--- title: "Electricity: CBSE Class 10 Science previous year questions" url: https://www.swavid.com/cbse/class-10/science/pyq/electricity --- # Electricity: CBSE Class 10 Science previous year questions 21 questions from CBSE Class 10 board papers, newest first, each with full working. ## CBSE Class 10 Science Question Paper 2026 (Set 31/1/1) with Solutions ### Question 34 *2 marks · Numerical* The resistance of a wire of $0.01\text{ cm}$ radius and $1.0\text{ cm}$ length is $7\ \Omega$. Calculate its resistivity. **Solution** 1. Given: radius $r = 0.01\text{ cm} = 10^{-4}\text{ m}$, length $l = 1.0\text{ cm} = 10^{-2}\text{ m}$, resistance $R = 7\ \Omega$ 2. Formula: $R = \rho \frac{l}{A}$, where $A = \pi r^2$, so resistivity $\rho = \frac{R \cdot \pi r^2}{l}$ 3. Substitution: $\rho = \frac{7 \times \frac{22}{7} \times (10^{-4}\text{ m})^2}{10^{-2}\text{ m}}$ 4. Result: $\rho = 2.2 \times 10^{-6}\ \Omega\text{ m}$ **Answer:** $2.2 \times 10^{-6}\ \Omega\text{ m}$ > Common mistake: Forgetting to convert centimeters to meters before substituting into the formula. ### Question 34 (OR) *2 marks · Numerical* An electric heater is rated $220\text{ V}; 11\text{ A}$. Calculate the power consumed if the heater is operated at $200\text{ V}$. **Solution** 1. Given: Rated voltage $V_1 = 220\text{ V}$, rated current $I = 11\text{ A}$. 2. Formula: Resistance of the heater $R = \frac{V_1}{I} = \frac{220}{20}= 20\ \Omega$. 3. When operated at $V_2 = 200\text{ V}$, the power consumed is $P = \frac{V_2^2}{R}$. 4. Substitution: $P = \frac{(200)^2}{20} = \frac{40000}{20} = 2000\text{ W}$. **Answer:** $2000\text{ W}$ > Common mistake: Using the initial power directly or assuming resistance changes with applied voltage. ### Question 39 *5 marks · Long answer* (a) (i) Due to change in length and area of cross-section of a conductor, resistance of conductor changes while resistivity does not change. Why ? (ii) Conductors of electric toasters and electric iron are made of an alloy rather than a pure metal. Why ? (iii) Define the S.I. unit of electric current. **Solution** 1. (i) Resistivity is a characteristic property of the material of the conductor and does not depend on its dimensions. 2. (ii) Alloys have higher resistivity than pure metals and do not oxidize (burn) easily at high temperatures. 3. (iii) 1 Ampere is defined as the flow of 1 Coulomb of electric charge through a cross-section of a conductor in 1 second. **Answer:** Resistivity is a material property; alloys are used for high resistance and oxidation resistance; 1 A = 1 C/s. > Common mistake: Confusing resistance with resistivity. ### Question 39 (OR) *5 marks · Long answer* (b) (i) How many bulbs of resistance $8\ \Omega$ each should be connected in parallel combination to draw a current of $2\text{ A}$ from a battery of $4\text{ V}$ ? (ii) Name the device used for measuring electric current. How is it connected in a circuit ? (iii) State Joule's law of heating. **Part (i)** 1. Given potential difference $V = 4\text{ V}$, current $I = 2\text{ A}$, and resistance of each bulb $R = 8\ \Omega$. 2. Let the total equivalent resistance of the parallel combination be $R_p$. 3. Using Ohm's law, $R_p = \frac{V}{I} = \frac{4\text{ V}}{2\text{ A}} = 2\ \Omega$. 4. Let $n$ bulbs of resistance $R$ be connected in parallel, so $\frac{R}{R_p} = n$ or $\frac{1}{R_p} = n \times \frac{1}{R}$. 5. Substituting the values, $\frac{1}{2} = n \times \frac{1}{8}$, which gives $n = \frac{8}{2} = 4$. Answer (i): 4 bulbs **Part (ii)** 1. The device used for measuring electric current is an ammeter. 