from infinity to a point A in space. The potential of the point A is given as $V$. What is the work done to move this charge from infinity in terms of $Q$ and $V$?
Answer
The work done $W$ in moving a charge $Q$ from infinity to a point at potential $V$ is given by:
$$W = Q \times V$$
It is the product of the charge and the potential at that point. The unit of work done is volt-coulomb (or joule).
(a) An electric iron consumes energy at a rate of 880 W when heating is at the maximum rate and 330 W when heating is at the minimum. If the source voltage is 220 V, calculate the current and resistance in each case.
(b) What is the heating effect of electric current?
(c) Find an expression for the amount of heat produced when a current passes through a resistor for some time.
(b) What is the heating effect of electric current?
(c) Find an expression for the amount of heat produced when a current passes through a resistor for some time.
Answer
(a) At maximum rate ($P = 880 \, \text{W}$): $$R = \frac{V^2}{P} = \frac{220^2}{880} = 55 \, \Omega, \quad I = \frac{P}{V} = \frac{880}{220} = 4 \, \text{A}$$ At minimum rate ($P = 330 \, \text{W}$): $$R = \frac{V^2}{P} = \frac{220^2}{330} \approx 146.66 \, \Omega, \quad I = \frac{P}{V} = \frac{330}{220} = 1.5 \, \text{A}$$ (b) The heating effect of electric current is the phenomenon where electrical energy is converted into heat when current flows through a conductor, due to collisions between electrons and atoms. This is governed by Joule’s Law:
$$H = I^2 R t$$ (c) Expression for heat produced:
When current $I$ flows through a resistor of resistance $R$ for time $t$, the heat produced is: $$H = I^2 R t$$ where $H$ is heat in joules, $I$ in amperes, $R$ in ohms, and $t$ in seconds.
When current $I$ flows through a resistor of resistance $R$ for time $t$, the heat produced is: $$H = I^2 R t$$ where $H$ is heat in joules, $I$ in amperes, $R$ in ohms, and $t$ in seconds.
(a) Three resistors $R_1, R_2, R_3$ are connected in parallel, and the combination is connected to a battery, ammeter, voltmeter, and key. Draw a suitable circuit diagram and obtain an expression for the equivalent resistance of the combination.
(b) Calculate the equivalent resistance of the following network:
(b) Calculate the equivalent resistance of the following network:
Answer
(a) Equivalent resistance in parallel:
In a parallel combination, the total current is $I = I_1 + I_2 + I_3$ and the voltage across each resistor is the same ($V$). Using $I_k = \dfrac{V}{R_k}$: $$\frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$$ $$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ (b) Equivalent resistance of the network:
$R_1$ and $R_2$ (both 20 Ω) are in parallel: $$\frac{1}{R_p} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} \Rightarrow R_p = 10 \, \Omega$$ $R_p$ is in series with $R_3 = 10 \, \Omega$: $$R_{\text{eq}} = 10 + 10 = 20 \, \Omega$$ The equivalent resistance of the network is 20 Ω.
In a parallel combination, the total current is $I = I_1 + I_2 + I_3$ and the voltage across each resistor is the same ($V$). Using $I_k = \dfrac{V}{R_k}$: $$\frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$$ $$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$ (b) Equivalent resistance of the network:
$R_1$ and $R_2$ (both 20 Ω) are in parallel: $$\frac{1}{R_p} = \frac{1}{20} + \frac{1}{20} = \frac{2}{20} \Rightarrow R_p = 10 \, \Omega$$ $R_p$ is in series with $R_3 = 10 \, \Omega$: $$R_{\text{eq}} = 10 + 10 = 20 \, \Omega$$ The equivalent resistance of the network is 20 Ω.
In the given circuit, A, B, C and D are four lamps connected with a battery of 60 V.
(i) What kind of combination are the lamps arranged in?
(ii) Write any two advantages of this combination.
(iii) Explain with calculations which lamp glows the brightest.
