Class 10 Science Chapter 11 Electricity NCERT Solutions 2026 PDF Download
📝 Introduction
Electricity is the flow of electric charge that powers our modern world. In this chapter, you will understand the fundamental concepts of electric current, potential difference, and the relationship between them defined by Ohm's Law. You will explore the factors affecting resistance and how resistors behave differently in Series and Parallel combinations. The chapter also covers the practical applications of the Heating Effect of Electric Current (Joule's Law) and calculations involving Electric Power and Energy consumption in households.
[Image of Electric Circuit Diagram Symbols]🔑 Key Concepts & Formulas
- Current ($I$): $I = \frac{Q}{t}$ (Unit: Ampere, A).
- Potential Difference ($V$): $V = \frac{W}{Q}$ (Unit: Volt, V).
- Ohm's Law: $V = IR$ (at constant temperature).
- Resistance ($R$): $R = \rho \frac{L}{A}$ ($\rho$ = Resistivity).
- Series Combination: $R_s = R_1 + R_2 + R_3$ (Current is same).
- Parallel Combination: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$ (Voltage is same).
- Joule's Law of Heating: $H = I^2Rt$ (Unit: Joule, J).
- Electric Power ($P$): $P = VI = I^2R = \frac{V^2}{R}$ (Unit: Watt, W).
- Commercial Unit of Energy: 1 kWh = $3.6 \times 10^6$ Joules.
📚 Part 1: NCERT Solutions (In-Text & Exercises)
Q1: What does an electric circuit mean?
Ans: A continuous and closed path of an electric current is called an electric circuit. It consists of electric devices (bulb), a source of electricity (battery), and wires connected together.
Q2: Define the unit of current.
Ans: The unit of current is Ampere (A). One ampere is defined as the flow of one coulomb of charge per second. ($1 A = 1 C/1 s$).
Q3: Calculate the number of electrons constituting one coulomb of charge.
Ans: Charge on 1 electron ($e$) = $1.6 \times 10^{-19} C$.
Total charge ($Q$) = $ne$.
$1 = n \times 1.6 \times 10^{-19}$
$n = \frac{1}{1.6 \times 10^{-19}} = 6.25 \times 10^{18}$ electrons.
Q4: Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Ans:
1. High Resistivity: Alloys (like Nichrome) have higher resistivity than pure metals, producing more heat.
2. No Oxidation: Alloys do not oxidize (burn) easily at high temperatures.
Q5: State Ohm's Law.
Ans: Ohm's law states that the potential difference ($V$) across the ends of a given metallic wire in an electric circuit is directly proportional to the current ($I$) flowing through it, provided its temperature remains the same.
Mathematically: $V \propto I \Rightarrow V = IR$.
Q6: Why is the series arrangement not used for domestic circuits?
Ans:
1. Single Switch: One switch controls all devices; you cannot turn them on/off individually.
2. Failure: If one appliance fails, the circuit breaks, and none work.
3. Voltage Division: The total voltage ($220V$) gets divided, so appliances don't work at full power.
4. Resistance: Total resistance increases, reducing current flow.
Q7: Explain the factors on which the resistance of a conductor depends.
Ans:
1. Length ($L$): Resistance is directly proportional to length ($R \propto L$).
2. Area of Cross-section ($A$): Resistance is inversely proportional to area ($R \propto 1/A$).
3. Material: Depends on the nature of the material (Resistivity $\rho$).
4. Temperature: Resistance increases with temperature for metals.
Q8: An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?
Ans:
Given: $V = 220 V, I = 0.50 A$.
Formula: $P = VI$.
$P = 220 \times 0.50 = 110 \text{ W}$.
⚡ Part 2: 15 Extra Practice Questions (PYQ Style)
Short Answer Type Questions
Q1: What determines the rate at which energy is delivered by a current?
Ans: The Electric Power determines the rate of energy consumption. ($P = W/t$).
Q2: Define Electric Potential Difference.
Ans: The electric potential difference between two points in an electric circuit is defined as the work done to move a unit charge from one point to the other. ($V = W/Q$).
