Science — CBSE Class X Sample Paper 1 (2025-26)
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આ MCQ મોડ્યુલ આના પર આધારિત છે: Combinations of Resistors, Electric Power and Heating Effect
આ મૂલ્યાંકન આના પર આધારિત હશે: Combinations of Resistors, Electric Power and Heating Effect
મૂલ્યાંકન બનાવવામાં તેમની સામગ્રી સામેલ કરવા ચિત્રો, PDF અથવા Word દસ્તાવેજ અપલોડ કરો.
Real appliances contain many resistors (bulbs, motors, heater coils) wired together. In this part we learn how resistors combine in series and in parallel, then turn to the heating effect of current (Joule's law) and to electric power. Finally we see how these ideas are applied in household wiring.
Two (or more) resistors can be wired in a circuit in two fundamental ways — series (end-to-end) or parallel (side-by-side).
Resistors are said to be connected in series when they are joined end-to-end so that the same current flows through each of them. The potential difference across the combination is the sum of the potential differences across each resistor.
If V is the total voltage and \(V_1, V_2, V_3\) are the voltages across \(R_1, R_2, R_3\), then \(V = V_1 + V_2 + V_3\). Applying \(V = IR\) to each:
\[ IR_s = IR_1 + IR_2 + IR_3 \quad\Rightarrow\quad \boxed{R_s = R_1 + R_2 + R_3} \]
Resistors are in parallel when they share the same two end-points, so the same potential difference acts across each. The total current is the sum of currents through each branch.
Total current: \(I = I_1 + I_2 + I_3\). Since each branch sees the same V:
\[ \dfrac{V}{R_p} = \dfrac{V}{R_1} + \dfrac{V}{R_2} + \dfrac{V}{R_3} \quad\Rightarrow\quad \boxed{\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}} \]
Q. Three resistors of 5 Ω, 10 Ω and 15 Ω are connected in series to a 12 V battery. Find (a) equivalent resistance, (b) current, (c) voltage across each resistor.
Solution. (a) \(R_s = 5 + 10 + 15 = 30\ \Omega\). (b) \(I = V/R_s = 12/30 = 0.4\) A. (c) \(V_1 = 0.4 \times 5 = 2\) V; \(V_2 = 0.4 \times 10 = 4\) V; \(V_3 = 0.4 \times 15 = 6\) V. Check: \(2 + 4 + 6 = 12\) V. ✓
Q. Two resistors of 6 Ω and 3 Ω are in parallel across a 9 V battery. Find (a) equivalent resistance, (b) total current, (c) current in each branch.
Solution. (a) \(R_p = \dfrac{6 \times 3}{6+3} = \dfrac{18}{9} = 2\ \Omega\). (b) \(I = 9/2 = 4.5\) A. (c) \(I_1 = 9/6 = 1.5\) A; \(I_2 = 9/3 = 3.0\) A. Check: \(1.5 + 3.0 = 4.5\) A. ✓
Q. Three resistors 2 Ω, 3 Ω and 6 Ω are in parallel across a 6 V battery. Find \(R_p\) and total current.
Solution. \(\dfrac{1}{R_p} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3+2+1}{6} = 1\). So \(R_p = 1\ \Omega\). \(I = V/R_p = 6/1 = 6\) A.
Q. A 4 Ω resistor is in series with a parallel combination of two resistors 6 Ω and 3 Ω. The source is 12 V. Find the current in the 4 Ω resistor and in the 6 Ω branch.
Solution. Parallel part: \(R_p = (6 \times 3)/(6+3) = 2\ \Omega\). Total: \(R = 4 + 2 = 6\ \Omega\). Main current: \(I = 12/6 = 2\) A (this is the current through the 4 Ω resistor). Voltage across the parallel part: \(V_p = I R_p = 2 \times 2 = 4\) V. Current through 6 Ω branch = \(4/6 = 0.67\) A; through 3 Ω branch = \(4/3 = 1.33\) A. Check: \(0.67 + 1.33 ≈ 2\) A. ✓
Q. Two resistors in parallel give 4 Ω. One of them is 12 Ω. Find the other.
Solution. \(1/4 = 1/12 + 1/R_2 \Rightarrow 1/R_2 = 1/4 - 1/12 = (3-1)/12 = 2/12 = 1/6\). So \(R_2 = 6\ \Omega\).