(iv) Find the total resistance of the circuit.
(i) What kind of combination are the lamps arranged in?
(ii) Write any two advantages of this combination.
(iii) Explain with calculations which lamp glows the brightest.
(iv) Find the total resistance of the circuit.
Answer
(i) The lamps are connected in parallel. Each lamp is connected across the same two terminal points of the battery, receiving the full 60 V.
(ii) Advantages of parallel connection:
1. Independent operation: If one lamp fails or is switched off, the others continue to operate normally.
2. Full voltage to each lamp: Every lamp receives the same voltage (60 V), ensuring each operates at its rated brightness.
(iii) Brightest lamp:
Power $P = V \times I$, with $V = 60 \, \text{V}$ for each:
Lamp A: $P_A = 3 \times 60 = 180 \, \text{W}$
Lamp B: $P_B = 4 \times 60 = 240 \, \text{W}$
Lamp C: $P_C = 5 \times 60 = 300 \, \text{W}$
Lamp D: $P_D = 3 \times 60 = 180 \, \text{W}$
Lamp C glows the brightest with 300 W (highest current = 5 A).
(iv) Total resistance:
$R_A = 60/3 = 20 \, \Omega$, $R_B = 60/4 = 15 \, \Omega$, $R_C = 60/5 = 12 \, \Omega$, $R_D = 60/3 = 20 \, \Omega$ $$\frac{1}{R_{\text{total}}} = \frac{1}{20} + \frac{1}{15} + \frac{1}{12} + \frac{1}{20} = 0.05 + 0.0667 + 0.0833 + 0.05 = 0.25$$ $$R_{\text{total}} = \frac{1}{0.25} = 4 \, \Omega$$ The total resistance of the circuit is 4 Ω.
(ii) Advantages of parallel connection:
1. Independent operation: If one lamp fails or is switched off, the others continue to operate normally.
2. Full voltage to each lamp: Every lamp receives the same voltage (60 V), ensuring each operates at its rated brightness.
(iii) Brightest lamp:
Power $P = V \times I$, with $V = 60 \, \text{V}$ for each:
Lamp A: $P_A = 3 \times 60 = 180 \, \text{W}$
Lamp B: $P_B = 4 \times 60 = 240 \, \text{W}$
Lamp C: $P_C = 5 \times 60 = 300 \, \text{W}$
Lamp D: $P_D = 3 \times 60 = 180 \, \text{W}$
Lamp C glows the brightest with 300 W (highest current = 5 A).
(iv) Total resistance:
$R_A = 60/3 = 20 \, \Omega$, $R_B = 60/4 = 15 \, \Omega$, $R_C = 60/5 = 12 \, \Omega$, $R_D = 60/3 = 20 \, \Omega$ $$\frac{1}{R_{\text{total}}} = \frac{1}{20} + \frac{1}{15} + \frac{1}{12} + \frac{1}{20} = 0.05 + 0.0667 + 0.0833 + 0.05 = 0.25$$ $$R_{\text{total}} = \frac{1}{0.25} = 4 \, \Omega$$ The total resistance of the circuit is 4 Ω.
An electric lamp of resistance 20 Ω and a conductor of resistance 4 Ω are connected to a 6 V battery. Calculate:
(a) The total resistance of the circuit.
(b) The current through the circuit.
(c) The potential difference across: (i) the electric lamp (ii) the conductor
(d) The power of the lamp.
(a) The total resistance of the circuit.
(b) The current through the circuit.
(c) The potential difference across: (i) the electric lamp (ii) the conductor
(d) The power of the lamp.
