This MCQ module is based on: Combinations of Resistors, Electric Power and Heating Effect
TOPIC 42 OF 50
Combinations of Resistors, Electric Power and Heating Effect
🎓 Class 10
Science
CBSE
Theory
Ch 11 — Electricity
⏱ ~26 min
🌐 Language: [gtranslate]
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Introduction — Building Real Circuits
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.
11.8 Combinations of Resistors
Two (or more) resistors can be wired in a circuit in two fundamental ways — series (end-to-end) or parallel (side-by-side).
11.8.1 Resistors in Series
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:
\[ IR_s = IR_1 + IR_2 + IR_3 \quad\Rightarrow\quad \boxed{R_s = R_1 + R_2 + R_3} \]
Key features of series combination
• Same current through each resistor.
• Total voltage is divided across the resistors in the ratio of their resistances.
• Equivalent resistance \(R_s = R_1 + R_2 + R_3 + \ldots\) — always greater than the largest individual R.
• Same current through each resistor.
• Total voltage is divided across the resistors in the ratio of their resistances.
• Equivalent resistance \(R_s = R_1 + R_2 + R_3 + \ldots\) — always greater than the largest individual R.
11.8.2 Resistors in Parallel
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:
\[ \dfrac{V}{R_p} = \dfrac{V}{R_1} + \dfrac{V}{R_2} + \dfrac{V}{R_3} \quad\Rightarrow\quad \boxed{\dfrac{1}{R_p} = \dfrac{1}{R_1} + \dfrac{1}{R_2} + \dfrac{1}{R_3}} \]
Key features of parallel combination
• Same voltage across each resistor.
• Total current is the sum of the branch currents.
• Equivalent resistance \(R_p\) is always less than the smallest individual R.
• For two resistors, \(R_p = \dfrac{R_1 R_2}{R_1 + R_2}\) (product over sum).
• Same voltage across each resistor.
• Total current is the sum of the branch currents.
• Equivalent resistance \(R_p\) is always less than the smallest individual R.
• For two resistors, \(R_p = \dfrac{R_1 R_2}{R_1 + R_2}\) (product over sum).
11.8.3 Why Household Appliances are Wired in Parallel
- Each appliance gets the full mains voltage (220 V in India), regardless of whether others are on.
- If one device fails, others keep working — the circuit is not broken.
- Each appliance can be switched on/off independently.
- Total resistance decreases → total current drawn from the mains is adequate for multiple devices.
11.9 Worked Examples — Series and Parallel
Example 1 — Simple Series
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. ✓
Example 2 — Simple Parallel
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. ✓
Example 3 — Three Resistors in Parallel
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.
Example 4 — Mixed (Series + Parallel)
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. ✓
Example 5 — Finding an Unknown
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 \(R_2 = 6\ \Omega\).
11.10 Heating Effect of Electric Current — Joule's Law
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.
Joule's law of heating: The heat produced in a resistor is:
\[ H = I^2 R t \]
where I is the current (A), R the resistance (Ω), t the time (s). H comes out in joules.
The heat produced is:
The heat produced is:
- Directly proportional to the square of the current,
- Directly proportional to the resistance,
- Directly proportional to the time for which the current flows.
11.10.1 Derivation of H = I²Rt
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:
\[ H = W = VIt = (IR)(I)(t) = I^2 R t \]
11.11 Electric Power
The rate at which electrical energy is dissipated (or consumed) in a circuit is called electric power.
Electric Power (P):
\[ P = \dfrac{W}{t} = VI = I^2 R = \dfrac{V^2}{R} \]
SI unit: watt (W). \(1\text{ W} = 1\text{ V} \times 1\text{ A} = 1\text{ J/s}\). Larger unit: kilowatt (kW) = 1000 W.
11.11.1 Commercial Unit of Electrical Energy — the kWh
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).
11.12 Worked Numericals on Power and Heating
Numerical 1 — Power of a Bulb
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.
Numerical 2 — Heat Produced
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.
Numerical 3 — Electrical Energy Bill
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.
Numerical 4 — Two Bulbs in Series vs Parallel
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.
Numerical 5 — Heater Wire Current
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.
Numerical 6 — Heat by Two Resistors
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.
Numerical 7 — Current Through a Fuse
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.
Numerical 8 — Comparing Energy (Joule ↔ kWh)
Q. Express 1.8 × 10⁷ J in kilowatt-hours.
Solution. \(1\ \text{kWh} = 3.6 \times 10^6\) J. So \(1.8 \times 10^7 / 3.6 \times 10^6 = 5\ \text{kWh}\).
Numerical 9 — Household Consumption
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.