When current flows through a resistor, electrical energy is converted to heat. The drifting electrons collide with metal atoms, giving up kinetic energy as thermal vibrations. This is the heating effect of current, used in heaters, geysers, irons and bulbs.
Work done by the battery to drive charge Q through potential difference V is \(W = VQ\). Since \(Q = It\), we have \(W = V I t\). For a resistor obeying Ohm's law, \(V = IR\). So work done appears as heat:
\[ H = W = VIt = (IR)(I)(t) = I^2 R t \]
The rate at which electrical energy is dissipated (or consumed) in a circuit is called electric power.
The joule is too tiny to measure household consumption. Electric companies use a bigger unit — the kilowatt-hour (kWh), popularly called a "unit".
\[ 1\ \text{kWh} = 1\ \text{kW} \times 1\ \text{h} = 1000\ \text{W} \times 3600\ \text{s} = 3.6 \times 10^{6}\ \text{J} \]
The electrical energy consumed by an appliance is: Energy (kWh) = Power (kW) × time (h).
Q. An electric bulb is rated 220 V, 100 W. Find (a) its resistance, (b) current drawn.
Solution. (a) From \(P = V^2/R\), \(R = V^2/P = 220 \times 220 / 100 = 484\) Ω. (b) \(I = P/V = 100/220 ≈ 0.45\) A.
Q. A current of 5 A flows through a 10 Ω resistor for 3 minutes. How much heat is produced?
Solution. \(H = I^2 R t = 5^2 \times 10 \times 180 = 25 \times 10 \times 180 = 45{,}000\) J = 45 kJ.
Q. An electric iron of 1000 W is used for 2 hours daily. Find the energy consumed in a month (30 days) and the cost at ₹5 per unit.
Solution. Daily energy = 1 kW × 2 h = 2 kWh. Monthly = 2 × 30 = 60 kWh. Cost = 60 × 5 = ₹300.
Q. Two bulbs of 40 W and 60 W (both rated 220 V) are connected in (i) series, (ii) parallel across 220 V mains. Compare brightness.
Solution. Resistances: \(R_{40} = 220^2/40 = 1210\) Ω; \(R_{60} = 220^2/60 ≈ 807\) Ω.
Series: Same current through both. Power dissipated in each = \(I^2 R\), so the bulb with higher R (the 40 W one) dissipates MORE power and glows brighter — opposite of what one might guess!
Parallel: Same voltage across both. Power = \(V^2/R\), so the lower-R bulb (60 W) is brighter. This is the normal case in households.
Q. An electric heater is rated 2 kW, 220 V. Find (a) current, (b) resistance.
Solution. (a) \(I = P/V = 2000/220 ≈ 9.09\) A. (b) \(R = V/I = 220/9.09 ≈ 24.2\) Ω.
Q. A 4 Ω and a 6 Ω resistor are in parallel across 12 V. Find the heat developed in 1 minute.
Solution. \(R_p = (4 \times 6)/(4+6) = 2.4\ \Omega\). Power \(P = V^2/R_p = 144/2.4 = 60\) W. Heat in 60 s: \(H = Pt = 60 \times 60 = 3600\) J.
Q. A fuse wire is rated at 5 A at 220 V. What is the maximum power of an appliance it can safely handle?
Solution. \(P_{max} = VI = 220 \times 5 = 1100\) W = 1.1 kW.
Q. Express 1.8 × 10⁷ J in kilowatt-hours.
Solution. \(1\ \text{kWh} = 3.6 \times 10^6\) J. So \(1.8 \times 10^7 / 3.6 \times 10^6 = 5\ \text{kWh}\).
Q. A house uses: 4 bulbs of 40 W for 5 h/day, 1 TV of 120 W for 4 h/day and 1 fridge of 200 W running 24 h/day. Find monthly energy (30 days) and cost at ₹6/unit.
Solution. Daily energy:
Bulbs: 4 × 40 × 5 = 800 Wh
TV: 120 × 4 = 480 Wh
Fridge: 200 × 24 = 4800 Wh
Total daily = 6080 Wh = 6.08 kWh. Monthly = 6.08 × 30 = 182.4 kWh. Cost = 182.4 × 6 = ₹1094.4.
