Electric current & circuit • Potential difference • Ohm's law • Resistance & resistivity • Series & parallel combinations • Heating effect of current • Electric power
Master Electricity — from charge flow to power bills!
Learn how electric current flows, verify Ohm's law, decode series & parallel circuits, apply Joule's law of heating, and calculate electric power & energy with complete board-level confidence.
🎓 Board Focus: Ohm's law numericals, series–parallel equivalent resistance, $P = VI = I^2R = \frac{V^2}{R}$ and Joule heating $H = I^2Rt$ appear EVERY YEAR. kWh cost-of-energy problems are near-guaranteed 3-mark questions.
NCERT-aligned concepts, circuit diagrams, worked numericals, and in-text questions with solutions
01
Electric Current and Electric Circuit (NCERT 11.1)
Direct Board Definition — Electric Current ($I$): Electric current is defined as the rate of flow of electric charge across a cross-section of a conductor per unit time. If a net charge $Q$ flows across any cross-section in time $t$, then the electric current $I$ is given by:
$$I = \frac{Q}{t}$$
SI Unit: The SI unit of electric current is the Ampere (A) (named after André-Marie Ampère).
$$1\text{ Ampere} = \frac{1\text{ Coulomb}}{1\text{ Second}} \quad (1\text{ A} = 1\text{ C/s})$$
Definition of 1 Ampere: One ampere is the current flowing when one coulomb of electric charge flows through a cross-section of a conductor in one second.
• Smaller units: 1 milliampere ($1\text{ mA} = 10^{-3}\text{ A}$), 1 microampere ($1\text{ }\mu\text{A} = 10^{-6}\text{ A}$).
Direction of Current vs Electron Flow:
Current Type
Direction of Motion
Carrier Particle
Conventional Current
From the positive terminal (+) to negative terminal (-) of the electric cell through the outer circuit.
Historical convention (treated as the direction of hypothetical positive charges).
Electronic Current (Actual Flow)
From the negative terminal (-) to positive terminal (+) of the cell.
Negatively charged conduction electrons drifting towards higher potential.
Quantization of Electric Charge:
Total charge $Q$ is quantized in integral multiples of the fundamental charge of an electron ($e = 1.6 \times 10^{-19}\text{ C}$):
$$Q = n \cdot e \implies n = \frac{Q}{e}$$
Number of Electrons in 1 Coulomb of Charge (Standard Board Question):
$$n = \frac{1\text{ C}}{1.6 \times 10^{-19}\text{ C}} = 6.25 \times 10^{18}\text{ electrons}$$
Continuous Closed Loop — The Electric Circuit: A continuous, closed conducting loop through which an electric current can flow between the terminals of an electric energy source (cell/battery) with an interrupting switch and load (bulb/resistor) constitutes an electric circuit.
02
Electric Potential & Potential Difference (NCERT 11.2)
Charges flow only when there is an electric pressure difference — the potential difference — along the conductor, just as water flows only when there is a pressure difference between two ends of a pipe.
A cell or battery produces this potential difference across its terminals by chemical action.
$$V = \frac{W}{Q}$$
V = potential difference in volts, W = work done in joules to move charge Q coulombs between two points
SI UnitVolt (V): 1 V is the PD between two points when 1 joule of work moves 1 coulomb of charge: $1\text{ V} = 1\text{ J C}^{-1}$. Named after Alessandro Volta.
Voltmeter Measures potential difference; always connected in PARALLEL across the two points (ideal voltmeter has very high resistance).
Cell/Battery The chemical action inside a cell maintains the potential difference even when no current is drawn. To maintain current, the cell spends its chemical energy.
NCERT Example 11.2How much work is done in moving a charge of 2 C across two points having a potential difference 12 V?
Solution
Given: $Q = 2C$, $V = 12V$.
Key Point: $W = VQ = 12 \times 2 = \mathbf{24J}$.
📌 In-Text Questions (NCERT Page 174)
Page 174 • Q1Name a device that helps to maintain a potential difference across a conductor.
Answer
Key Point: A cell or a battery (combination of cells) maintains a potential difference across a conductor through chemical action within it.
Page 174 • Q2What is meant by saying that the potential difference between two points is 1 V?
Answer
Key Point: It means that 1 joule of work is done to move a charge of 1 coulomb from one point to the other ($V = \frac{W}{Q} = \frac{1J}{1C} = 1V$).
Page 174 • Q3How much energy is given to each coulomb of charge passing through a 6 V battery?
Answer
Key Point: $W = V \times Q = 6V \times 1C = \mathbf{6J}$ of energy is given to each coulomb of charge.
03
Circuit Diagram & Standard Symbols (NCERT 11.3)
A real circuit with wires running everywhere is messy. So we draw a schematic diagram in which each component is shown by a standard symbol — the universal language of electricians and physicists.
Table 11.1 — Symbols of Commonly Used Components
Component
Symbol Description / Key Point
An electric cell
Long line (+ terminal) | short thick line (− terminal)
A battery (combination of cells)
Several cells joined in series; + of one to − of next
Plug key / switch (open)
Broken line with lifted lever — current cannot pass
Plug key / switch (closed)
Lever resting on contact — continuous path
A wire joint
Dot at the junction of two wires
Wires crossing without joining
Crossing lines with NO dot
Electric bulb
Circle with a cross inside
Resistor of resistance R
Zig-zag / rectangular box labelled R
Variable resistance / rheostat
Resistor with a diagonal arrow through it
Ammeter
Circle with letter A — connect IN SERIES
Voltmeter
Circle with letter V — connect IN PARALLEL
🎯 Exam Tip: In every circuit diagram, draw arrowheads showing the direction of current (from + terminal of battery around to −). CBSE marking schemes deduct for missing arrows or wrong ammeter/voltmeter placement!
04
Ohm's Law (NCERT 11.4)
Key Point: In 1827, Georg Simon Ohm established: the potential difference V across the ends of a given metallic wire is directly proportional to the current I flowing through it, provided its temperature remains constant. The V–I graph is a straight line through the origin.
$$V \propto I \;\;\implies\;\; V = IR$$
R = resistance, constant for the wire at a given temperature; SI unit ohm (Ω)
1 Ohm $R = 1~\Omega$ when 1 V across the ends drives a current of 1 A: $1~\Omega = 1ext{ V/A}$.
Inverse Form $I = \frac{V}{R}$ → current is inversely proportional to resistance. Double R ⇒ current halves.
Rheostat A variable resistance device used to change current in a circuit without changing the voltage source.
V–I Graph: Plotting I on x-axis and V on y-axis gives a straight line through the origin. Slope = $\frac{V}{I} = R$. A steeper slope means higher resistance.
🧠 "V = IR — Vir": Voltage = Current × Resistance. Rearrange as needed: $I = \frac{V}{R}$ and $R = \frac{V}{I}$.
🎯 Exam Tip: Always write the validity condition "temperature remaining constant" while stating Ohm's law — CBSE deducts ½ mark for omitting it!
NCERT Example 11.3(a) How much current will an electric bulb draw from a 220 V source if its filament resistance is 1200 Ω? (b) How much current will an electric heater coil draw from a 220 V source if its resistance is 100 Ω?
Key Point: Note the huge difference in currents drawn by a bulb and a heater from the same 220 V source!
NCERT Example 11.4The potential difference between the terminals of an electric heater is 60 V when it draws a current of 4 A. What current will the heater draw if the potential difference is increased to 120 V?
Experiments (Activity 11.3) show that the resistance of a uniform metallic conductor depends on its length, area of cross-section and the nature of its material. Doubling length halves the current; a thicker wire carries more current.
$$R \propto \frac{l}{A} \;\;\implies\;\; R = \frac{\rho\, l}A$$
ρ (rho) = electrical resistivity of the material — SI unit ohm-metre (Ω m)
Length $R \propto l$ — longer wire ⇒ more resistance.
Area $R \propto \frac{1}{A}$ — thicker wire ⇒ less resistance (easy path for electrons).
Resistivity ρ A characteristic property of the material, independent of length/area. Metals & alloys: $10^{-8}$ to $10^{-6}~\Omega\cdot\text{m}$; insulators like rubber/glass: $10^{12}$–$10^{17}~\Omega\cdot\text{m}$.
Temperature Both R and ρ of metals increase with temperature.
Table 11.2 — Resistivity at 20 °C (Key Values)
Class
Material
Resistivity (Ω m)
Conductors
Silver (best conductor)
$1.60 \times 10^{-8}$
Copper
$1.62 \times 10^{-8}$
Aluminium
$2.63 \times 10^{-8}$
Tungsten / Iron / Mercury
$5.20$–$94.0 \times 10^{-8}$
Alloys
Manganin (Cu, Mn, Ni)
$44 \times 10^{-6}$
Constantan (Cu, Ni)
$49 \times 10^{-6}$
Nichrome (Ni, Cr, Mn, Fe)
$100 \times 10^{-6}$
Insulators
Glass, Hard rubber, Ebonite, Diamond
$10^{10}$–$10^{17}$
🧠 "Alloy = High ρ + No Oxidise": Alloys have HIGHER resistivity than their parent metals and do not oxidise (burn) readily at high temperature — perfect for heaters, irons, toasters.
NCERT Example 11.5Resistance of a metal wire of length 1 m is 26 Ω at 20 °C. If the diameter of the wire is 0.3 mm, what will be the resistivity of the metal at that temperature? Using Table 11.2, predict the material of the wire.
NCERT Example 11.6A wire of given material having length l and area of cross-section A has a resistance of 4 Ω. What would be the resistance of another wire of the same material having length l/2 and area of cross-section 2A?
