NCERT Solutions Class 10 Science Chapter 8 Electricity

Master your physics concepts with complete NCERT Solutions Class 10 Science Chapter 8 Electricity. This comprehensive guide provides accurate, step-by-step solutions for all in-text questions and textbook exercises aligned with the latest CBSE syllabus. Electricity is one of the highest-weightage chapters in Class 10 Physics, carrying 7 to 9 marks in the board examinations. Whether you are revising fundamental concepts like electric current, Ohm’s law, and Joule’s heating effect, or looking for NCERT Solutions step by step to solve numericals on resistor combinations, this chapter-wise guide covers every derivation, formula, and conceptual problem thoroughly.

Chapter Overview: Electricity in Class 10 Science

Chapter 8 (Electricity) builds the bedrock of modern electrical physics. It shifts focus from static charges to dynamic electric circuits, potential differences, and electrical energy dissipation. Mastering this chapter requires a solid grasp of theoretical definitions, circuit diagram interpretations, and algebraic manipulation of physical formulas.

The chapter is divided into six core thematic areas:

• Electric Current and Circuit: Definition of electric current ($I = Q/t$), direction of conventional current vs. electronic current, and unit definitions (Ampere and Coulomb).
• Electric Potential and Potential Difference: Work done in moving a unit positive charge ($V = W/Q$), definition of 1 Volt, and the function of a cell/battery.
• Ohm’s Law and Electrical Resistance: Linear relationship between voltage and current ($V = IR$), definition of resistance ($1\ \Omega$), and factors governing conductor resistance ($R = \rho l/A$).
• System of Resistors: Series ($R_s = R_1 + R_2 + \dots$) and parallel ($1/R_p = 1/R_1 + 1/R_2 + \dots$) combinations, current distribution, and voltage division.
• Heating Effect of Electric Current: Joule’s Law of Heating ($H = I^2Rt$), working of heating appliances, electric fuses, and tungsten bulb filaments.
• Electric Power and Energy: Rate of electrical energy consumption ($P = VI = I^2R = V^2/R$), commercial units of electrical energy ($1\text{ kWh} = 3.6 \times 10^6\text{ J}$), and electricity bill calculations.

Important Formulas and Derivations

Before attempting the exercise questions, ensure that you have memorized these essential mathematical relations and their standard SI units.

Physical Quantity	Formula	Standard SI Unit	Key Variables
Electric Current ($I$)	$I = \frac{Q}{t} = \frac{ne}{t}$	Ampere ($\text{A}$)	$Q =$ Charge ($\text{C}$), $t =$ Time ($\text{s}$), $n =$ No. of electrons, $e = 1.6 \times 10^{-19}\text{ C}$
Potential Difference ($V$)	$V = \frac{W}{Q}$	Volt ($\text{V}$)	$W =$ Work Done ($\text{J}$), $Q =$ Charge ($\text{C}$)
Ohm’s Law	$V = IR \implies R = \frac{V}{I}$	Ohm ($\Omega$)	$V =$ Voltage ($\text{V}$), $I =$ Current ($\text{A}$), $R =$ Resistance ($\Omega$)
Resistance of a Conductor	$R = \rho \frac{l}{A}$	Ohm ($\Omega$)	$\rho =$ Resistivity ($\Omega\cdot\text{m}$), $l =$ Length ($\text{m}$), $A =$ Cross-sectional Area ($\text{m}^2$)
Equivalent Resistance (Series)	$R_s = R_1 + R_2 + R_3 + \dots$	Ohm ($\Omega$)	Current remains constant across all series components.
Equivalent Resistance (Parallel)	$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3} + \dots$	Ohm ($\Omega$)	Potential difference remains constant across all parallel branches.
Joule’s Law of Heating ($H$)	$H = I^2Rt = VIt = \frac{V^2}{R}t$	Joule ($\text{J}$)	$H =$ Heat energy generated ($\text{J}$), $t =$ Time duration ($\text{s}$)
Electric Power ($P$)	$P = VI = I^2R = \frac{V^2}{R}$	Watt ($\text{W}$)	$1\text{ kW} = 1000\text{ W}$
Commercial Unit of Energy ($E$)	$E = P \times t$ ($1\text{ kWh} = 3.6 \times 10^6\text{ J}$)	Kilowatt-hour ($\text{kWh}$ or "Units")	$P$ in Kilowatts ($\text{kW}$), $t$ in hours ($\text{h}$)

Derivation 1: Equivalent Resistance for Resistors in Series

Consider three resistors $R_1$, $R_2$, and $R_3$ connected end-to-end in series across a battery maintaining a potential difference $V$.