2. It is always connected in series in the electric circuit through which the current is to be measured. Answer (ii): Ammeter, connected in series **Part (iii)** 1. Joule's law of heating states that the heat produced in a resistor is directly proportional to the square of current for a given resistance, directly proportional to resistance for a given current, and directly proportional to the time for which the current flows through the resistor. 2. Mathematically, $H = I^2Rt$. Answer (iii): $H = I^2Rt$ **Answer:** Refer to the sub-parts for detailed answers. > Common mistake: Connecting the ammeter in parallel instead of in series, or forgetting to write the mathematical expression for Joule's law of heating. ## CBSE Class 10 Science Question Paper 2025 (Set 31/1/1) with Solutions ### Question 25 *2 marks · Numerical* (a) A wire of resistance $R$ is cut into three equal parts. If these three parts are then joined in parallel, calculate the total resistance of the combination so formed. **Solution** 1. Given: Resistance of original wire is $R$. It is cut into three equal parts, so resistance of each part is $R' = \frac{R}{3}$. 2. Formula: For parallel combination, $\frac{1}{R_p} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$ 3. Substitution: $\frac{1}{R_p} = \frac{1}{R/3} + \frac{1}{R/3} + \frac{1}{R/3} = \frac{3}{R} + \frac{3}{R} + \frac{3}{R}$ 4. Working: $\frac{1}{R_p} = \frac{9}{R}$ 5. Result: $R_p = \frac{R}{9}$ **Answer:** Total resistance of the combination is $\frac{R}{9}$. > Common mistake: Taking resistance of each part as $3R$ instead of $R/3$. ### Question 25 (OR) *2 marks · Very short answer* (b) Define electric power. When do we say that the power consumed in an electric circuit is 1 watt ? **Solution** 1. Electric power is defined as the rate at which electric energy is dissipated or consumed in an electric circuit, given by $P = VI$. 2. The power consumed is 1 watt when a potential difference of 1 volt causes a current of 1 ampere to flow through the circuit. **Answer:** Electric power is the rate of consumption of electric energy ($P = VI$). Power is 1 watt when 1 A current flows through a potential difference of 1 V. > Common mistake: Forgetting to mention the units (volt and ampere) while defining 1 watt. ### Question 32 *3 marks · Short answer* Consider the following electric circuit: Calculate the values of the following : (a) The total resistance of the circuit (b) The total current drawn from the source (c) Potential difference across the parallel combination of $10\ \Omega$ and $15\ \Omega$ resistors **Part (a)** 1. The $10\ \Omega$ and $15\ \Omega$ resistors are connected in parallel, so their equivalent resistance $R_1$ is given by $\frac{1}{R_1} = \frac{1}{10} + \frac{1}{15} = \frac{3+2}{30} = \frac{5}{30}$, which gives $R_1 = 6\ \Omega$. 2. The $60\ \Omega$ and $40\ \Omega$ resistors are connected in parallel, so their equivalent resistance $R_2$ is given by $\frac{1}{R_2} = \frac{1}{60} + \frac{1}{40} = \frac{2+3}{120} = \frac{5}{120}$, which gives $R_2 = 24\ \Omega$. 3. The total resistance of the circuit is $R = R_1 + R_2 = 6\ \Omega + 24\ \Omega = 30\ \Omega$. Answer (a): $30\ \Omega$ **Part (b)** 1. Using Ohm's law, the total current drawn from the source is $I = \frac{V}{R}$. 2. Substituting the values, $I = \frac{15\ \text{V}}{30\ \Omega} = 0.5\ \text{A}$. Answer (b): $0.5\ \text{A}$ **Part (c)** 1. The potential difference across the parallel combination of $10\ \Omega$ and $15\ \Omega$ resistors is the voltage across $R_1$. 