Answer
(a) Total resistance (series): $$R_{\text{total}} = 20 + 4 = 24 \, \Omega$$ (b) Current: $$I = \frac{V}{R} = \frac{6}{24} = 0.25 \, \text{A}$$ (c) Potential difference:
(i) Across lamp: $V_{\text{lamp}} = 0.25 \times 20 = 5 \, \text{V}$
(ii) Across conductor: $V_{\text{cond}} = 0.25 \times 4 = 1 \, \text{V}$
(d) Power of the lamp: $$P = V \times I = 5 \times 0.25 = 1.25 \, \text{W}$$
(i) Across lamp: $V_{\text{lamp}} = 0.25 \times 20 = 5 \, \text{V}$
(ii) Across conductor: $V_{\text{cond}} = 0.25 \times 4 = 1 \, \text{V}$
(d) Power of the lamp: $$P = V \times I = 5 \times 0.25 = 1.25 \, \text{W}$$
A bulb is rated 40 W, 220 V. Find the current drawn by it when connected to a 220 V supply. Also find its resistance. If the bulb is replaced by a bulb of rating 25 W, 220 V, will there be any change in the value of current and resistance? Justify your answer and determine the change.
Answer
For 40 W, 220 V bulb: $$I = \frac{P}{V} = \frac{40}{220} \approx 0.18 \, \text{A}$$
$$R = \frac{V^2}{P} = \frac{220^2}{40} = 1210 \, \Omega$$ For 25 W, 220 V bulb: $$I = \frac{P}{V} = \frac{25}{220} \approx 0.113 \, \text{A}$$
$$R = \frac{V^2}{P} = \frac{220^2}{25} = 1936 \, \Omega$$ Yes, there is a change. When the 40 W bulb is replaced by the 25 W bulb:
Change in current: $\Delta I = 0.18 – 0.113 = 0.067 \, \text{A}$ (current decreases)
Change in resistance: $\Delta R = 1936 – 1210 = 726 \, \Omega$ (resistance increases)
Change in current: $\Delta I = 0.18 – 0.113 = 0.067 \, \text{A}$ (current decreases)
Change in resistance: $\Delta R = 1936 – 1210 = 726 \, \Omega$ (resistance increases)
Explore More CBSE Class 10 Physics Chapters
Frequently Asked Questions
What does the Electricity chapter cover in CBSE Class 10 Physics? ⌄
The Electricity chapter in CBSE Class 10 covers key concepts including electric charge and current, potential difference, Ohm’s Law, resistance and resistivity, series and parallel combinations of resistors, and the heating effect of electric current (Joule’s Law). It also includes practical applications such as household circuits, fuses, and the power consumption of electrical devices.
How many marks does the Electricity chapter carry in the CBSE Class 10 board exam? ⌄
Electricity is one of the most heavily tested chapters in CBSE Class 10 Physics. It falls under the Physics section, which carries approximately 27 marks in the board paper. Questions from Electricity typically appear as 1-mark MCQs, 2–3 mark short answers, and 5-mark long answer or case-study questions, making it a high-value chapter to master.
What are the most important topics in the Electricity chapter for board exams? ⌄
The most frequently tested topics are: Ohm’s Law and its verification, calculating resistance using $R = \rho l/A$, series and parallel resistor combinations with numerical problems, Joule’s heating law ($H = I^2Rt$), electric power ($P = VI = I^2R = V^2/R$), and household circuit safety devices like fuses. Numerical problems based on these formulas are almost guaranteed in every board paper.
What common mistakes do students make when solving Electricity questions? ⌄
A very common mistake is confusing series and parallel formulas — students sometimes apply the series formula $R = R_1 + R_2$ in parallel circuits and vice versa. Another frequent error is forgetting to convert units (e.g., milliamperes to amperes, centimetres² to metres²) before substituting into formulas. Practising previous year questions regularly helps your child avoid these costly errors under exam pressure.
How does Angle Belearn help students score well in the Electricity chapter? ⌄
Angle Belearn’s CBSE specialists curate chapter-wise question banks drawn directly from real board papers, each paired with clear, step-by-step solutions that show exactly how marks are earned. Students who practise with Angle Belearn develop the habit of showing structured, formula-first working — the approach that scores full marks in board exams. Regular practice builds both speed and accuracy so your child walks into the exam confident.