1. In a series circuit, the same electric current $I$ passes through each resistor.
2. The total potential difference $V$ across the combination is equal to the sum of potential differences across each individual resistor:
   $$V = V_1 + V_2 + V_3$$
3. Applying Ohm’s Law ($V = IR$) to each individual resistor:
   $$V_1 = IR_1,\quad V_2 = IR_2,\quad V_3 = IR_3$$
4. If $R_s$ is the equivalent resistance of the series combination:
   $$V = IR_s$$
5. Substituting these values into the total potential difference equation:
   $$IR_s = IR_1 + IR_2 + IR_3$$
   Dividing both sides by $I$:
   $$R_s = R_1 + R_2 + R_3$$

Derivation 2: Equivalent Resistance for Resistors in Parallel

Consider three resistors $R_1$, $R_2$, and $R_3$ connected in parallel between common junctions $A$ and $B$ across a battery of voltage $V$.

1. In a parallel circuit, the potential difference $V$ across each resistor is identical.
2. The total current $I$ drawn from the source divides across the individual branches:
   $$I = I_1 + I_2 + I_3$$
3. Applying Ohm’s Law to each branch:
   $$I_1 = \frac{V}{R_1},\quad I_2 = \frac{V}{R_2},\quad I_3 = \frac{V}{R_3}$$
4. If $R_p$ is the equivalent resistance of the parallel combination:
   $$I = \frac{V}{R_p}$$
5. Substituting these into the total current equation:
   $$\frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$$
   Dividing both sides by $V$:
   $$\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$

NCERT In-Text Questions and Solutions (Step-by-Step)

In-Text Questions (Page Reference: Section 8.1 - Electric Current and Circuit)

Question 1: What does an electric circuit mean?

Answer: An electric circuit is a continuous and closed conducting loop or path through which an electric current can flow. It typically comprises electrical components such as a source of electric current (cell/battery), a switch or plug key, conducting connecting wires, and an electrical load (such as a bulb or resistor). If the circuit is broken or the switch is turned off, the circuit becomes open and current stops flowing.

Question 2: Define the unit of current.

Answer: The SI unit of electric current is the Ampere ($\text{A}$).

One Ampere is defined as the flow of electric current through a conductor when one Coulomb of electric charge passes through any cross-section of the conductor in one second.

$$\text{1 Ampere } (1\text{ A}) = \frac{1\text{ Coulomb } (1\text{ C})}{1\text{ Second } (1\text{ s})}$$

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

Solution:

• Given: Total charge $Q = 1\text{ C}$, Elementary charge of an electron $e = 1.6 \times 10^{-19}\text{ C}$
• To Find: Number of electrons ($n$)
• Formula: $Q = n \cdot e \implies n = \frac{Q}{e}$
• Substitution & Calculation:
  $$n = \frac{1\text{ C}}{1.6 \times 10^{-19}\text{ C}} = \frac{10^{19}}{1.6} = \frac{10 \times 10^{18}}{1.6} = 6.25 \times 10^{18}$$
• Final Answer: $6.25 \times 10^{18}\text{ electrons}$ constitute one coulomb of charge.

In-Text Questions (Page Reference: Section 8.2 - Electric Potential & Potential Difference)

Question 4: Name a device that helps to maintain a potential difference across a conductor.

Answer: An electric cell or a battery (a group of cells) helps maintain a potential difference across a conductor through chemical reactions occurring inside its electrolyte.

Question 5: How much energy is given to each coulomb of charge passing through a $6\text{ V}$ battery?