2. $V_1 = I \times R_1 = 0.5\ \text{A} \times 6\ \Omega = 3\ \text{V}$. Answer (c): $3\ \text{V}$ **Answer:** Total resistance is $30\ \Omega$, total current is $0.5\ \text{A}$, and potential difference across the parallel combination is $3\ \text{V}$. > Common mistake: Adding all resistor values directly without identifying series and parallel combinations correctly. ### Question 33 *3 marks · Short answer* (a) Write the relationship between resistivity and resistance of a cylindrical conductor of length $l$ and area of cross-section $A$. Hence derive the SI unit of resistivity. (b) Why are alloys used in electrical heating devices ? **Part (a)** 1. The resistance $R$ of a cylindrical conductor is directly proportional to its length $l$ and inversely proportional to its cross-sectional area $A$, written as $R = \rho \frac{l}{A}$, where $\rho$ is resistivity. 2. Rearranging the formula for resistivity gives $\rho = \frac{R \times A}{l}$. 3. Substituting the SI units, $\text{SI unit of } \rho = \frac{\Omega \times \text{m}^2}{\text{m}} = \Omega\text{m}$. Answer (a): $\rho = \frac{RA}{l}$, SI unit is $\Omega\text{m}$ **Part (b)** 1. Alloys are used in electrical heating devices because their resistivity is much higher than that of pure metals, and they do not burn (oxidize) easily at high temperatures. Answer (b): High resistivity and high melting point without oxidation **Answer:** Resistivity $\rho = \frac{RA}{l}$ with SI unit $\Omega\text{m}$, and alloys are used because they have high resistivity and do not oxidize at high temperatures. > Common mistake: Writing the unit of resistivity incorrectly as $\Omega/\text{m}$ instead of $\Omega\text{m}$. ### Question 39 *4 marks · Case-based* In our homes, we receive the supply of electric power through a main supply also called mains, either supported through overhead electric poles or by underground cables. In our country the potential difference between the two wires (live wire and neutral wire) of this supply is 220 V. (a) Write the colours of the insulation covers of the line wires through which supply comes to our homes. (b) What should be the current rating of the electric circuit (220 V) so that an electric iron of 1 kW power rating can be operated ? (c) (i) What is the function of the earth wire ? State the advantage of the earth wire in domestic electric appliances such as electric iron. **Part (a)** 1. 1. State the colour of the insulation cover of the live wire in domestic wiring. 2. 2. Red (or brown according to the new international convention). Answer (a): Red (or brown) **Part (b)** 1. 1. State the given values: Power $P = 1 \text{ kW} = 1000 \text{ W}$, Voltage $V = 220 \text{ V}$. 2. 2. Use the formula $P = V \times I$ to find current $I = \frac{P}{V}$. 3. 3. Substitute values: $I = \frac{1000}{220} = 4.55 \text{ A}$. Answer (b): $4.55 \text{ A}$ **Part (c) (i)** 1. 1. State that the earth wire provides a low-resistance conducting path to the earth. 2. 2. State that it ensures any leakage of current to the metallic body of an appliance keeps its potential to that of the earth, thus preventing severe electric shocks to the user. Answer (c) (i): Earth wire prevents electric shocks by sending any leakage current from the metallic body of the appliance to the earth. **Answer:** Live wire is red, current rating is 4.55 A, and earth wire protects against electric shocks. > Common mistake: Forgetting to convert kW to W when calculating current. ### Question 39 (OR) *2 marks · Case-based* (c) (ii) List two precautions to be taken to avoid electrical accidents. State how these precautions prevent possible damage to the circuit/appliance. **Part (i)** 1. 