Numerical 10 — Heat vs Current
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.
11.13 Applications of the Heating Effect
- Electric heaters, irons, geysers, toasters, kettles — coils of nichrome wound on ceramic, which glow red-hot.
- Incandescent bulbs — a tungsten filament (melting point 3380 °C) heats to ~2500 °C in an inert gas (argon + a little nitrogen) and emits white light. Most of the energy goes out as heat, which is why LEDs are far more efficient.
- Electric fuse — a short piece of wire made of a tin-lead alloy with low melting point. If the current exceeds a safe value, the fuse heats up and melts, breaking the circuit and protecting the appliance and wiring.
11.14 Household Electric Wiring (Brief)
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.
11.14.1 Short-Circuit and Overloading
- Short circuit — if the live and neutral wires touch directly (due to damaged insulation), the resistance becomes almost zero and the current becomes very large. This melts the fuse (or trips the MCB) instantly.
- Overloading — if too many high-power appliances are plugged into the same socket, the wire carries more current than it is rated for. It heats up, possibly enough to cause a fire.
Activity 11.3 — Series vs Parallel BrightnessL3 Apply
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:
- Connect the two bulbs in series with the battery and switch. Observe brightness.
- Now connect the same two bulbs in parallel across the battery. Observe brightness.
- Unscrew one bulb in each arrangement — does the other keep glowing?
Predict: Will the bulbs appear brighter in the series or the parallel arrangement? What happens to the second bulb when one is unscrewed?
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.
Competency-Based Questions
A student sets up a circuit with three resistors — 2 Ω, 4 Ω and 6 Ω — a 12 V battery, an ammeter and a voltmeter. She first connects them all in series, then rebuilds them all in parallel.
Q1. Find the equivalent resistance in (a) series and (b) parallel. L3 Apply
(a) \(R_s = 2+4+6 = 12\) Ω. (b) \(1/R_p = 1/2+1/4+1/6 = (6+3+2)/12 = 11/12\). So \(R_p = 12/11 ≈ 1.09\) Ω.
Q2. (MCQ) For the parallel arrangement, the total current drawn from the battery is L2 Understand
(c) \(I = V/R_p = 12 \times 11/12 = 11\) A.
Q3. State Joule's law of heating. L1 Remember
The heat produced in a resistor is directly proportional to (i) the square of the current, (ii) its resistance, and (iii) the time: H = I²Rt.
Q4. (Numerical) A 4 kW heater is used for 2 h/day for a month of 30 days. Calculate the electrical energy consumed in kWh and the cost at ₹5 per unit. L3 Apply
Daily: 4 × 2 = 8 kWh. Monthly: 8 × 30 = 240 kWh. Cost = 240 × 5 = ₹1200.
Q5. (HOT) Why do appliances such as electric iron, heater and toaster have their heating elements made of an alloy rather than a pure metal? L4 Analyse
Alloys such as nichrome have (i) high resistivity — more heat is produced per unit current; (ii) high melting point — they can glow red-hot without melting; (iii) low temperature coefficient — resistance changes little with temperature; (iv) they don't oxidise easily — they last long. Pure metals like copper have low resistivity and oxidise quickly at high temperatures.
Assertion–Reason Questions
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.
Assertion (A): The equivalent resistance of resistors in parallel is always less than the smallest individual resistance.
Reason (R): In parallel combination, extra current paths reduce overall opposition to current flow.
(A) — Both true; R correctly explains A.
Assertion (A): An electric fuse is made of a metal with high melting point.
Reason (R): The fuse must not melt easily during normal operation.
(D) — A is false, R is true. A fuse is made of a low-melting-point alloy (tin-lead) so that it melts quickly when current exceeds the rated value. The reason statement is correct in general but does not apply to a fuse.
Assertion (A): Domestic appliances are connected in parallel in household wiring.
Reason (R): Parallel connection supplies the same voltage to every appliance and allows them to operate independently.
(A) — Both true; R correctly explains A.
Frequently Asked Questions — Resistors in Series/Parallel, Power & Heating Effect
What is resistors in series/parallel, power & heating effect in Class 10 Science (CBSE board)?
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.
Why is resistors in series important in NCERT Class 10 Science?
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.
How is resistors in series/parallel, power & heating effect tested in the Class 10 Science CBSE board exam?
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.
What are the key terms to remember for resistors in series/parallel, power & heating effect in Class 10 Science?
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.
Is Resistors in Series/Parallel, Power & Heating Effect included in the Class 10 Science syllabus for 2025–26 CBSE 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.
How should I prepare resistors in series/parallel, power & heating effect for the CBSE Class 10 Science board exam?
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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