Q3: Two resistors of $10\Omega$ and $20\Omega$ are connected in series. Calculate the equivalent resistance.
Ans:
In series, $R_s = R_1 + R_2$.
$R_s = 10 + 20 = 30 \Omega$.
Q4: Why is Tungsten used almost exclusively for filament of electric lamps?
Ans:
1. It has a very high melting point ($3380^\circ C$), so it does not melt at high temperatures.
2. It emits light when heated to incandescence.
Q5: Draw a schematic diagram of a circuit consisting of a battery of three cells of 2V each, a $5\Omega$ resistor, an $8\Omega$ resistor, and a $12\Omega$ resistor, and a plug key, all connected in series.
Ans:
Long Answer Type Questions
Q6: (a) State Joule's Law of Heating. (b) Calculate the heat generated while transferring 96000 coulomb of charge in one hour through a potential difference of 50 V.
Ans:
(a) Joule's Law states that heat produced in a resistor is directly proportional to the square of current ($I^2$), resistance ($R$), and time ($t$). ($H \propto I^2Rt$).
(b) Given: $Q = 96000 C, t = 1hr = 3600s, V = 50V$.
Heat $H = V \times Q$ (Since $V=W/Q \Rightarrow W=VQ$).
$H = 50 \times 96000 = 4,800,000 \text{ J} = 4.8 \times 10^6 \text{ J}$.
Q7: Three resistors $R_1, R_2, R_3$ are connected in parallel. Derive the expression for equivalent resistance.
Ans:
In parallel, $V$ is same, $I$ divides.
$I = I_1 + I_2 + I_3$
From Ohm's Law: $I = V/R_p$, $I_1 = V/R_1$, etc.
$\frac{V}{R_p} = \frac{V}{R_1} + \frac{V}{R_2} + \frac{V}{R_3}$
Cancel $V$: $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.
Q8: An electric refrigerator rated 400 W operates 8 hours/day. What is the cost of energy to operate it for 30 days at ₹3.00 per kWh?
Ans:
Energy per day = $P \times t = 400 \text{ W} \times 8 \text{ h} = 3200 \text{ Wh} = 3.2 \text{ kWh}$.
Total Energy for 30 days = $3.2 \times 30 = 96 \text{ kWh}$.
Cost = $96 \times 3 = ₹288$.
Competency Based Questions
Q9: Two wires of the same material and same length have radii 1 mm and 2 mm respectively. Compare their resistances.
Ans:
$R \propto \frac{1}{A}$ and $A = \pi r^2$. So $R \propto \frac{1}{r^2}$.
Ratio $R_1 : R_2 = r_2^2 : r_1^2$.
$R_1 : R_2 = (2)^2 : (1)^2 = 4 : 1$.
The thinner wire (1mm) has 4 times the resistance of the thicker wire.
Q10: Why does the cord of an electric heater not glow while the heating element does?
Ans: The heating element is made of an alloy (like Nichrome) which has very high resistance. According to $H=I^2Rt$, high resistance produces large heat, making it glow. The cord is made of Copper (low resistance), so negligible heat is produced, and it does not glow.
Q11: Name the instrument used to measure (i) Current (ii) Potential Difference. How are they connected?
Ans:
(i) Ammeter: Connected in Series (Low resistance).
(ii) Voltmeter: Connected in Parallel (High resistance).
Q12: What is the function of a fuse in an electric circuit?
Ans: A fuse is a safety device with a low melting point. If current exceeds the safe limit (due to short circuit or overload), the fuse wire melts and breaks the circuit, protecting appliances from damage.
Q13: How does the resistance of a wire change if its length is doubled?
Ans: Since $R \propto L$, if length is doubled, the resistance also doubles.
Q14: Define 1 Volt.
Ans: One volt is the potential difference between two points in a current-carrying conductor when 1 Joule of work is done to move a charge of 1 Coulomb from one point to the other ($1V = 1J/1C$).
Q15: What represents the slope of a V-I graph?
Ans: The slope of a V-I graph (V on y-axis, I on x-axis) represents the Resistance (R) of the conductor.