1. List two precautions such as using an appropriate fuse in every circuit and avoiding the connection of too many appliances to a single socket. 2. 2. State how they prevent damage: fuses melt and break the circuit when current exceeds the safe limit, preventing overheating and short-circuiting. Answer (i): Precautions: use a fuse and avoid overloading; they break the circuit or prevent overheating during excessive current flow. **Answer:** Precautions include using an electric fuse and not overloading circuits to prevent damage. > Common mistake: Failing to explain the mechanism of how the precaution prevents damage. ## CBSE Class 10 Science Question Paper 2024 (Set 31/1/1) with Solutions ### Question 25 *2 marks · Numerical* (A) Show how you would connect three resistors each of resistance $6\text{ }\Omega$, so that the combination has a resistance of $9\text{ }\Omega$. Also justify your answer. **Solution** 1. To get an equivalent resistance of $9\text{ }\Omega$ using three $6\text{ }\Omega$ resistors, connect two $6\text{ }\Omega$ resistors in parallel and the third $6\text{ }\Omega$ resistor in series with this parallel combination. 2. For the two resistors in parallel: $\frac{1}{R_p} = \frac{1}{6} + \frac{1}{6} = \frac{2}{6} = \frac{1}{3}$, so $R_p = 3\text{ }\Omega$. 3. For the series combination with the third resistor: $R_{eq} = R_p + 6 = 3 + 6 = 9\text{ }\Omega$. 4. Justification: The calculated equivalent resistance matches the required $9\text{ }\Omega$. **Answer:** Two resistors in parallel connected in series with the third resistor. > Common mistake: Connecting all three in series or all three in parallel. ### Question 25 (OR) *2 marks · Numerical* (B) In the given circuit calculate the power consumed in watts in the resistor of $2\text{ }\Omega$ : **Solution** 1. Given: Resistance $R_1 = 1\text{ }\Omega$, $R_2 = 2\text{ }\Omega$, Voltage $V = 6\text{ V}$ connected in series. 2. Formula: Total resistance $R_s = R_1 + R_2$, current $I = \frac{V}{R_s}$, and power $P = I^2 R_2$. 3. Substitution: $R_s = 1 + 2 = 3\text{ }\Omega$, $I = \frac{6}{3} = 2\text{ A}$, and $P = (2)^2 \times 2$. 4. Result: $P = 4 \times 2 = 8\text{ W}$. **Answer:** 8 W > Common mistake: Using the total voltage across the circuit instead of finding the current first, or taking total resistance incorrectly. ### Question 36 *5 marks · Long answer* (A) (i) Define electric power. Express it in terms of potential difference (V) and resistance (R). (ii) An electric oven is designed to work on the mains voltage of $220\text{ V}$. This oven consumes $11\text{ units}$ of electrical energy in $5\text{ hours}$. Calculate : (a) power rating of the oven (b) current drawn by the oven (c) resistance of the oven when it is red hot **Part (i)** 1. Electric power is defined as the rate at which electric energy is dissipated or consumed in an electric circuit: $P = \frac{W}{t} = VI$. 2. Using Ohm's law ($V = IR$), electric power in terms of potential difference ($V$) and resistance ($R$) is expressed as: $P = \frac{V^2}{R}$. Answer (i): Electric power $P = VI = \frac{V^2}{R}$. **Part (ii)(a)** 1. Given: Energy $E = 11\text{ units} = 11\text{ kWh}$, Time $t = 5\text{ hours}$. 