Solution:

• Given: Charge $Q = 1\text{ C}$, Potential Difference $V = 6\text{ V}$
• To Find: Energy given (Work done, $W$)
• Formula: $V = \frac{W}{Q} \implies W = V \times Q$
• Substitution & Calculation:
  $$W = 6\text{ V} \times 1\text{ C} = 6\text{ J}$$
• Final Answer: The energy transferred is $6\text{ Joules (J)}$.

In-Text Questions (Page Reference: Section 8.4 - Resistance and Factors Affecting Conductor Resistance)

Question 6: On what factors does the resistance of a conductor depend?

Answer: The electrical resistance ($R$) of a metallic conductor depends on four primary factors:

1. Length of the conductor ($l$): Resistance is directly proportional to length ($R \propto l$). Doubling the length doubles the resistance.
2. Area of cross-section ($A$): Resistance is inversely proportional to cross-sectional area ($R \propto \frac{1}{A}$). A thicker wire offers less resistance.
3. Nature of the material: Quantified by the electrical resistivity ($\rho$) of the substance. Good conductors like silver and copper have very low resistivity, while insulators and alloys (like nichrome) have high resistivity.
4. Temperature of the conductor: For pure metallic conductors, resistance increases linearly with an increase in temperature.

Question 7: A wire of given material having length $l$ and area of cross-section $A$ has a resistance of $4\ \Omega$. What would be the resistance of another wire of the same material having length $l/2$ and area of cross-section $2A$?

Solution:

• Given:
  First wire: Length $= l$, Area $= A$, Resistance $R_1 = \rho \frac{l}{A} = 4\ \Omega$
  Second wire: Length $l_2 = \frac{l}{2}$, Area $A_2 = 2A$, Material is identical so resistivity remains $\rho$.
• To Find: Resistance of the second wire ($R_2$)
• Formula: $R_2 = \rho \frac{l_2}{A_2}$
• Substitution:
  $$R_2 = \rho \frac{\frac{l}{2}}{2A} = \rho \frac{l}{4A} = \frac{1}{4} \left(\rho \frac{l}{A}\right)$$
• Calculation:
  $$R_2 = \frac{1}{4} \times R_1 = \frac{1}{4} \times 4\ \Omega = 1\ \Omega$$
• Final Answer: The resistance of the new wire is $1\ \Omega$.

In-Text Questions (Page Reference: Section 8.5 & 8.6 - Resistor Networks & Heating Effect)

Question 8: How can three resistors of resistances $2\ \Omega$, $3\ \Omega$, and $6\ \Omega$ be connected to give a total resistance of (a) $4\ \Omega$, (b) $1\ \Omega$?

Solution:

(a) To obtain $4\ \Omega$:

• Connect the $3\ \Omega$ and $6\ \Omega$ resistors in parallel, and connect this parallel combination in series with the $2\ \Omega$ resistor.
• Equivalent resistance of parallel branch ($R_p$):
  $$\frac{1}{R_p} = \frac{1}{3} + \frac{1}{6} = \frac{2 + 1}{6} = \frac{3}{6} = \frac{1}{2} \implies R_p = 2\ \Omega$$
• Total resistance ($R_{total}$):
  $$R_{total} = 2\ \Omega + R_p = 2\ \Omega + 2\ \Omega = 4\ \Omega$$
• Final Answer (a): Connect $3\ \Omega$ and $6\ \Omega$ in parallel, then in series with $2\ \Omega$.

(b) To obtain $1\ \Omega$:

• Connect all three resistors ($2\ \Omega$, $3\ \Omega$, and $6\ \Omega$) in parallel.
• Equivalent resistance ($R_p$):
  $$\frac{1}{R_p} = \frac{1}{2} + \frac{1}{3} + \frac{1}{6} = \frac{3 + 2 + 1}{6} = \frac{6}{6} = 1 \implies R_p = 1\ \Omega$$
• Final Answer (b): Connect all three resistors in parallel.

Question 9: Compute the heat generated while transferring $96,000\text{ C}$ of charge in two hours through a potential difference of $40\text{ V}$.