Page 181 • Q1On what factors does the resistance of a conductor depend?
Answer
Resistance depends on: (i) its length ($l$), (ii) its area of cross-section ($A$), (iii) the nature of its material (resistivity ρ), and (iv) temperature. $R = \frac{\rho l}{A}$.
Page 181 • Q2Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Answer
Through a thick wire. A thick wire has larger cross-sectional area A, so its resistance ($R \propto \frac{1}{A}$) is smaller and more current flows ($I = \frac{V}{R}$).
Page 181 • Q3Let the resistance of an electrical component remain constant while the potential difference across the two ends decreases to half of its former value. What change will occur in the current through it?
Answer
By Ohm's law, $I = \frac{V}{R}$. With R constant and V halved, the current is also reduced to half of its former value.
Page 181 • Q4Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Answer
(i) Alloys have higher resistivity than pure metals, producing more heat. (ii) Alloys do not oxidise (burn) readily at high temperatures, so the coils last longer.
Page 181 • Q5Use the data in Table 11.2 to answer: (a) Which among iron and mercury is a better conductor? (b) Which material is the best conductor?
Answer
(a) Iron ($10.0 \times 10^{-8}~\Omega$m) has lower resistivity than mercury ($94.0 \times 10^{-8}~\Omega$m), so iron is a better conductor.
(b) Silver ($1.60 \times 10^{-8}~\Omega$m) is the best conductor.
06
Resistors in Series (NCERT 11.6.1)
Resistors joined end to end between two points form a series combination.
The same current I flows through every resistor (ammeter reading is same at any position), while the total potential difference divides: $V = V_1 + V_2 + V_3$.
$$R_s = R_1 + R_2 + R_3$$
Equivalent series resistance = sum of individual resistances — always GREATER than the largest individual resistance
Derivation: $V = IR$ (whole circuit) and $V = V_1 + V_2 + V_3 = IR_1 + IR_2 + IR_3$ with common current I ⇒ $IR = I(R_1+R_2+R_3)$ ⇒ $R_s = R_1+R_2+R_3$.
🧠 "Series = Same current, Sum of voltages, Sum of resistances."
🎯 Why series is avoided at home: (i) all appliances get the same current though they need different currents; (ii) if ONE component fails, the whole circuit breaks (dead fairy lights!); (iii) equivalent resistance becomes too large, cutting the current.
NCERT Example 11.7An electric lamp whose resistance is 20 Ω, and a conductor of 4 Ω resistance are connected to a 6 V battery. Calculate (a) the total resistance of the circuit, (b) the current through the circuit, and (c) the potential difference across the electric lamp and conductor.
Page 185 • Q1Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω, an 8 Ω and a 12 Ω resistor, and a plug key, all connected in series.
Answer
The diagram shows three cells in series (+ to −, giving 6 V total), followed in one loop by the plug key (closed), the 5 Ω resistor, 8 Ω resistor and 12 Ω resistor, connected by plain wires.
Draw arrowheads showing conventional current leaving the + terminal.
Page 185 • Q2Redraw the circuit putting in an ammeter to measure current through the resistors and a voltmeter across the 12 Ω resistor. What would be the readings?
Answer
Key Point: Ammeter A goes in series; voltmeter V goes in parallel with the 12 Ω resistor.
Key Point: Total resistance $= 5 + 8 + 12 = 25~\Omega$, Total voltage $= 3 \times 2 = 6V$.
Resistors connected between the same two common points (X and Y) form a parallel combination.
Every resistor gets the same potential difference V, while the total current divides: $I = I_1 + I_2 + I_3$.
$$\frac1{R_p} = \frac1{R_1} + \frac1{R_2} + \frac1{R_3}$$
Reciprocal of equivalent resistance = sum of reciprocals — Rp is LESS than the smallest individual resistance
Derivation: $I = \frac{V}{R_p}$ and $I = I_1+I_2+I_3 = \frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}$ with common V ⇒ divide by V ⇒ $\frac{1}{R_p} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$.
🧠 "Parallel = Same voltage, Divided current, Reduced resistance." For TWO resistors shortcut: $R_p = \dfrac{R_1 R_2}{R_1 + R_2}$ (product over sum).
🎯 Why homes use parallel wiring: (i) every appliance gets the full 220 V; (ii) each appliance draws its own current and works independently; (iii) if one appliance fails or is switched off, others keep working; (iv) total resistance decreases so appliances can draw the large currents they need.
NCERT Example 11.8Resistors R₁ = 5 Ω, R₂ = 10 Ω and R₃ = 30 Ω are connected to a battery of 12 V. Calculate (a) the current through each resistor, (b) the total current in the circuit, and (c) the total circuit resistance.
NCERT Example 11.9In Fig. 11.12, R₁ = 10 Ω, R₂ = 40 Ω, R₃ = 30 Ω, R₄ = 20 Ω, R₅ = 60 Ω, and a 12 V battery is connected to the arrangement. Calculate (a) the total resistance, and (b) the total current flowing in the circuit.
Rule: in parallel, the combination is always slightly less than the SMALLEST resistance.
Page 188 • Q2An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?
Total current $I = \frac{220}{31.25} = \mathbf{7.04A}$.
So the iron must be $R = \frac{220}{7.04} = \mathbf{31.25~\Omega}$ drawing 7.04 A.
Page 188 • Q3What are the advantages of connecting electrical devices in parallel with the battery instead of connecting them in series?
Answer
Key Point: (i) Each device gets the full battery voltage and can draw the current it needs (same V, different I). (ii) Each device has an independent switch and operation — if one fails, the others keep working. (iii) Overall resistance decreases, so large required currents are possible (impossible in series where R adds up).
Page 188 • Q4How can three resistors of resistances 2 Ω, 3 Ω and 6 Ω be connected to give a total resistance of (a) 4 Ω, (b) 1 Ω?
Answer
Key Point: (a) Connect 3 Ω and 6 Ω in parallel ($\frac{3\times6}{3+6} = 2~\Omega$), then in series with 2 Ω: $2+2 = \mathbf{4~\Omega}$.
Key Point: (b) Connect all three in parallel: $\frac{1}{R} = \frac{1}{2}+\frac{1}{3}+\frac{1}{6} = 1$ ⇒ $R = \mathbf{1~\Omega}$.
Page 188 • Q5What is (a) the highest, (b) the lowest total resistance that can be secured by combinations of four coils of resistance 4 Ω, 8 Ω, 12 Ω, 24 Ω?
Answer
Key Point: (a) Highest — all in series: $4+8+12+24 = \mathbf{48~\Omega}$.
Heating Effect of Electric Current — Joule's Law (NCERT 11.7)
In a purely resistive circuit, the source energy is continually dissipated entirely as heat — the heating effect of current.
The work done in moving charge Q through potential difference V is $W = VQ = VIt$; using Ohm's law V = IR: .
$$H = I^2Rt$$
Joule's law of heating: heat produced in joules; double effect of each factor below
Law Part 1 H ∝ $I^2$ — directly proportional to the square of current (double I ⇒ 4× heat!).
Law Part 2 H ∝ R — directly proportional to resistance for a given current.
Law Part 3 H ∝ t — directly proportional to time of current flow.
🧠 "I square R tee": $H = I^2Rt$ — the only heating formula you need. Convert everything to SI units FIRST (minutes → seconds).
NCERT Example 11.10An electric iron consumes energy at a rate of 840 W when heating is at maximum and 360 W at minimum. The voltage is 220 V. What are the current and resistance in each case?
Applications of Heating Effect: Filaments & Fuses (NCERT 11.7.1)
Joule heating has many useful applications — and some unavoidable disadvantages (wasted energy, hot components).
Two classic exam favourites: .
Electric Bulb Filament of tungsten: very high resistivity + extremely high melting point (3380 °C) so it gets white-hot without melting. Bulbs are filled with chemically inactive nitrogen/argon to prolong filament life. Most power appears as heat, a small part as light.
Heating Devices Iron, toaster, oven, kettle, heater — all use nichrome-type alloy coils (high resistivity, no oxidation at high T).
Electric Fuse A safety wire of metal/alloy with low melting point, placed in series with the live wire. If current exceeds the rated value, the fuse heats up (H = I²Rt), melts and breaks the circuit — protecting appliances and preventing fires. Domestic ratings: 1 A, 2 A, 3 A, 5 A, 10 A...
Fuse sizing example: A 1 kW iron on 220 V draws $\frac{1000}{220} = 4.54A$ ⇒ use a 5 A fuse (slightly above normal operating current).
🎯 Classic contrast question: Heater element wire = HIGH resistivity + HIGH melting point (must stay solid while producing heat); Fuse wire = HIGH resistance in small length but LOW melting point (must melt quickly). Don't mix these up!
10
Electric Power & Commercial Unit of Energy (NCERT 11.8)
Electric power is the rate at which electrical energy is dissipated or consumed. The SI unit is the watt (W) — power consumed by a device carrying 1 A at 1 V.
$$P = VI = I^2R = \frac{V^2}R$$
All three forms are equivalent — pick the one matching the given data!
Energy = P × t $W = Pt$. Unit: watt-hour (Wh). Energy consumed when 1 W runs for 1 hour.
Commercial Unit 1 kilowatt-hour (kWh) = 1 'unit' = 1000 W × 3600 s = $\mathbf{3.6 \times 10^6J}$.
Bigger Units 1 kW = 1000 W; 1 MW = 10⁶ W; 1 hp = 746 W (extra knowledge).
🧠 "Power Triangle PIV": P = V×I covers most problems; if only R and I known use I²R; if only V and R known use V²/R. For bills: units (kWh) = $\frac{P(W) \times t(h)}{1000}$, Cost = units × rate.
