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આ MCQ મોડ્યુલ આના પર આધારિત છે: Electricity – NCERT Exercises
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Electricity – NCERT Exercises
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આ મૂલ્યાંકન આના પર આધારિત હશે: Electricity – NCERT Exercises
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Chapter 11 — Summary
Electric charge & current
Charge is quantised: \(Q = ne\), \(e = 1.6 \times 10^{-19}\) C. Current \(I = Q/t\); unit ampere (1 A = 1 C/s).
Conventional current
Flows from + terminal through the external circuit to − terminal; opposite to electron drift.
Potential difference
\(V = W/Q\); unit volt (1 V = 1 J/C). Measured with a voltmeter connected in parallel.
Ohm's law
\(V = IR\) at constant temperature. Resistance \(R = V/I\); unit ohm (Ω). V–I graph: straight line for ohmic conductors.
Factors affecting R
R ∝ L, R ∝ 1/A, depends on material and rises with temperature for metals. \(R = \rho L/A\).
Resistivity ρ
Material constant, unit Ω·m. Conductors ~10⁻⁸, insulators ~10¹². Alloys such as nichrome have high ρ and stable R.
Series combination
\(R_s = R_1+R_2+R_3+\ldots\). Same current, voltage splits.
Parallel combination
\(1/R_p = 1/R_1+1/R_2+\ldots\). Same voltage, current splits. Used in household wiring.
Heating effect (Joule's law)
\(H = I^2 R t\) joules. Basis of heaters, irons, bulbs, fuses.
Electric power
\(P = VI = I^2R = V^2/R\); unit watt. 1 kW = 1000 W.
Commercial unit
1 kWh = 3.6 × 10⁶ J. Energy (kWh) = Power (kW) × time (h).
Household wiring
Three-wire system (live, neutral, earth). Appliances in parallel. Fuses/MCBs protect against overload and short-circuit.
Key Formulae at a Glance
\(I = Q/t\) • \(V = W/Q\) • \(V = IR\) • \(R = \rho L/A\)
Series: \(R_s = R_1+R_2+R_3\) • Parallel: \(1/R_p = 1/R_1+1/R_2+1/R_3\)
\(H = I^2Rt\) • \(P = VI = I^2R = V^2/R\) • \(1\ \text{kWh} = 3.6\times10^6\) J
Key Terms
Electric chargeProperty of matter causing electric force; unit coulomb (C).
Electric currentRate of flow of charge; I = Q/t; unit ampere.
Potential differenceWork done per unit charge; V = W/Q; unit volt.
Ohm's lawV = IR at constant temperature.
Resistance (R)Opposition to current flow; unit ohm (Ω).
Resistivity (ρ)Material's intrinsic resistance; unit Ω·m.
Ohmic conductorObeys V = IR; V–I graph is a straight line.
Series circuitOne path; same current everywhere.
Parallel circuitMultiple paths; same voltage across each.
Joule's lawHeat H = I²Rt in a resistor.
Electric powerRate of energy consumption; P = VI.
Kilowatt-hourCommercial unit of electrical energy; 1 kWh = 3.6 × 10⁶ J.
FuseLow-melting-point wire for overload protection.
Short circuitLive and neutral touching; near-zero R; huge current.
NichromeAlloy used in heating coils; high ρ, high m.p.
NCERT Exercises — Step-by-Step Solutions
Q1. 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', find the ratio R/R'.
Each cut piece has resistance \(R/5\). Five such resistors in parallel:
\[ \dfrac{1}{R'} = 5 \times \dfrac{5}{R} = \dfrac{25}{R} \Rightarrow R' = \dfrac{R}{25}. \]
So \(R/R' = 25\).
Q2. Which of the following terms does not represent electrical power in a circuit?
(a) I²R (b) IR² (c) VI (d) V²/R
Options (a), (c) and (d) are valid expressions for power. Option (b) IR² has units of V·Ω = W·Ω/A, which is not power. Answer: (b).
Q3. An electric bulb is rated 220 V and 100 W. When it is operated on 110 V, the power consumed will be
(a) 100 W (b) 75 W (c) 50 W (d) 25 W
At 220 V, \(R = V^2/P = 220^2/100 = 484\) Ω (assumed constant). At 110 V, \(P' = V'^2/R = 110^2/484 = 12100/484 = 25\) W.
Answer: (d) 25 W.