2. Power rating $P = \frac{E}{t} = \frac{11\text{ kWh}}{5\text{ h}} = 2.2\text{ kW} = 2200\text{ W}$. Answer (ii)(a): $2.2\text{ kW}$ (or $2200\text{ W}$) **Part (ii)(b)** 1. Given: Voltage $V = 220\text{ V}$, Power $P = 2200\text{ W}$. 2. Formula: $P = VI \implies I = \frac{P}{V} = \frac{2200\text{ W}}{220\text{ V}} = 10\text{ A}$. Answer (ii)(b): $10\text{ A}$ **Part (ii)(c)** 1. Formula: $R = \frac{V}{I}$ or $R = \frac{V^2}{P}$. 2. Substitution: $R = \frac{220\text{ V}}{10\text{ A}} = 22\text{ }\Omega$. Answer (ii)(c): $22\text{ }\Omega$ **Answer:** Power definitions, power rating = $2.2\text{ kW}$, current = $10\text{ A}$, resistance = $22\text{ }\Omega$. > Common mistake: Forgetting to convert kilowatts to watts or confusing electrical units (kWh) with power. ### Question 36 (OR) *5 marks · Long answer* (B) (i) Write the relation between resistance $R$ and electrical resistivity $\rho$ of the material of a conductor in the shape of cylinder of length $l$ and area of cross-section $A$. Hence derive the SI unit of electrical resistivity. (ii) The resistance of a metal wire of length $3\text{ m}$ is $60\text{ }\Omega$. If the area of cross-section of the wire is $4 \times 10^{-7}\text{ m}^2$, calculate the electrical resistivity of the wire. (iii) State how would electrical resistivity be affected if the wire (of part 'ii') is stretched so that its length is doubled. Justify your answer. **Part (i)** 1. Relation: $R = \rho \frac{l}{A}$, where $\rho$ is electrical resistivity. 2. Rearranging for resistivity: $\rho = \frac{R A}{l}$. 3. Substituting SI units: $\rho = \frac{\Omega \cdot \text{m}^2}{\text{m}} = \Omega \cdot \text{m}$. Hence, SI unit is ohm-metre ($\Omega\text{m}$). Answer (i): $\rho = \frac{RA}{l}$, SI unit is $\Omega\text{m}$. **Part (ii)** 1. Given: $l = 3\text{ m}$, $R = 60\text{ }\Omega$, $A = 4 \times 10^{-7}\text{ m}^2$. 2. Formula: $\rho = \frac{R \cdot A}{l}$. 3. Substitution: $\rho = \frac{60 \times 4 \times 10^{-7}}{3} = 20 \times 4 \times 10^{-7} = 8 \times 10^{-6}\text{ }\Omega\text{m}$. Answer (ii): $8 \times 10^{-6}\text{ }\Omega\text{m}$ **Part (iii)** 1. Electrical resistivity does not change when the wire is stretched. 2. Justification: Resistivity is a characteristic property of the material of the conductor and depends only on the nature of the material and temperature, not on its dimensions. Answer (iii): Remains unchanged as resistivity depends only on material and temperature. **Answer:** Resistivity formula and SI unit derived, resistivity = $8 \times 10^{-6}\text{ }\Omega\text{m}$, resistivity remains unchanged on stretching. > Common mistake: Stating that resistivity changes when a wire is stretched (confusing it with resistance). ## CBSE Class 10 Science Question Paper 2023 (Set 31/1/1) with Solutions ### Question 13 *1 mark · MCQ* If four identical resistors, of resistance $8\ \Omega$, are first connected in series so as to give an effective resistance $R_s$, and then connected in parallel so as to give an effective resistance $R_p$, then the ratio $\frac{R_s}{R_p}$ is - 32 - 2 - 0.5 - 16 **Solution** 1. For four identical resistors of resistance $R = 8\ \Omega$ connected in series, the equivalent resistance is $R_s = 4R = 4 \times 8\ \Omega = 32\ \Omega$. 2. For the same resistors connected in parallel, the equivalent resistance is $R_p = \frac{R}{4} = \frac{8}{4} = 2\ \Omega$. 