🎯 Exam Tip: Writing P = V/I instead of P = VI loses a full mark in CBSE marking schemes! Also, energy answers asked "in kWh" must NOT be left in joules.
💡 More to Know: Electrons are NOT consumed in a circuit. We pay the electricity board for the energy that moves electrons through our gadgets — not for the electrons themselves!
NCERT Example 11.12An electric bulb is connected to a 220 V generator. The current is 0.50 A. What is the power of the bulb?
NCERT Example 11.13An electric refrigerator rated 400 W operates 8 hour/day. What is the cost of the energy to operate it for 30 days at ₹3.00 per kWh?
Solution
Key Point: Total energy $= 400W \times 8ext{ h/day} \times 30days = 96000Wh = 96kWh$.
All 18 exercise questions • Step-by-step numericals • MCQs with reasoning
NCERT Official In-Text Questions — Page 200 (Electric Current & Circuit)
In-Text Q1. What does an electric circuit mean?
Board-Standard Model Answer:
A continuous and closed conducting path through which an electric current flows is called an electric circuit. It typically consists of a source of electric energy (cell or battery), conducting connecting wires, a load (such as an electric bulb or resistor), and a switching device (plug key) to open or close the circuit.
In-Text Q2. Define the unit of current.
Board-Standard Model Answer:
The SI unit of electric current is the Ampere (A). Definition: One ampere is constituted by the flow of one coulomb of charge per second through any cross-section of a conducting wire.
$$1\text{ A} = \frac{1\text{ C}}{1\text{ s}} = 1\text{ C s}^{-1}$$
In-Text Q3. Calculate the number of electrons constituting one coulomb of charge.
Board-Standard Model Answer: Given: Net charge $Q = 1\text{ C}$, elementary electronic charge $e = 1.6 \times 10^{-19}\text{ C}$. Formula: By quantization of electric charge:
$$Q = n \cdot e \implies n = \frac{Q}{e}$$
Calculation:
$$n = \frac{1\text{ C}}{1.6 \times 10^{-19}\text{ C}} = \frac{10^{19}}{1.6} = 0.625 \times 10^{19} = \mathbf{6.25 \times 10^{18}\text{ electrons}}$$
Conclusion: Exactly $6.25 \times 10^{18}$ electrons constitute one coulomb of negative charge.
NCERT Official In-Text Questions — Page 202 (Potential Difference)
In-Text Q4. Name a device that helps to maintain a potential difference across a conductor.
Board-Standard Model Answer:
An electric cell, battery (group of cells), or DC generator maintains a constant potential difference across a conductor through spontaneous internal chemical redox reactions.
In-Text Q5. What is meant by saying that the potential difference between two points is 1 V?
Board-Standard Model Answer:
The potential difference between two points is 1 Volt ($1\text{ V}$) if one joule of work is done in moving an electric charge of one coulomb from one point to the other against the electrostatic field.
$$1\text{ Volt} = \frac{1\text{ Joule}}{1\text{ Coulomb}} \quad (1\text{ V} = 1\text{ J C}^{-1})$$
In-Text Q6. How much energy is given to each coulomb of charge passing through a 6 V battery?
Board-Standard Model Answer: Given: Charge $Q = 1\text{ C}$, Potential difference $V = 6\text{ V}$. Formula: Work done (energy imparted) $W = Q \times V$. Calculation:
$$W = 1\text{ C} \times 6\text{ V} = \mathbf{6\text{ Joules}}$$
Conclusion:$6\text{ J}$ of energy is imparted to each coulomb of charge.
NCERT Official In-Text Questions — Page 209 (Ohm's Law & Resistance)
In-Text Q7. On what factors does the resistance of a conductor depend?
Board-Standard Model Answer:
The electrical resistance ($R$) of a conductor depends on four primary parameters:
Length of the Conductor ($l$): Resistance is directly proportional to length: $R \propto l$.
Area of Cross-Section ($A$): Resistance is inversely proportional to cross-sectional area: $R \propto \frac{1}{A}$.
Nature of Material: Represented by electrical resistivity $\rho$, determined by atomic electronic structure.
Temperature: For metallic conductors, resistance increases linearly with rising temperature.
Combined formula: $R = \rho \frac{l}{A}$.
In-Text Q8. Will current flow more easily through a thick wire or a thin wire of the same material, when connected to the same source? Why?
Board-Standard Model Answer:
Electric current flows more easily through a thick wire.
Scientific Reason:
Resistance of a conductor is inversely proportional to its cross-sectional area ($R \propto \frac{1}{A}$). A thick wire has a substantially larger cross-sectional area ($A$) than a thin wire of identical length and material, resulting in a lower electrical resistance ($R$). According to Ohm's law ($I = \frac{V}{R}$), for a given voltage source $V$, lower resistance allows a larger current to flow with fewer electron collisions.
In-Text Q9. Let the resistance of an electrical component remains constant while the potential difference across the two ends of the component decreases to half of its former value. What change will occur in the current through it?
Board-Standard Model Answer:
According to Ohm's law:
$$I = \frac{V}{R}$$
When the potential difference is halved ($V' = \frac{V}{2}$) while resistance $R$ remains constant:
$$I' = \frac{V'}{R} = \frac{V/2}{R} = \frac{1}{2} \left(\frac{V}{R}\right) = \frac{I}{2}$$
Conclusion: The electric current flowing through the component will also decrease to exactly half of its former value.
In-Text Q10. Why are coils of electric toasters and electric irons made of an alloy rather than a pure metal?
Board-Standard Model Answer:
Heating coils of domestic appliances (toasters, geysers, irons) are fabricated from alloys such as Nichrome ($Ni, Cr, Mn, Fe$) rather than pure metals due to two decisive reasons:
Much Higher Resistivity: The resistivity of alloys is orders of magnitude higher than that of their constituent pure metals (Nichrome $\rho \approx 100 \times 10^{-8}\ \Omega\cdot\text{m}$ vs Copper $\rho \approx 1.62 \times 10^{-8}\ \Omega\cdot\text{m}$), producing copious Joule heat ($H = I^2Rt$) for a given current.
High Resistance to Oxidation (No Burning): Alloys do not oxidize or melt easily even at red-hot incandescent temperatures ($> 800^\circ\text{C}$), ensuring structural integrity and long operating life.
NCERT Official In-Text Questions — Page 216 (Series vs Parallel Combinations)
In-Text Q11. Draw a schematic diagram of a circuit consisting of a battery of three cells of 2 V each, a 5 Ω resistor, an 8 Ω resistor, and a 12 Ω resistor, and a plug key, all connected in series.
Board-Standard Model Answer: Circuit Parameters: Total voltage $V = 2\text{ V} + 2\text{ V} + 2\text{ V} = 6\text{ V}$. Resistors in series $R_1 = 5\ \Omega, R_2 = 8\ \Omega, R_3 = 12\ \Omega$. Total Resistance: $R_s = 5 + 8 + 12 = 25\ \Omega$. Circuit Description: The positive terminal of the $6\text{ V}$ battery connects to the plug key, which connects sequentially to $5\ \Omega$, $8\ \Omega$, and $12\ \Omega$ resistors in series, returning to the negative terminal.
In-Text Q12. Put in an ammeter to measure the current through the resistors and a voltmeter to measure the potential difference across the 12 Ω resistor. What would be the readings in the ammeter and the voltmeter?
Board-Standard Model Answer: 1. Reading of Ammeter ($I$):
The ammeter is connected in series. Total resistance $R = 25\ \Omega$, total potential difference $V = 6\text{ V}$.
$$I = \frac{V}{R} = \frac{6\text{ V}}{25\ \Omega} = \mathbf{0.24\text{ A}}$$
Ammeter reading is $0.24\text{ A}$.
2. Reading of Voltmeter ($V_{12}$):
The voltmeter is connected in parallel across the $12\ \Omega$ resistor.
$$V_{12} = I \times R_3 = 0.24\text{ A} \times 12\ \Omega = \mathbf{2.88\text{ V}}$$
Voltmeter reading is $2.88\text{ V}$.
In-Text Q13. Why does the cord of an electric heater not glow while the heating element does?
Board-Standard Model Answer:
The connecting power cord and the heating element carry the identical electric current ($I$) in series. According to Joule's law of heating ($H = I^2Rt$):
• Connecting Cord: Fabricated from high-conductivity copper or aluminum having negligible resistance ($R_{\text{cord}} \approx 0$). Consequently, heat produced is minuscule, so it remains cool and does not glow.
• Heating Element: Made of Nichrome alloy having exceptionally high resistance ($R_{\text{element}}$ is very large). Huge quantities of Joule heat are dissipated, elevating its temperature to red-hot incandescence, causing it to glow brightly.
In-Text Q14. Compute the heat generated while transferring 96000 coulombs of charge in one hour through a potential difference of 50 V.
In-Text Q15. An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 s.
Board-Standard Model Answer: Given: Resistance $R = 20\ \Omega$, Current $I = 5\text{ A}$, Time $t = 30\text{ s}$. Formula: By Joule's law of heating:
$$H = I^2 R t$$
Calculation:
$$H = (5\text{ A})^2 \times 20\ \Omega \times 30\text{ s} = 25 \times 20 \times 30 = \mathbf{15,000\text{ J}} \quad (\text{or } 1.5 \times 10^4\text{ J})$$
Conclusion: Heat developed in 30 seconds is $15,000\text{ Joules}$ ($15\text{ kJ}$).