Q4. Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then in parallel in a circuit across the same potential difference. The ratio of heat produced in series to parallel combinations would be
(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1
Let each wire's resistance = R. Series \(R_s = 2R\); parallel \(R_p = R/2\). For the same V, heat in time t is \(H = V^2 t/R\). Ratio \(H_s : H_p = (1/R_s) : (1/R_p) = (1/2R) : (2/R) = 1:4\).
Answer: (c) 1:4.
Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
A voltmeter is connected in parallel across the two points between which the potential difference is to be measured. An ideal voltmeter has very high resistance so it draws negligible current from the circuit.
Q6. A copper wire has diameter 0.5 mm and resistivity 1.6 × 10⁻⁸ Ω·m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?
Radius r = 0.25 mm = 2.5 × 10⁻⁴ m. Area \(A = \pi r^2 = 3.14 \times (2.5 \times 10^{-4})^2 = 1.96 \times 10^{-7}\) m².
Length \(L = RA/\rho = (10)(1.96 \times 10^{-7})/(1.6 \times 10^{-8}) = 122.5\) m.
If diameter doubles, area becomes 4× larger, so R becomes 1/4. New R = 10/4 = 2.5 Ω.
Q7. The values of current I flowing in a given resistor for the corresponding values of potential difference V across the resistor are given below.
V (V): 0.5, 1.0, 2.0, 3.0, 4.0 ; I (A): 0.1, 0.2, 0.4, 0.6, 0.8. Plot a graph between V and I and calculate the resistance of that resistor.
The data shows V/I = 5 Ω in every case. The V–I graph is a straight line through the origin, and its slope corresponds to 1/R. Therefore R = 5 Ω.
Q8. When a 12 V battery is connected across an unknown resistor, there is a current of 2.5 mA in the circuit. Find the value of the resistance of the resistor.
\(R = V/I = 12 / (2.5 \times 10^{-3}) = 4800\) Ω = 4.8 kΩ.
Q9. A battery of 9 V is connected in series with resistors of 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω. How much current would flow through the 12 Ω resistor?
Total resistance = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω. Current \(I = 9/13.4 ≈ 0.671\) A. In a series circuit, the same current flows through every resistor, including the 12 Ω one.
I ≈ 0.67 A through the 12 Ω resistor.
Q10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Required equivalent resistance \(R_p = V/I = 220/5 = 44\) Ω. If n resistors of 176 Ω are in parallel, \(R_p = 176/n\). So \(n = 176/44 = 4\).
Four 176 Ω resistors in parallel.
Q11. Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.
(i) Connect two 6 Ω resistors in parallel: \(R_p = (6 \times 6)/12 = 3\) Ω. Put the third 6 Ω in series: \(3 + 6 = 9\) Ω. ✓
(ii) Connect two 6 Ω in series: \(6 + 6 = 12\) Ω. Put the third 6 Ω in parallel with this: \(R = (12 \times 6)/(12+6) = 72/18 = 4\) Ω. ✓
Q12. Several electric bulbs designed to be used on a 220 V electric supply line, are rated 10 W. How many lamps can be connected in parallel with each other across the two wires of a 220 V line if the maximum allowable current is 5 A?
Current drawn by one bulb = \(P/V = 10/220 = 0.04545\) A. Number of bulbs = \(5/0.04545 ≈ 110\).
110 bulbs can be connected.
Q13. A hot plate of an electric oven connected to a 220 V line has two resistance coils A and B, each of 24 Ω resistance, which may be used separately, in series or in parallel. What are the currents in the three cases?
Separately (single coil): \(I = 220/24 ≈ 9.17\) A.
In series: \(R_s = 48\) Ω; \(I = 220/48 ≈ 4.58\) A.
In parallel: \(R_p = 12\) Ω; \(I = 220/12 ≈ 18.33\) A.
Q14. Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.
(i) Series: total R = 1 + 2 = 3 Ω; I = 6/3 = 2 A; P₂ = I²R = 2² × 2 = 8 W.
(ii) Parallel: voltage across 2 Ω = 4 V; P₂ = V²/R = 16/2 = 8 W.
Both cases give the same power (8 W) in the 2 Ω resistor.
Q15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to the electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Total power = 100 + 60 = 160 W. Current drawn \(I = P/V = 160/220 ≈ 0.727\) A.
Q16. Which uses more energy, a 250 W TV set in 1 hr, or a 1200 W toaster in 10 minutes?
TV energy: \(250 \times 3600 = 9.0 \times 10^5\) J.
Toaster energy: \(1200 \times 600 = 7.2 \times 10^5\) J.
The TV uses more energy (9 × 10⁵ J vs 7.2 × 10⁵ J).