3. The ratio of $R_s$ to $R_p$ is $\frac{R_s}{R_p} = \frac{32}{2} = 16$. **Answer:** (d) 16 > Common mistake: Students often invert the ratio and calculate $R_p/R_s$ instead of $R_s/R_p$, leading to the incorrect option 0.5. ### Question 14 *1 mark · MCQ* In domestic electric circuits the wiring with $15\text{ A}$ current rating is for the electric devices which have - higher power ratings such as geyser. - lower power ratings such as fan. - metallic bodies and low power ratings. - non-metallic bodies and low power ratings. **Solution** 1. Two separate circuits are used in domestic wiring: a $5\text{ A}$ current rating for lower power appliances like fans and bulbs, and a $15\text{ A}$ current rating for higher power appliances like geysers, air conditioners, and electric irons. **Answer:** "(a) higher power ratings such as geyser." > Common mistake: Confusing the $5\text{ A}$ circuit with the $15\text{ A}$ circuit. ### Question 15 *1 mark · MCQ* In the following diagram, the position of the needle is shown on the scale of a voltmeter. The least count of the voltmeter and the reading shown by it respectively are : [Diagram showing voltmeter scale with needle between 1.5 and 2] - $0.15\text{ V}$ and $1.6\text{ V}$ - $0.05\text{ V}$ and $1.6\text{ V}$ - $0.15\text{ V}$ and $1.8\text{ V}$ - $0.05\text{ V}$ and $1.8\text{ V}$ **Solution** 1. The voltmeter scale has $10$ divisions between $0$ and $1.5\text{ V}$ or between $1.5$ and $2.0\text{ V}$. Between $1.5$ and $2.0$, there are $10$ subdivisions, so the least count is $\frac{2.0 - 1.5}{10} = \frac{0.5}{10} = 0.05\text{ V}$. 2. The needle is at $2$ divisions after $1.5\text{ V}$, which gives a reading of $1.5 + 2 \times 0.05 = 1.6\text{ V}$. **Answer:** "(b) $0.05\text{ V}$ and $1.6\text{ V}$" > Common mistake: Dividing by the wrong number of divisions while calculating the least count. ### Question 36 *5 marks · Long answer* (a) An electric iron consumes energy at a rate of $880\text{ W}$ when heating is at the maximum rate and $330\text{ W}$ when the heating is at the minimum. If the source voltage is $220\text{ V}$, calculate the current and resistance in each case. (b) What is heating effect of electric current? (c) Find an expression for the amount of heat produced when a current passes through a resistor for some time. **Part (a)** 1. Given: Voltage $V = 220\text{ V}$. Case 1 power $P_1 = 880\text{ W}$. 2. Current $I_1 = \frac{P_1}{V} = \frac{880}{220} = 4\text{ A}$. 3. Resistance $R_1 = \frac{V}{I_1} = \frac{220}{4} = 55\text{ }\Omega$. 4. Case 2 power $P_2 = 330\text{ W}$. Current $I_2 = \frac{P_2}{V} = \frac{330}{220} = 1.5\text{ A}$. 5. Resistance $R_2 = \frac{V}{I_2} = \frac{220}{1.5} = 146.67\text{ }\Omega$. Answer (a): Case 1: $4\text{ A}$, $55\text{ }\Omega$; Case 2: $1.5\text{ A}$, $146.67\text{ }\Omega$. **Part (b)** 1. When an electric current is passed through a high resistance wire, the resistance wire becomes very hot and produces heat. This is called the heating effect of electric current. Answer (b): The generation of heat in a resistor due to the flow of electric current through it. **Part (c)** 1. Let a current $I$ flow through a resistor of resistance $R$ for time $t$, with potential difference $V$ across its ends. 2. Work done $W$ in moving charge $Q$ is $W = V \times Q = V \times I \times t$. 