NCERT Chapter-End Exercises (Q1 to Q18)
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′, then the ratio R/R′ is –
(a) 1/25 (b) 1/5 (c) 5 (d) 25
Solution
Each part has resistance $\frac{R}{5}$. Five equal parts in parallel: $\frac{1}{R'} = \frac{5}{R/5} = \frac{25}{R}$ ⇒ $R' = \frac{R}{25}$.
At 110 V: $P = \frac{(110)^2}{484} = \mathbf{25W}$ → Correct option (d). (Quarter voltage ⇒ quarter power since P ∝ V².)
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 and parallel combinations would be –
(a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1
Solution
Let each wire be R. Series: $H_s = \frac{V^2 t}{2R}$. Parallel: $R_p = \frac{R}{2}$ ⇒ $H_p = \frac{2V^2t}{R}$.
Q5. How is a voltmeter connected in the circuit to measure the potential difference between two points?
Answer
A voltmeter is connected in parallel across the two points between which the potential difference is to be measured (it has very high resistance so it draws negligible current).
Q6. A copper wire has diameter 0.5 mm and resistivity of 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?
If diameter doubles, area becomes 4×, so $R' = \frac{R}{4} = \mathbf{2.5~\Omega}$ — resistance falls to one-fourth.
Q7. Plot a graph between V and I and calculate the resistance of that resistor. I (amperes): 0.5, 1.0, 2.0, 3.0, 4.0 V (volts): 1.6, 3.4, 6.7, 10.2, 13.2
Solution
The graph is a straight line through the origin (ohmic conductor). Resistance = slope of V–I line:
Using first and last points: $R = \frac{\Delta V}{\Delta I} = \frac{13.2 - 1.6}{4.0 - 0.5} = \frac{11.6}{3.5} \approx \mathbf{3.3~\Omega}$.
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.
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 respectively. How much current would flow through the 12 Ω resistor?
Solution
In series, the SAME current flows through every resistor.
$R_{total} = 0.2+0.3+0.4+0.5+12 = 13.4~\Omega$ ⇒ $I = \dfrac{9}{13.4} = \mathbf{0.67A}$ through each, including the 12 Ω resistor.
Q10. How many 176 Ω resistors (in parallel) are required to carry 5 A on a 220 V line?
Solution
Current through one resistor: $i = \frac{220}{176} = 1.25A$.
Number needed: $n = \dfrac{5}{1.25} = \mathbf{4}$ 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.
Solution
(i) Connect two in parallel ($\frac{6\times6}{12} = 3~\Omega$), then in series with the third: $3 + 6 = \mathbf{9~\Omega}$.
(ii) Connect two in series ($6+6 = 12~\Omega$), then in parallel with the third: $\frac{6\times12}{18} = \mathbf{4~\Omega}$.
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 220 V line if the maximum allowable current is 5 A?
Solution
Current per lamp: $i = \frac{P}{V} = \frac{10}{220} = \frac{1}{22}A$.
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?
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.
(ii) The 2 Ω resistor gets the full 4 V: $P_2 = \frac{V^2}{R} = \frac{16}{2} = \mathbf{8W}$.
Comparison: The 2 Ω resistor uses the same power (8 W) in both circuits.
Q15. Two lamps, one rated 100 W at 220 V, and the other 60 W at 220 V, are connected in parallel to electric mains supply. What current is drawn from the line if the supply voltage is 220 V?
Solution
In parallel, each lamp operates at rated voltage, so each draws rated current:
Q17. An electric heater of resistance 44 Ω draws 5 A from the service mains for 2 hours. Calculate the rate at which heat is developed in the heater.
Solution
Rate of heat development = power $P = I^2R = (5)^2 \times 44 = \mathbf{1100ext{ J/s}\ (= 1100W)}$.
(The 2 hours is extra data — rate does not depend on time.)
Q18. Explain the following: (a) Why is tungsten used almost exclusively for filament of electric lamps? (b) Why are conductors of electric heating devices made of an alloy rather than pure metal? (c) Why is the series arrangement not used for domestic circuits? (d) How does resistance vary with area of cross-section? (e) Why are copper and aluminium wires usually employed for electricity transmission?
Answer
(a) Tungsten has a very high melting point (3380 °C) and high resistivity — its filament gets white-hot and emits light without melting.
(b) Alloys have higher resistivity (more heat per current) and do not oxidise/burn readily at high temperatures.
(c) In series all devices get the same current though they need different currents; total resistance increases reducing current; and if one device fails, the entire circuit breaks.
(d) Resistance is inversely proportional to area: $R \propto \frac{1}{A}$ — thicker wire ⇒ lower resistance.
(e) Copper and aluminium have very low resistivity, so transmission losses ($H = I^2Rt$) are minimal.
🏆 Topic-wise PYQ Bank — CBSE Board Questions (2008–2026)
A boy records that 4000 J of work is required to transfer 10 coulombs of charge between two points of a resistor of 50 Ω. The current passing through it is:
Assertion (A): In an open circuit, the current passes from one terminal of the electric cell to another. Reason (R): Generally, the metal disc of a cell acts as the positive terminal.
Explanation
An OPEN circuit has no closed path, so no current flows — A is false. In a dry cell the metal disc (flat bottom) is the negative terminal (the metal cap is +) — R is also false → (D).
Q5 • MCQ A-R — 1M | CBSE Recurring 2023–2026
Assertion (A): Conventional current flows from the positive terminal of a battery through the external circuit to the negative terminal. Reason (R): Electrons (negative charges) move from the negative terminal to the positive terminal, so conventional current is opposite to electron flow.
Explanation
Both statements are true and R correctly explains why conventional direction is opposite to electron flow → (A).
Q6 • SA — 2M | CBSE Recurring 2015–2025
Define electric current. Write its SI unit. How is the direction of conventional current related to the direction of flow of electrons?
Answer
Electric current is the amount of charge flowing through a particular area in unit time: $I = \frac{Q}{t}$. SI unit: ampere (A). Conventional current flows opposite to electron flow (electrons are negative).
Q7 • SA — 2M | CBSE 2011; 2012; 2015; Recurring
Define potential difference between two points. Write its SI unit. Name the instrument used to measure it and state how it is connected.
Answer
Potential difference = work done per unit charge in moving charge between two points: $V = \frac{W}{Q}$. SI unit: volt (V), 1 V = 1 J/C. Instrument: voltmeter, in parallel.
Q8 • VSA — 1M | CBSE 2008; 2011; 2012; 2015
Name the instrument used to measure electric current in a circuit. How is it connected?
Answer
Ammeter — connected in series (low resistance so it doesn't alter the current).
Q9 • VSA — 1M | CBSE Recurring 2015–2024
Name a device that helps to maintain a potential difference across a conductor.
Draw the circuit symbols for: (i) electric cell, (ii) battery, (iii) open key, (iv) resistor, (v) rheostat, (vi) ammeter, (vii) voltmeter, (viii) wire junction.
Answer
(i) Long line (+) | short thick line (−); (ii) several cells in series; (iii) broken line with lifted lever; (iv) zig-zag/rectangle labelled R; (v) resistor with diagonal arrow through it; (vi) circle 'A' in series; (vii) circle 'V' in parallel; (viii) dot at junction of two wires. (Draw as per Table 11.1.)
Q13 • Case Study — 4M | CBSE 2023; 2024; 2025; 2026
Case Study — Electric Current and Potential Difference: Current is rate of flow of charge (I = Q/t, unit A); PD is work done per unit charge (V = W/Q, unit V). Ammeter (low resistance) in series; voltmeter (high resistance) in parallel.
(i) What is the PD if 40 J of work moves a charge of 8 C between two points?
(ii) A charge of 150 C flows through a wire in 1 minute. Find the current.
(iii) Why is an ammeter connected in series and a voltmeter in parallel?
(iv) What happens if an ammeter is mistakenly connected in parallel?
(ii) $I = \frac{Q}{t} = \frac{150}{60} = \mathbf{2.5A}$.
(iii) The ammeter has very low resistance and must carry the full circuit current → series. The voltmeter has very high resistance and must compare two points without drawing appreciable current → parallel.
(iv) In parallel the low-resistance ammeter acts as a short circuit — a dangerously large current flows through it, likely burning out the meter/circuit.
A resistance wire is stretched so as to double its length. Its new resistivity will:
Explanation
Resistivity ρ is a material property — stretching changes length and area (resistance becomes 4×), but NOT resistivity.
Q17 • MCQ — 1M | CBSE PYQ; Recurring 2019–2025
A cylindrical conductor of length l and uniform area of cross-section A has resistance R. Another conductor of the same material has length 2.5l and resistance 0.5R. Its area of cross-section is:
Which one among a bar of an alloy of mass 2 kg and a 3 kg iron bar of the same dimensions has greater resistivity?
Explanation
Alloys generally have higher resistivity than pure metals — independent of mass/dimensions → alloy bar.
Q19 • MCQ A-R — 1M | CBSE 2020
Assertion (A): Alloys are commonly used in electrical heating devices like electric iron and heaters. Reason (R): The resistivity of an alloy is generally higher than that of its constituent metals and alloys have high melting points compared to their constituent metals.
Explanation
High resistivity ⇒ more heat ($H = I^2Rt$); high melting point + no oxidation ⇒ coils last long → (A).
Q20 • MCQ A-R — 1M | CBSE Recurring 2023–2026
Assertion (A): The resistance of a metallic conductor increases with increase in temperature. Reason (R): As temperature increases, the number of free electrons in the conductor decreases, reducing the current for the same applied voltage.
Explanation
A is true. R is FALSE — the number of free electrons stays essentially constant; increased lattice vibrations obstruct electron flow and raise resistance → (C).