Q17. An electric heater of resistance 8 Ω draws 15 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
The rate at which heat is developed is the electric power: \(P = I^2 R = 15^2 \times 8 = 225 \times 8 = 1800\) W = 1.8 kW. (The time in the question is given to test comprehension; it is not needed for the rate.)
Q18. Explain the following. (a) Why is tungsten used almost exclusively for filament of electric lamps? (b) Why are the conductors of electric heating devices, such as bread-toasters and electric irons, made of an alloy rather than a pure metal? (c) Why is the series arrangement not used for domestic circuits? (d) How does the resistance of a wire vary with its area of cross-section? (e) Why are copper and aluminium wires usually employed for electricity transmission?
(a) Tungsten has a very high melting point (~3380 °C), high resistivity and low rate of evaporation. It can be heated to about 2500 °C to glow white-hot without melting, making it ideal for bulb filaments.
(b) Alloys like nichrome have higher resistivity than pure metals (more heat for same current), don't oxidise readily at high temperatures and have high melting points, so they last long as heating elements.
(c) In series, all appliances share the same current and the voltage divides, so each appliance gets less than mains voltage. If one appliance fails or is switched off, the entire circuit breaks. Also, different appliances need different currents. These problems are avoided by parallel wiring.
(d) Resistance is inversely proportional to area of cross-section: \(R \propto 1/A\).
(e) Copper and aluminium have low resistivity (good conductors), are ductile (drawn into long wires easily), and relatively inexpensive. Aluminium, being lighter, is preferred for long-distance overhead transmission lines; copper, being more conductive, is preferred for household wiring.
Frequently Asked Questions — NCERT Exercises & Intext Questions
How do I solve NCERT Class 10 Science Chapter 11 (Electricity) exercise questions for the CBSE board exam?
Solve NCERT Chapter 11 — Electricity — exercise questions by first reading the question carefully, writing down the given data, recalling the relevant concepts like current, voltage, resistance, and applying them step by step. This Part 4 covers every intext and end-of-chapter exercise from the NCERT textbook. Write balanced equations, label diagrams clearly and show each step — CBSE Class 10 board examiners award step marks even if the final answer has a small slip. Practising these solutions strengthens conceptual clarity and builds speed for the board exam.
Are the NCERT intext questions from Electricity important for the Class 10 board exam?
Yes, NCERT intext questions for Chapter 11 Electricity are highly important for the CBSE Class 10 Science board exam. Many board questions are directly lifted or only slightly modified from these intext questions, and they test the foundational concepts — current, voltage, resistance — that chapter-end questions build on. Attempt every intext question first, then move on to the exercises. This practice ensures complete NCERT coverage, which is the CBSE exam's primary source.
What types of questions from Electricity are asked in the CBSE Class 10 Science board exam?
The CBSE Class 10 board paper asks a mix of question types from Electricity: 1-mark MCQ and assertion-reason, 2-mark short answers, 3-mark explanations, 5-mark long answers with diagrams or derivations, and 4-mark competency-based / case-study questions. These test understanding of current, voltage, resistance, Ohm's law. Practising every NCERT exercise and intext question prepares you to answer all of these formats with confidence.
How many marks does Chapter 11 — Electricity — carry in the Class 10 Science CBSE paper?
Chapter 11 — Electricity — is part of the Class 10 Science syllabus and typically contributes 5–9 marks in the CBSE board paper, depending on the annual weightage. Questions are drawn from definitions, reasoning, numerical/descriptive problems and diagrams on topics like current, voltage, resistance. Solving the NCERT exercises in this part is essential because CBSE directly references NCERT for question design.
Where can I find step-by-step NCERT solutions for Chapter 11 Electricity Class 10 Science?
You can find complete, step-by-step NCERT solutions for Chapter 11 Electricity Class 10 Science on MyAiSchool. Every intext and end-of-chapter exercise question is solved with full working, labelled diagrams and CBSE-aligned mark distribution. Solutions highlight key points about current, voltage, resistance that examiners look for. This makes revision quick and exam-focused for Class 10 CBSE board students.
What is the best way to revise Electricity before the Class 10 Science board exam?
The best way to revise Electricity for the CBSE Class 10 Science board exam is a three-pass approach. First pass: skim the chapter and note down key terms like current, voltage, resistance in a one-page mind map. Second pass: solve every NCERT intext and exercise question without looking at the solution, then self-check. Third pass: attempt previous CBSE board questions and competency-based questions under timed conditions. This structured revision secures full marks for this chapter.
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