Solution:

• Given: Charge $Q = 96,000\text{ C}$, Potential difference $V = 40\text{ V}$, Time $t = 2\text{ h} = 7200\text{ s}$
• To Find: Heat generated ($H$)
• Formula: $H = W = V \times Q$ (Note: When total charge and voltage are given, heat produced is simply the total electrical work done).
• Substitution & Calculation:
  $$H = 40\text{ V} \times 96,000\text{ C} = 3,840,000\text{ J} = 3.84 \times 10^6\text{ J} = 3.84\text{ MJ}$$
• Final Answer: The heat generated is $3.84 \times 10^6\text{ Joules}$ ($3.84\text{ MJ}$).

NCERT Textbook Exercise Solutions (Complete Working)

Exercise Question 1: A piece of wire of resistance $R$ is cut into five equal parts. These parts are then connected in parallel. If the equivalent resistance of this combination is $R'$, then the ratio $R/R'$ is:

(a) $1/25$ (b) $1/5$ (c) $5$ (d) $25$

Solution:

1. When a wire of uniform resistance $R$ is cut into $5$ equal segments, the resistance of each individual part becomes:
   $$r = \frac{R}{5}$$
2. When these $5$ segments are connected in parallel, the equivalent resistance $R'$ is given by:
   $$\frac{1}{R'} = \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} + \frac{1}{r} = \frac{5}{r}$$
3. Substitute $r = \frac{R}{5}$ into the expression:
   $$\frac{1}{R'} = \frac{5}{\frac{R}{5}} = \frac{25}{R} \implies \frac{R}{R'} = 25$$

Correct Option: (d) $25$

Exercise Question 2: Which of the following terms does not represent electrical power in a circuit?

(a) $I^2R$ (b) $IR^2$ (c) $VI$ (d) $V^2/R$

Solution:

Electric power is defined as $P = VI$. From Ohm’s law ($V = IR$ and $I = V/R$):

• Substituting $V = IR \implies P = (IR)I = I^2R$
• Substituting $I = V/R \implies P = V(V/R) = V^2/R$

The expression $IR^2$ has no physical equivalence in electrical circuits.

Correct Option: (b) $IR^2$

Exercise Question 3: An electric bulb is rated $220\text{ V}$ and $100\text{ W}$. When it is operated on $110\text{ V}$, the power consumed will be:

(a) $100\text{ W}$ (b) $75\text{ W}$ (c) $50\text{ W}$ (d) $25\text{ W}$

Solution:

• Step 1: Calculate the internal filament resistance ($R$) of the bulb from its rated specifications:
  Given: $V_{rated} = 220\text{ V}$, $P_{rated} = 100\text{ W}$
  $$P = \frac{V^2}{R} \implies R = \frac{V_{rated}^2}{P_{rated}} = \frac{(220)^2}{100} = \frac{48400}{100} = 484\ \Omega$$
• Step 2: Calculate actual power consumption at operating voltage $V' = 110\text{ V}$ (resistance remains constant):
  $$P' = \frac{(V')^2}{R} = \frac{(110)^2}{484} = \frac{12100}{484} = 25\text{ W}$$

Correct Option: (d) $25\text{ W}$

Exercise Question 4: Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be:

(a) $1:2$ (b) $2:1$ (c) $1:4$ (d) $4:1$

Solution:

• Let the resistance of each individual wire be $R$.
• In Series: Equivalent resistance $R_s = R + R = 2R$.
  Heat produced in time $t$ at voltage $V$:
  $$H_s = \frac{V^2}{R_s}t = \frac{V^2}{2R}t$$
• In Parallel: Equivalent resistance $\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R} \implies R_p = \frac{R}{2}$.
  Heat produced in time $t$ at the same voltage $V$:
  $$H_p = \frac{V^2}{R_p}t = \frac{V^2}{\frac{R}{2}}t = \frac{2V^2}{R}t$$
• Ratio of heat produced:
  $$\frac{H_s}{H_p} = \frac{\frac{V^2}{2R}t}{\frac{2V^2}{R}t} = \frac{1}{4}$$

Correct Option: (c) $1:4$

Exercise Question 5: Why does the cord of an electric heater not glow while the heating element does?

Answer: According to Joule’s law of heating ($H = I^2Rt$), the heat produced in a conductor is directly proportional to its electrical resistance for a given current.