Q21 • MCQ A-R — 1M | CBSE Recurring 2023–2026
Assertion (A): Copper and aluminium wires are usually used for electricity transmission. Reason (R): The resistivities of copper and aluminium are very low, so current flows easily with minimum energy loss.
Explanation
Low ρ ⇒ low R ⇒ minimum $I^2R$ heating loss during long-distance transmission → (A).
Q22 • VSA — 1M | CBSE 2010–2015; Recurring
Define Ohm's law. State the condition under which it is valid. Write its mathematical form.
Answer
The potential difference across the ends of a conductor is directly proportional to the current through it, provided the temperature remains constant: $V = IR$.
Q23 • VSA — 1M | CBSE Board Term I, 2014; Recurring
What is meant by 1 ohm resistance? Define SI unit of resistance.
Answer
Resistance is 1 ohm when a PD of 1 volt drives a current of 1 ampere: $1~\Omega = \frac{1V}{1A}$.
Q24 • SA — 2M | CBSE Board Term I, 2015
(i) List three factors on which the resistance of a conductor depends. (ii) Write the SI unit of resistivity.
Answer
(i) Length ($R \propto l$), area of cross-section ($R \propto \frac{1}{A}$), nature of material (ρ); temperature also matters. (ii) ohm-metre (Ω m).
Q25 • VSA — 1M | CBSE Board Term I, 2017
How does the resistivity of alloys compare with those of the pure metals from which they may have been formed?
Answer
Alloys have generally HIGHER resistivity than their constituent pure metals.
Q26 • VSA — 1M | CBSE 2012; 2013
Why are copper or aluminium wires usually used for electricity transmission and distribution purposes?
Answer
They have very low resistivity, so transmission heat loss ($P = I^2R$) is minimum.
Q27 • VSA — 1M | CBSE 2016
Why do electricians wear rubber hand gloves while working with electrical installations?
Answer
Rubber is an excellent insulator — gloves prevent current flowing through the body to earth, protecting against electric shock.
Q28 • SA — 3M | CBSE Board Term I, 2016
State Ohm's law. Draw a labelled circuit diagram to verify it in the laboratory. What kind of curve do you get between V and I? How would you use this graph to determine resistance?
Answer
Statement: V ∝ I at constant temperature ⇒ V = IR.
Circuit: Battery + plug key + ammeter in series with nichrome wire XY; voltmeter in parallel across XY. Use 1, 2, 3, 4 cells; tabulate V and I each time; plot V vs I.
Graph: Straight line through origin. Resistance = slope of line: pick two convenient points and compute $R = \dfrac{\Delta V}{\Delta I}$.
Q29 • SA — 3M | CBSE Board Term I, 2014
State and explain Ohm's law. Define resistance and give its SI unit. Draw V-I graph for an ohmic conductor and list two important features.
Answer
Ohm's law: At constant temperature V ∝ I ⇒ V = IR. Resistance is the property of a conductor to resist the flow of charges; SI unit ohm (Ω).
Features of V-I graph: (i) Straight line through the origin; (ii) Slope = R is constant for all points (R independent of current at constant temperature).
Q30 • SA — 3M | CBSE Recurring 2015–2025
Derive the relation R = ρl/A. On what factors does resistance depend? Write SI unit of resistivity.
Answer
Experiments show $R \propto l$ and $R \propto \frac{1}{A}$. Combining: $R \propto \frac{l}{A}$ ⇒ $\mathbf{R = \rho\frac{l}{A}}$ where ρ = resistivity (constant of proportionality).
Resistance depends on length, cross-section area, material and temperature. SI unit of ρ: ohm-metre (Ω m).
Q31 • Numerical — 2M | CBSE Recurring
Calculate the resistance of 2 km long copper wire of radius 2 mm. (Resistivity of copper = 1.72 × 10⁻⁸ Ω m)
A piece of wire having resistance R is cut into five equal parts. (i) How will the resistance of each part compare with the original? (ii) If the five parts are placed in parallel, how does the combination compare with the original resistance?
Solution
(i) Each part has resistance $\frac{R}{5}$ (one-fifth of original, since R ∝ l).
(ii) $\frac{1}{R'} = 5 \times \frac{5}{R} = \frac{25}{R}$ ⇒ $R' = \frac{R}{25}$ — one twenty-fifth of the original.
Q33 • Case Study — 4M | CBSE 2023; 2024; 2025; 2026
Case Study — Ohm's Law and Resistance: V ∝ I at constant temperature ⇒ V = IR. R depends on length, area, material and temperature: R = ρl/A (ρ in Ω m). For metals, resistance rises with temperature. Alloys like nichrome have higher resistivity than pure metals.
(i) A student plots V-I graphs for two resistors A and B. Slope of A is steeper than B. Which has higher resistance?
(ii) A wire of resistivity ρ, length l, area A — what happens to R if length is doubled AND area is halved?
(iii) Two wires of the same material have lengths 2 m and 4 m with equal cross-sections. Find the ratio of their resistances.
(iv) Why is constantan used for making standard resistors?
Solution
(i) Resistor A — steeper slope means larger $\frac{V}{I}$ = larger R.
(ii) $R' = \dfrac{\rho\,(2l)}{A/2} = 4\dfrac{\rho l}{A} = \mathbf{4R}$.
(iii) R ∝ l ⇒ ratio = $\mathbf{1 : 2}$.
(iv) Constantan's resistivity changes very little with temperature, so its resistance stays nearly constant — ideal for standard resistors.
📌 Topic 3 — Series & Parallel Combinations of Resistors (Q34–Q51)
Two resistors connected in series give an equivalent resistance of 10 Ω. When the same two are connected in parallel, they give 2.4 Ω. The individual resistances are:
Explanation
Sum = 10; $\frac{R_1R_2}{R_1+R_2} = 2.4$ ⇒ product $= 24$. Numbers with sum 10 and product 24 → 6 and 4.
Q35 • MCQ — 1M | CBSE BYJU'S MCQ; Recurring
If n identical resistors each of resistance R are connected in parallel, their equivalent resistance is:
In an electrical circuit, two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be:
In a parallel combination of resistors, which of the following is the same across all resistors?
Explanation
All parallel branches join the same two nodes ⇒ same potential difference; currents differ inversely with R.
Q38 • MCQ A-R — 1M | CBSE 2023; 2024; 2025
Assertion (A): In a series circuit, the same current passes through all the resistors, but the potential difference across each is different. Reason (R): In a series connection there is only one path for current; the PD distributes across each resistor proportional to its resistance (V = IR).
Explanation
Single path ⇒ common I; each $V_i = IR_i \propto R_i$. Both true, R explains A → (A).
Q39 • MCQ A-R — 1M | CBSE 2023; 2024; 2025; 2026
Assertion (A): In a parallel circuit, the potential difference across each resistor is the same but the current through each is different. Reason (R): Each resistor is directly connected between the two terminals of the supply. The current through each branch depends on its resistance (I = V/R).
Explanation
Common PD across branches; branch current $I = \frac{V}{R}$ differs with R → (A).
Q40 • MCQ A-R — 1M | CBSE Recurring 2023–2026
Assertion (A): The equivalent resistance in a parallel combination is always less than the smallest individual resistance. Reason (R): Adding more parallel paths provides more routes for current, effectively reducing the total resistance of the network.
Explanation
$\frac{1}{R_p} = \frac{1}{R_1}+\frac{1}{R_2}+...$ exceeds every individual reciprocal, so $R_p$ is below even the smallest R. More paths = easier flow → (A).
Q41 • SA — 3M | CBSE Recurring 2015–2025
Derive the expression for equivalent resistance when three resistors R₁, R₂ and R₃ are connected in series.
Answer
In series, the same current I flows through all; battery voltage divides: $V = V_1+V_2+V_3$. By Ohm's law $IR_s = IR_1+IR_2+IR_3$ ⇒ $\mathbf{R_s = R_1+R_2+R_3}$ (greater than any individual resistance).
Q42 • SA — 3M | CBSE Recurring 2015–2025
Derive the expression for equivalent resistance when three resistors R₁, R₂ and R₃ are connected in parallel.
Answer
In parallel, each resistor gets the same PD V; currents divide: $I = I_1+I_2+I_3$. So $\dfrac{V}{R_p} = \dfrac{V}{R_1}+\dfrac{V}{R_2}+\dfrac{V}{R_3}$ ⇒ $\mathbf{\dfrac{1}{R_p} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}$ (Rp less than the smallest individual resistance).
Q43 • SA — 3M | CBSE 2024
State Ohm's law. Write the formula for the equivalent resistance of a parallel combination of three resistors. Find the total resistance of a network where 10 Ω and 15 Ω are in parallel, and this combination is in series with 5 Ω.
Solution
Ohm's law: V = IR (V ∝ I at constant temperature).
A circuit has a 5 Ω resistor in series with a parallel combination of a 12 Ω resistor and a 17 Ω resistor, connected to a 5 V battery. Find (i) equivalent resistance, (ii) current through each branch, (iii) potential difference across the 12 Ω resistor.
(iii) $V_{12} = \mathbf{2.92V}$ (same as across the parallel block).
Q45 • Numerical — 2M | CBSE Recurring 2016–2025
A 9 Ω resistance is cut into 3 equal parts and all three parts are connected in parallel. Find the equivalent resistance of the combination.
Solution
Each part: $\frac{9}{3} = 3~\Omega$. Three equal parts in parallel: $R_p = \frac{3}{3} = \mathbf{1~\Omega}$.
Q46 • Numerical — 2M | CBSE Recurring
You take two resistors of resistance 2R and 3R and connect them in parallel in an electric circuit. Calculate the ratio of the electrical power consumed by 2R to that consumed by 3R.