Q. If the current through a resistor is doubled keeping R and t the same, by what factor does the heat produced change?
Solution. From \(H = I^2 R t\), doubling I quadruples H. Heat becomes 4 times.
Electricity reaches Indian homes at 220 V ± 10%, 50 Hz. Three wires enter the main switch: live (red or brown), neutral (black or blue) and earth (green or yellow-green). The live wire carries current; the neutral completes the circuit; the earth wire is connected to a metal plate buried in the ground and is connected to metal casings of appliances for safety.
Aim: To compare the brightness of two identical bulbs connected first in series and then in parallel with the same battery.
Materials: two identical 2.5 V torch bulbs in holders, two 1.5 V cells (3 V battery), switch, connecting wires.
Procedure:
In series, each bulb gets only half the battery voltage (1.5 V), so both glow dimly; unscrewing one breaks the circuit and the other goes off.
In parallel, each bulb gets the full 3 V and glows brightly; unscrewing one does not affect the other. This is exactly why domestic appliances are wired in parallel — independent operation and full voltage to every device.
Options: (A) Both A & R true, R correctly explains A. (B) Both A & R true, R does NOT explain A. (C) A true, R false. (D) A false, R true.
Resistors in Series/Parallel, Power & Heating Effect is a key topic in NCERT Class 10 Science Chapter 11 — Electricity. It explains combinations of resistors, joule's law of heating and electric power with numerical problems. Core ideas covered include resistors in series, resistors in parallel, equivalent resistance, heating effect of current. Mastering this subtopic is essential for scoring well in the CBSE Class 10 Science board exam because board papers repeatedly test these concepts through MCQs, short answers and long-answer questions. This part gives a complete, exam-ready explanation with activities, diagrams and competency-based practice aligned to NCERT.
Resistors in series is important in NCERT Class 10 Science because it forms the foundation for understanding resistors in series/parallel, power & heating effect in Chapter 11 — Electricity. Without a clear idea of resistors in series, students cannot answer higher-order CBSE board questions involving resistors in parallel, equivalent resistance, heating effect of current. Board papers regularly include 2-mark and 3-mark questions on this concept, and competency-based questions often link resistors in series to real-life situations. Building clarity here pays off directly in board marks.
The CBSE Class 10 Science board exam tests resistors in series/parallel, power & heating effect through a mix of 1-mark MCQs, 2-mark short answers, 3-mark explanations with examples, 5-mark descriptive questions (often with diagrams or balanced equations) and 4-mark competency-based questions. Expect direct questions on resistors in series, resistors in parallel, equivalent resistance and application-based questions drawn from NCERT activities. Students who follow NCERT thoroughly and practice this chapter's questions consistently score in the 90%+ range.
The key terms to remember for resistors in series/parallel, power & heating effect in NCERT Class 10 Science Chapter 11 are: resistors in series, resistors in parallel, equivalent resistance, heating effect of current, Joule's law of heating, electric power. Each of these concepts carries exam weightage and regularly appears in the CBSE board paper. Write clear one-line definitions of every term in your revision notes and revisit them before the exam. Linking these terms visually through a flowchart or concept map makes recall easier during the Class 10 Science board exam.
Yes, Resistors in Series/Parallel, Power & Heating Effect is a part of the NCERT Class 10 Science syllabus (2025–26) prescribed by CBSE. It falls under Chapter 11 — Electricity — and is examined in the annual board paper. The current syllabus retains the full treatment of resistors in series, resistors in parallel, equivalent resistance as per the NCERT textbook. Because CBSE bases every board question on NCERT, studying this part thoroughly ensures complete syllabus coverage and guarantees marks from this chapter.
Prepare resistors in series/parallel, power & heating effect for the CBSE Class 10 Science board exam in three steps. First, read this NCERT part carefully, highlighting definitions and diagrams of resistors in series, resistors in parallel, equivalent resistance. Second, solve every in-text question and end-of-chapter exercise — CBSE questions often come directly from NCERT. Third, practice competency-based and assertion-reason questions to sharpen reasoning. Write answers in the exam-style format (point-wise with diagrams) and time yourself. This method delivers confidence and full marks in the board exam.
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