• The connecting cord of a heater is made of highly conductive copper or aluminium wire having extremely low resistance. Consequently, negligible heat is produced, and the cord remains cool without glowing.
• In contrast, the heating element is fabricated from high-resistivity alloys such as Nichrome. Because of its large resistance, a significant amount of heat is generated ($I^2R$), causing the element to become red-hot and glow brightly.

Exercise Question 6: An electric motor takes $5\text{ A}$ from a $220\text{ V}$ line. Determine the power of the motor and the energy consumed in $2\text{ hours}$.

Solution:

• Given: Current $I = 5\text{ A}$, Voltage $V = 220\text{ V}$, Time $t = 2\text{ h} = 2 \times 3600\text{ s} = 7200\text{ s}$
• Part A: Power of the motor ($P$)
  $$P = V \times I = 220\text{ V} \times 5\text{ A} = 1100\text{ W} = 1.1\text{ kW}$$
  Power = $1100\text{ W}$
• Part B: Energy Consumed ($E$)
  In Joules:
  $$E = P \times t = 1100\text{ W} \times 7200\text{ s} = 7,920,000\text{ J} = 7.92 \times 10^6\text{ J}$$
  In Commercial Units ($\text{kWh}$):
  $$E = 1.1\text{ kW} \times 2\text{ h} = 2.2\text{ kWh}$$
  Energy Consumed = $7.92 \times 10^6\text{ J}$ ($2.2\text{ kWh}$)

Common Mistakes to Avoid in Class 10 Electricity

Based on CBSE board examiner reports, students frequently lose marks in Chapter 8 due to predictable numerical and conceptual slip-ups. Keep the following checkpoints in mind:

• Forgetting Standard SI Unit Conversions: Time must always be converted into seconds ($\text{s}$) when calculating heat in Joules ($H = I^2Rt$) or charge in Coulombs ($Q = It$). However, when calculating commercial electrical units ($\text{kWh}$), time must be kept in hours ($\text{h}$) and power in kilowatts ($\text{kW}$).
• Misapplying Power Formulas in Resistor Combinations:
  • Use $P = I^2R$ when comparing components connected in series (since current $I$ is constant). Higher resistance dissipates more power.
  • Use $P = \frac{V^2}{R}$ when comparing appliances connected in parallel across a domestic line (since voltage $V$ is constant). Lower resistance draws more current and consumes higher power.
• Reciprocal Neglect in Parallel Equivalent Resistance: When calculating $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2}$, students frequently forget to take the reciprocal of the final fraction. Always invert your result to state $R_p$.
• Wire Stretching vs. Cutting Problems: If a wire is stretched to double its length ($l' = 2l$), its cross-sectional area automatically halves ($A' = A/2$) because volume remains constant. Therefore, its new resistance increases by a factor of $4$ ($R' = 4R$).
• Ammeter and Voltmeter Connections: In circuit diagrams, always remember that an ammeter is connected in series (low resistance) and a voltmeter is connected in parallel (high resistance) across the component.

CBSE Board Exam Relevance and Weightage

In the CBSE Class 10 Science board paper, Chapter 8 Electricity is a core contributor to Section B, C, and E. The chapter consistently carries 7 to 9 marks distributed across multiple question formats:

Question Type	Marks	Typical Focus Topics
Multiple Choice (MCQ / Assertion-Reason)	1 Mark	Unit conversions, bulb brightness comparisons, $V-I$ graph slope interpretations.
Short Answer (SA-I & SA-II)	2 – 3 Marks	Factors affecting resistance, Joule’s law explanations, equivalent resistance calculations.
Long Answer / Case-Based	4 – 5 Marks	Circuit analysis combining series-parallel resistors, heating appliance power ratings, electricity bill calculations.

More NCERT Solutions and Practice

Solving textbook exercises is the essential first step toward scoring a centum in CBSE Class 10 Science. To build speed and absolute accuracy for board numericals, you must test yourself against mixed circuit problems, previous years' board questions (PYQs), and competency-based case studies.

Access comprehensive NCERT Solutions Science chapter wise, structured sample test papers, and instant answer keys on Theorify QPTool. Generate customized, board-pattern practice test papers for Class 10 Physics at qptool.theorify.in to reinforce your preparation today.