Solution
In parallel, same V: $P = \frac{V^2}{R}$ ⇒ $\dfrac{P_{2R}}{P_{3R}} = \dfrac{V^2/2R}{V^2/3R} = \mathbf{3:2}$.
Q47 • SA — 3M | CBSE Recurring 2016–2025
Why is the parallel combination preferred over series for connecting domestic electrical appliances? Give three reasons.
Answer
(i) Every appliance gets the full 220 V and draws the current it needs. (ii) Each appliance works independently — one failing/switched off does not affect others. (iii) Equivalent resistance decreases, so appliances needing large currents can operate properly.
Q48 • Numerical — 3M | CBSE Recurring 2017–2025
Three resistors of 5 Ω, 10 Ω and 30 Ω are connected in parallel across a 6 V battery of negligible internal resistance. (i) Find the current through each resistor. (ii) Find the total current. (iii) Find the equivalent resistance.
Two resistors R₁ = 6 Ω and R₂ = 12 Ω are connected (a) in series, (b) in parallel. Find the equivalent resistance in each case. If the combination is connected to a 12 V battery, find the total current in each case.
Case Study — Domestic circuit with LED bulbs: Five LED bulbs are wired in PARALLEL across a 220 V source; keys K₁ and K₂ control separate groups of bulbs.
(a) State what happens when key K₁ is closed.
(b) What changes when K₂ is also closed?
(c) If one bulb fuses, will the others continue to glow? Why?
(d) Why is parallel wiring used in domestic circuits?
Answer
(a) The bulbs controlled by K₁ light at full brightness (each receives 220 V); K₂'s group stays off.
(b) The K₂ group also lights up independently — all bulbs now get full 220 V.
(c) Yes — the remaining bulbs keep glowing because each branch carries its own current; a fused bulb doesn't break other branches.
(d) Parallel wiring gives every appliance the full mains voltage and independent operation via separate switches; one fault doesn't affect everything.
Q51 • Case Study — 4M | CBSE 2023; 2024; 2025; 2026
Case Study — Series vs Parallel: Series: Rs = R₁+R₂+R₃ (same current, PD divides). Parallel: $\frac{1}{R_p} = \frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}$ (same PD, current divides).
(i) Two resistors 4 Ω and 6 Ω in series with a 10 V battery — find circuit current and PD across each.
(ii) Same two in parallel with 10 V — find branch currents and total current.
(iii) Why is equivalent resistance in parallel always less than the smallest individual resistance?
(iv) Give one application each of series and parallel connections.
📌 Topic 4 — Heating Effect of Electric Current (Q52–Q69)
Q52 • MCQ — 1M | CBSE Recurring 2016–2026
The heating effect of electric current is used in:
Explanation
Heaters, irons, toasters and kettles all work on $H = I^2Rt$; motors/fans/generators convert electrical ↔ mechanical energy.
Q53 • MCQ — 1M | CBSE Recurring 2017–2026
Fuse wire is always connected in the:
Explanation
The fuse is placed in series with the live wire so that when it melts, the appliance is fully disconnected from the high-potential supply.
Q54 • MCQ — 1M | CBSE Recurring 2018–2026
The property of fuse wire which makes it suitable for use as a safety device is:
Explanation
High resistivity makes it heat quickly ($H=I^2Rt$); low melting point lets it melt fast to break the circuit on overload.
Q55 • MCQ — 1M | CBSE 2011; 2014; 2015; Recurring
Why is tungsten used for making the filament of an electric bulb?
Explanation
Tungsten's melting point is ~3380 °C — the filament glows white-hot without melting; bulbs are filled with inactive N₂/Ar gases to prevent oxidation.
Q56 • MCQ — 1M | CBSE BYJU'S MCQ; Recurring
In an electric circuit, two resistors of 2 Ω and 4 Ω respectively are connected in series to a 6 V battery. The heat dissipated by the 4 Ω resistor in 5 s will be:
Assertion (A): Nichrome is used to make the heating element of an electric heater. Reason (R): Nichrome is an alloy with high resistivity and very high melting point, and it does not oxidise easily even at high temperatures.
Explanation
All three nichrome properties (high ρ, high m.p., no oxidation) directly make it ideal for heating elements → (A).
Q58 • MCQ A-R — 1M | CBSE Recurring 2023–2026
Assertion (A): The fuse wire is connected in series with the live wire in a household circuit. Reason (R): In the event of short circuit or overloading, the fuse wire melts due to excessive heating (high resistivity, low melting point), breaking the circuit and protecting the appliances.
Explanation
Fuse in series with the live wire ensures that when it melts, the appliance is cut off from supply entirely → (A).
Q59 • SA — 2M | CBSE 2020; Recurring 2016–2025
Write the mathematical expression for Joule's law of heating. State its three parts.
Answer
$\mathbf{H = I^2Rt}$: heat produced is (i) directly proportional to the square of the current for a given R, (ii) directly proportional to resistance for a given I, and (iii) directly proportional to the time for which current flows.
Q60 • Numerical — 2M | CBSE 2020
Compute the heat generated while transferring 96,000 coulombs of charge in two hours through a potential difference of 40 V.
Solution
Heat = work done $= V \times Q = 40 \times 96000 = \mathbf{3.84 \times 10^6J}$. (Time is extra information.)
Q61 • SA — 2M | CBSE 2017
Nichrome is used to make the element of an electric heater. Give two reasons.
Answer
(i) Nichrome has high resistivity, so it produces large heat for the same current ($H = I^2Rt$). (ii) It has a high melting point and does not oxidise readily at high temperatures — the element lasts long without burning out.
Q62 • SA — 2M | CBSE Foreign 2005; 2016
Should the heating element of an electric iron be made of iron, silver or nichrome wire? Give two reasons.
Answer
Nichrome. (i) Its high resistivity produces much more heat than low-resistivity iron or silver. (ii) It withstands high temperatures without melting or oxidising, unlike pure metals which would burn/rust quickly.
Q63 • SA — 2M | CBSE thestudypath; Recurring 2016–2024
State the difference between the wire used in the element of an electric heater and in a fuse wire.
Answer
Heater element wire: an alloy (nichrome) with high resistivity and HIGH melting point — it becomes red-hot but must NOT melt. Fuse wire: metal/alloy with LOW melting point and appropriate resistance — it deliberately melts when current exceeds the rated value, breaking the circuit.
Q64 • SA — 3M | CBSE Recurring 2016–2025
Explain the role of a fuse in series with any electrical appliance in an electric circuit. Why should a fuse with a defined rating not be replaced by one with a larger rating?
Answer
The fuse protects circuits and appliances from excessive current: if I exceeds its rating, heat ($H = I^2Rt$) melts the low-melting-point fuse wire and breaks the circuit, preventing damage/fire.
A higher-rated fuse would carry currents beyond the appliance's safe limit WITHOUT melting — the wiring/appliance could overheat and catch fire before the fuse acts. The fuse must be slightly above the normal operating current but below the damaging current.
Q65 • SA — 2M | CBSE 2011; 2014; 2015; Recurring
(a) Why is tungsten used for the filament of an electric bulb? (b) Name two electrical appliances based on the heating effect of electric current.
Answer
(a) Tungsten has a very high melting point (~3380 °C) and high resistivity — it glows white-hot and emits light without melting; inert N₂/Ar filling prevents oxidation.
(b) Electric heater, electric iron / toaster / geyser / electric bulb filament (any two).
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.
Solution
Rate of heat production = power $P = I^2R = (15)^2 \times 8 = \mathbf{1800ext{ J/s}\ (= 1.8kW)}$.
(The 2 hours is extra data — 'rate' does not depend on time.)
Q67 • Numerical — 3M | CBSE Recurring
A small bulb has a resistance of 2 Ω when cold. It takes a current of 0.4 A from a source of 4 V and then starts glowing. (i) Calculate the resistance of the bulb when it is glowing. (ii) Elaborate on the reason for the difference in resistance.
Solution
(i) When glowing: $R = \dfrac{V}{I} = \dfrac{4}{0.4} = \mathbf{10~\Omega}$.
(ii) When current flows, the filament heats up to a very high temperature. Resistance of metals increases with temperature (greater lattice vibrations obstruct electron flow), so the hot filament has much higher resistance than when cold.
Q68 • Numerical — 3M | CBSE Recurring 2018–2025
An electric iron consumes energy at a rate of 420 W when heating is at maximum and 180 W when at minimum. The voltage is 220 V. What is the current and resistance in each case?
Q69 • Case Study — 4M | CBSE 2023; 2024; 2025; 2026
Case Study — Heating Effect: Heat produced by current: H = I²Rt. Rate of heating P = I²R = V²/R = VI. A fuse operates on this principle — excessive current melts its wire and breaks the circuit.
(i) Write Joule's law of heating. Name three appliances based on it.
(ii) An electric heater coil of resistance 110 Ω draws 2 A. Find the heat produced in 5 minutes.
(iii) Distinguish between fuse wire and heating element wire.
(iv) A fuse of 5 A rating protects a 1000 W, 250 V heater. Will it work safely? Justify.
Solution
(i) $H = I^2Rt$ — H ∝ I², ∝ R, ∝ t. Appliances: electric iron, toaster, electric kettle / heater / geyser.
(ii) $H = I^2Rt = 4\times110\times300 = \mathbf{132000J\ (= 1.32\times10^5J)}$.
(iii) Fuse: low melting point, appropriate resistivity — MELTS to break circuit. Heating element: high resistivity, HIGH melting point — must NOT melt while producing heat.
(iv) Heater current $= \frac{1000}{250} = 4A < 5A$ rating → the fuse carries normal current safely but will melt if current rises beyond 5 A. Yes, safe.
📌 Topic 5 — Electric Power & P-V-I-R Relations (Q70–Q84)
Q70 • MCQ — 1M | CBSE Recurring 2016–2026
The SI unit of electric power is:
Explanation
Power = rate of energy consumption = watt (W); 1 W = 1 J/s = 1 V × 1 A.
In order to reduce electricity consumption at home, what kind of appliances should one purchase?
Explanation
Energy consumed E = P × t — a lower power rating means less energy consumed per hour of use.
Q75 • MCQ — 1M | CBSE NCERT; Recurring 2017–2025
Which uses more energy — a 250 W TV set in 1 hour or a 1200 W toaster in 10 minutes?
Explanation
TV: $250 \times 1 = 250$ Wh. Toaster: $1200 \times \frac{1}{6} = 200$ Wh → TV uses more.
Q76 • MCQ A-R — 1M | CBSE Recurring 2023–2026
Assertion (A): When two identical bulbs are connected in series, the power consumed is less than when connected in parallel. Reason (R): In series, the equivalent resistance increases and total power $P = V^2/R$ decreases. In parallel, equivalent resistance decreases and total power increases.
Explanation
Rs = 2R vs Rp = R/2 for identical bulbs ⇒ series power $\frac{V^2}{2R}$ is one-fourth of parallel power $\frac{V^2}{R/2}$ → (A).
Q77 • MCQ A-R — 1M | CBSE 2024; 2025; 2026
Assertion (A): A 60 W bulb glows brighter than a 100 W bulb when both are connected in series. Reason (R): In series both carry the same current; Power = I²R and the 60 W bulb has higher resistance (R = V²/P), so it dissipates more power.
Explanation
$R_{60} > R_{100}$ (same rated V); with common series current, higher R dissipates more power ($P = I^2R$) → 60 W glows brighter → (A).
Q78 • SA — 2M | CBSE Recurring 2015–2025
Define electric power. Write the formula for electric power in terms of: (i) V and I, (ii) I and R, (iii) V and R.
Answer
Electric power is the rate at which electrical energy is consumed/dissipated in a circuit.
(i) $P = VI$ (ii) $P = I^2R$ (iii) $P = \dfrac{V^2}{R}$. SI unit: watt (W).
Q79 • Numerical — 3M | CBSE Recurring 2016–2025
A bulb is rated 220 V, 100 W. (i) What is its resistance? (ii) Find the current drawn by the bulb. (iii) What is the energy consumed in 2 hours?
A bulb is rated 330 V, 110 W. (i) What is its resistance? (ii) Three such bulbs burn for 5 hours at a stretch. What is the energy consumed? (iii) Calculate the cost if the rate is ₹7 per unit (kWh).
An electric heater rated 1 kW, 220 V is used for 3 hours daily. Find the energy consumed in 20 days in kWh.
Solution
$E = P \times t = 1kW \times 3ext{ h/day} \times 20days = \mathbf{60kWh}$.
Q82 • Numerical — 2M | CBSE Recurring
Two resistors of resistance 2R and 3R are connected in parallel to a supply. Calculate the ratio of the power consumed by 2R to that consumed by 3R.
Solution
In parallel, same V: $\dfrac{P_{2R}}{P_{3R}} = \dfrac{V^2/2R}{V^2/3R} = \mathbf{3:2}$ — smaller resistance consumes more power.
Q83 • LA — 5M | CBSE Recurring 2017–2025
Establish the relation P = V²/R. A household uses: one electric fan (80 W) for 12 hours, two electric lamps (60 W each) for 6 hours and one electric press (1000 W) for 2 hours daily. Find the total electrical energy consumed per day and the cost at ₹6 per kWh.
Solution
Derivation: $P = VI$; by Ohm's law $I = \frac{V}{R}$ ⇒ $P = V \cdot \frac{V}{R} = \mathbf{\dfrac{V^2}{R}}$.
Total $= 3680Wh = \mathbf{3.68ext{ kWh/day}}$; Cost $= 3.68\times6 = \mathbf{₹22.08}$ per day.
Q84 • Case Study — 4M | CBSE 2023; 2024; 2025; 2026
Case Study — Electric Power and Energy: P = VI = I²R = V²/R (watt). Energy E = Pt; commercial unit 1 kWh = 1 unit = 3.6 × 10⁶ J. In series, same I ⇒ P ∝ R. In parallel, same V ⇒ P ∝ 1/R.
(i) An electric bulb is rated 220 V, 60 W. What is its resistance? What current does it draw?
(ii) A 2 kW geyser is used for 1 hour daily. Find (a) energy consumed in 30 days, (b) cost at ₹8/unit.
(iii) Two bulbs rated 40 W and 100 W are connected in parallel to a 220 V supply. Which draws more current? Which is brighter? Why?
(iv) The same bulbs are now connected in series. Which glows brighter and why?
(iii) In parallel both get 220 V, so the 100 W bulb draws more current ($I = P/V$) and is brighter (dissipates more power).
(iv) In series the 40 W bulb glows brighter: it has higher R, and with common current $P = I^2R$ favours higher resistance.
📌 Topic 6 — Mixed Long Answer Questions (Q85–Q101)
Q85 • LA — 5M | CBSE Recurring 2016–2025
(a) State Ohm's law. (b) Draw a labelled circuit diagram to verify it in the laboratory. (c) A circuit consists of a 6 V battery, an ammeter, a variable resistor and three resistors R₁ = 5 Ω, R₂ = 10 Ω and R₃ = 20 Ω in series. Find: (i) total resistance, (ii) circuit current, (iii) PD across each resistor.
Solution
(a) V ∝ I at constant temperature ⇒ V = IR.
(b) Battery–key–ammeter–resistors in series; voltmeter in parallel across the resistors; rheostat varies current. (Draw with arrowheads.)
(a) Derive the expression for equivalent resistance when three resistors are connected in parallel. (b) Three resistors of 5 Ω, 10 Ω and 30 Ω are connected in parallel across a 6 V battery. Find: (i) equivalent resistance, (ii) current through each, (iii) total current drawn.
Solution
(a) Same V across branches; currents add: $\frac{V}{R_p} = \frac{V}{R_1}+\frac{V}{R_2}+\frac{V}{R_3}$ ⇒ $\mathbf{\dfrac{1}{R_p} = \dfrac{1}{R_1}+\dfrac{1}{R_2}+\dfrac{1}{R_3}}$.
(a) Write Joule's law of heating. (b) An electric iron of resistance 20 Ω takes a current of 5 A. Calculate the heat developed in 30 seconds. (c) Explain why tungsten is used for bulb filaments and nichrome for heating elements. (d) Why is fuse wire always connected in series with the live wire?
(c) Tungsten: very high melting point (~3380 °C) + high resistivity → glows white-hot without melting. Nichrome: high resistivity + does not oxidise at high temperature → durable heat production.
(d) In series with live wire, so when it melts under excess current, the appliance is fully disconnected from the high-potential supply — making it dead and safe.
Q88 • LA — 5M | CBSE 2017; 2020; 2023; Recurring
(a) With a circuit diagram, state and explain Ohm's law. Define resistance. (b) What is resistivity? On what factors does it depend? SI unit? (c) A metallic wire of length 2 m has resistance 4 Ω, area of cross-section 2 × 10⁻⁷ m². Find its resistivity. What is the new resistance if stretched to double length?
Solution
(a) V ∝ I at constant temperature ⇒ V = IR; resistance opposes charge flow (unit Ω).
(b) ρ is a material property independent of dimensions ($R = \rho l/A$); depends only on nature of material & temperature. Unit: Ω m.
(a) List three advantages of connecting appliances in parallel over series in domestic circuits. (b) An electric lamp of 20 Ω and a conductor of 4 Ω are in series with a 6 V battery. Find: (i) total resistance, (ii) circuit current, (iii) PD across lamp, (iv) PD across conductor, (v) power consumed by the lamp.
Solution
(a) Full voltage per appliance; independent operation/failure; lower equivalent resistance allowing needed currents.
(b)(i) $\mathbf{24~\Omega}$; (ii) $I = \frac{6}{24} = \mathbf{0.25A}$.
(a) Define electric power. Establish P = I²R. (b) Two lamps rated 100 W at 220 V and 60 W at 220 V are connected: (i) in parallel across 220 V — find total current and power; (ii) in series across 220 V — find total power and state which glows brighter.
Solution
(a) Power = rate of energy consumption. $P = VI$; with $V = IR$: $\mathbf{P = I^2R}$.
(b)(i) Parallel: each runs at rating ⇒ $P = \mathbf{160W}$; $I = \frac{160}{220} \approx \mathbf{0.73A}$.
(ii) Series: $R_{100} = 484~\Omega$, $R_{60} \approx 806.7~\Omega$, $R_s \approx 1290.7~\Omega$; $I \approx 0.17A$; $P_s \approx \mathbf{37.5W}$. The 60 W lamp glows brighter ($P = I^2R$, higher R).
(a) Explain why series arrangement is not used for domestic circuits. (b) A house uses: 3 bulbs of 100 W for 4 h/day; 2 fans of 50 W for 8 h/day; 1 TV of 200 W for 3 h/day. Find (i) total energy per day in kWh, (ii) monthly cost (30 days) at ₹6 per unit.
Solution
(a) Same current through all appliances though they need different currents; total resistance rises reducing current; one fault breaks the whole circuit.
(ii) $2.6\times30=78$ kWh ⇒ Cost $= 78\times6 = \mathbf{₹468}$.
Q92 • LA — 5M | CBSE PYQ; 2019; 2022; Recurring
(a) State the factors on which resistance of a cylindrical conductor depends; write R in terms of ρ, l and A. (b) A conductor of length l, area A has resistance R. Another of the same material has length 2.5l and resistance 0.5R — find its area. (c) If this second wire is stretched to three times its original length, find the new resistance.
Solution
(a) Length (∝), area (∝1/A), material (ρ), temperature: $R = \dfrac{\rho l}{A}$.
(c) Stretch to 3l: area becomes $\frac{5A}{3}$ ⇒ $R' = \rho\dfrac{3l}{5A/3} = \dfrac{9}{5}R = \mathbf{1.8R}$.
Q93 • LA — 5M | CBSE 2019; 2021; 2024; Recurring
(a) An electric room heater draws 2.4 A from a 120 V line. What is the rate of energy transformation? (b) An electric kettle rated 2 kW converts 1500 g of water at 20 °C to steam at 100 °C (specific heat = 4200 J/kg°C, latent heat = 2.26 × 10⁶ J/kg). How long does it take?
Solution
(a) $P = VI = 120\times2.4 = \mathbf{288W\ (=288ext{ J/s})}$.
(b) $Q = mc\Delta T + mL = 1.5\times4200\times80 + 1.5\times2.26\times10^6 = 3.894\times10^6J$.
(a) State Ohm's law and describe an experiment to verify it. (b) How does the V-I graph differ for an ohmic conductor versus a non-ohmic device like a bulb? (c) A bulb has resistance 2 Ω when cold and takes 0.4 A from a 4 V source when glowing. (i) Find resistance when glowing. (ii) Why does resistance change? (iii) Why is a bulb not used to verify Ohm's law?
Solution
(a) V ∝ I at constant temperature. Circuit: battery, key, ammeter, rheostat and wire in series; voltmeter across wire; record V-I pairs; plot graph — straight line through origin verifies the law.
(b) Ohmic: straight line through origin (constant slope). Bulb: curved line — slope decreases as filament heats up.
(c)(i) $R = \frac{4}{0.4} = \mathbf{10~\Omega}$. (ii) Filament heats to very high temperature; metal resistance increases with temperature.
(iii) Because its temperature does NOT remain constant — the validity condition of Ohm's law fails (V/I not constant).
Q95 • LA — 5M | CBSE Recurring 2017–2025
(a) Define: (i) 1 volt, (ii) 1 ampere, (iii) 1 ohm, (iv) 1 watt, (v) 1 kilowatt-hour. (b) A household uses an electric heater of 1000 W for 4 h, an electric fan of 100 W for 12 h, and a refrigerator of 200 W for 24 h per day. Calculate: (i) energy consumed per day in kWh, (ii) monthly bill at ₹7 per unit.
Solution
(a)(i) 1 V = PD when 1 J of work moves 1 C between two points. (ii) 1 A = current when 1 C flows per second. (iii) 1 Ω = resistance when 1 V drives 1 A. (iv) 1 W = power when 1 A flows at 1 V (= 1 J/s). (v) 1 kWh = energy used by 1 kW appliance in 1 hour = 3.6 × 10⁶ J.
(a) State Ohm's law. Describe an experiment to verify it. Plot the V-I graph. (b) What is resistivity? Write its SI unit. Give order of values for a good conductor, an alloy and an insulator. (c) A wire of resistance R is bent in the shape of a circle. Find equivalent resistance between two ends of its diameter.
Solution
(a) V ∝ I at constant temperature (V = IR). Experiment as in Q94 — V-I straight line through origin verifies law.
(b) Resistivity ρ is the resistance offered by a conductor of unit length and unit cross-section — a material property. SI unit Ω m. Good conductors ~10⁻⁸ Ω m; alloys ~10⁻⁶ Ω m; insulators ~10¹²–10¹⁷ Ω m.
(c) Each semicircular half has $\frac{R}{2}$; between diameter ends the two halves are in PARALLEL: $R_{eq} = \dfrac{(R/2)}{2} = \mathbf{\dfrac{R}{4}}$.
Q97 • LA — 5M | CBSE 2016; 2019; 2022; Recurring
(a) Explain with reason why parallel combination is preferred in domestic wiring. (b) Draw a circuit diagram showing three resistors R₁, R₂, R₃ in parallel with a battery and switch; write expression for total current. (c) Three resistors of 3 Ω, 4 Ω and 12 Ω are connected in parallel across a 12 V battery. Find: (i) equivalent resistance, (ii) total current, (iii) current through each resistor.
Solution
(a) Full voltage per appliance, independent switching/failure, reduced equivalent resistance permitting large operating currents.
A bulb of 60 W-220 V is connected in series with a bulb of 100 W-220 V across a 220 V supply. Find: (i) current in circuit, (ii) power consumed by each bulb, (iii) which glows brighter and why? Repeat for both bulbs in parallel and compare.
Series: $I = \frac{220}{1290.7} \approx \mathbf{0.17A}$; $P_{60} = I^2R_{60} \approx \mathbf{23.3W}$, $P_{100} = I^2R_{100} \approx \mathbf{14W}$ → 60 W glows brighter (higher R dissipates more at same current).
Parallel: Each gets 220 V ⇒ consumes rated powers 60 W & 100 W; 100 W glows brighter (draws more current, more power).
Q99 • LA — 5M | CBSE 2017; 2020; 2022; Recurring
(a) Explain briefly the applications of Joule's heating effect in daily life (three examples). (b) Should a fuse have high or low resistance and high or low melting point? Why? (c) Why is the fuse always connected in the live wire? (d) An electric fuse is rated 5 A. What happens if a 10 A current flows through it?
Answer
(a) Electric bulb filament (tungsten glows to emit light); electric iron/heater/toaster (heating element produces heat); electric fuse (melts on overload to protect circuits).
(b) HIGH resistance (so it heats quickly for small excess currents, H = I²Rt) and LOW melting point (so it melts fast and breaks the circuit).
(c) So that when it melts, the appliance is disconnected from the high-potential supply and becomes completely safe/dead.
(d) The fuse wire heats rapidly ($H \propto I^2$), melts and breaks the circuit instantly — protecting appliances from damage/fire.
Q100 • LA — 5M | CBSE cbseguidanceweb; 2018–2025; Recurring
(a) A battery of emf 4 V and internal resistance 2 Ω is connected to an external resistance of 14 Ω. Calculate: (i) current in the circuit, (ii) terminal voltage of the battery, (iii) power dissipated in the external resistance. (b) A battery of 10 V carries 20,000 C through a resistance of 20 Ω. Find the work done. If this takes 10 minutes, find the power.
(a) What are the characteristics of a fuse wire? How does it protect a circuit? (b) Compare fuse wire and nichrome wire in terms of: (i) resistivity, (ii) melting point, (iii) use. (c) In a household circuit an accidental short circuit occurs — explain what happens and how the fuse protects the appliances. (d) Why should fuses not be replaced by wires of higher current rating?
Answer
(a) Fuse wire: alloy of appropriate composition with HIGH resistivity and LOW melting point, rated 1 A, 2 A, 3 A, 5 A, 10 A etc. When current exceeds rating, I²R heating melts it and breaks the circuit, protecting appliances and wiring from overheating/fire.
(b) (i) Fuse: high resistivity (in its small length); nichrome: high resistivity too but as element wire. (ii) Fuse: LOW melting point; nichrome: VERY HIGH melting point. (iii) Fuse: safety device that melts; nichrome: heating element that must never melt.
(c) In a short circuit, live and neutral wires touch directly — resistance falls almost to zero, current shoots up enormously. The huge current melts the fuse instantly, disconnecting the supply before wiring or appliances are damaged by heat/fire.
(d) A higher-rated fuse allows currents beyond the circuit's safe limit to flow without melting — wiring/appliances can overheat and catch fire. Fuse rating must match the circuit's safe maximum.
📊 Topic-Wise Summary & CBSE Exam Blueprint
Topic-Wise Question Breakdown (Q.1 to Q.101)
Topic Area
MCQ (1M)
A-R (1M)
VSA/SA (2–3M)
Case Study (4M)
LA (5M)
Total Qs
1. Electric Current & Potential Difference
3
2
8
1
—
Q1–Q13 (13)
2. Ohm's Law, Resistance & Resistivity
5
3
10
1
—
Q14–Q33 (20)
3. Series & Parallel Combinations
4
3
8
2
—
Q34–Q51 (18)
4. Heating Effect of Current & Applications
5
2
9
1
—
Q52–Q69 (18)
5. Electric Power & P-V-I-R Relations
6
2
6
1
—
Q70–Q84 (15)
6. Mixed Long Answers (5M)
—
—
—
—
17
Q85–Q101 (17)
TOTAL CHAPTER 11 PYQ BANK
23
12
41
6
17
101 Questions
📋 Year-Wise Most Asked Question Types (2015–2026)
Exam Year
High-Yield Key Topics
2026
Ohm's law numericals, series-parallel networks, case studies on power/energy cost
2025
Parallel domestic circuits, Joule heating H = I²Rt, fuse working
2024
A-R on bulb brightness, resistivity factors, kWh billing problems
2023
Case study on current/PD basics, V-I graph interpretation
2022
Equivalent resistance derivations, power rating numericals
2021 (T1)
MCQ heavy — units, Ohm's law, symbols, fuse properties
2020
Joule's law statement, heat numericals, alloys in heating devices
2019
Series vs parallel advantages, circuit diagrams
2018
Ohm's law verification experiment, wire-resistance numericals