3. Since $V = I \times R$ according to Ohm's law, substituting this gives $H = (I \times R) \times I \times t = I^2Rt$. Answer (c): Expression: $H = I^2Rt$. **Answer:** Calculated current and resistance for both cases, defined heating effect, and derived Joule's law expression. > Common mistake: Using incorrect formulas for resistance calculation like $R = \frac{V^2}{P}$ without deriving or properly substituting values. ## CBSE Class 10 Science Question Paper 2022 (Set 31/1/1) with Solutions ### Question 11 *3 marks · Short answer* (a) Three resistors $R_1$, $R_2$ and $R_3$ are connected in parallel and the combination is connected to a battery, an ammeter, a voltmeter and a key. Draw suitable circuit diagram to show the arrangement of these circuit components along with the direction of current flowing. (b) Calculate the equivalent resistance of the following network: **Part (a)** 1. Draw a circuit diagram showing resistors $R_1$, $R_2$, and $R_3$ connected in parallel across a battery with a key, an ammeter in series, and a voltmeter across the parallel combination. 2. Indicate the direction of current from the positive to the negative terminal of the battery. Answer (a): Circuit diagram showing parallel combination with ammeter, voltmeter, and key. **Part (b)** 1. From the figure, $R_1 = 5\ \Omega$ and $R_4 = 5\ \Omega$ are in series with a parallel combination of $R_2 = 10\ \Omega$ and $R_3 = 10\ \Omega$. 2. The equivalent resistance of $R_2$ and $R_3$ in parallel is $\frac{10 \times 10}{10 + 10} = 5\ \Omega$. 3. Total equivalent resistance $R = R_1 + R_{\text{parallel}} + R_4 = 5 + 5 + 5 = 15\ \Omega$. Answer (b): $15\ \Omega$ **Answer:** Circuit diagram drawn with parallel resistors and series resistor. Equivalent resistance = $20\ \Omega$. > Common mistake: Misinterpreting the series and parallel connections in the given diagram network. ### Question 12 *3 marks · Numerical* (a) (i) Define Electric Power and write its SI unit. (ii) Two bulbs rated $100\text{ W}; 220\text{ V}$ and $60\text{ W}; 220\text{ V}$ are connected in parallel to an electric mains of $220\text{ V}$. Find the current drawn by the bulbs from the mains. **Solution** 1. Given: Power of first bulb $P_1 = 100\text{ W}$, Voltage $V = 220\text{ V}$. 2. Given: Power of second bulb $P_2 = 60\text{ W}$, Voltage $V = 220\text{ V}$. 3. Formula: Total power in parallel $P = P_1 + P_2$. 4. Substitution: $P = 100\text{ W} + 60\text{ W} = 160\text{ W}$. 5. Formula: Current drawn $I = \frac{P}{V}$. 6. Substitution: $I = \frac{160\text{ W}}{220\text{ V}} = \frac{8}{11}\text{ A} \approx 0.73\text{ A}$. 7. Result: $0.73\text{ A}$. **Answer:** $0.73\text{ A}$ > Common mistake: Calculating currents separately and making calculation errors in fractions. ### Question 12 (OR) *3 marks · Numerical* (b) (i) State Joule's law of heating. Express it mathematically when an appliance of resistance R is connected to a source of voltage V and the current I flows through the appliance for a time t. (ii) A $5\;\Omega$ resistor is connected across a battery of 6 volts. Calculate the energy that dissipates as heat in $10\text{ s}$. **Part (i)** 1. Joule's law of heating states that the heat produced in a resistor is directly proportional to the square of current for a given resistance, proportional to resistance for a given current, and proportional to the time for which the current flows through the resistor. 2. Mathematically, it is expressed as $H = I^2Rt$. Answer (i): $H = I^2Rt$ **Part (ii)** 1. Given: Resistance $R = 5\;\Omega$, Voltage $V = 6\text{ V}$, Time $t = 10\text{ s}$. 2. Formula: Current $I = \frac{V}{R} = \frac{6}{5} = 1.2\text{ A}$, and Heat energy $H = I^2Rt$. 3. Substitution: $H = (1.2)^2 \times 5 \times 10 = 1.44 \times 5 \times 10$. 4. Result: $72\text{ J}$. Answer (ii): 72 J **Answer:** The heat produced is $H = I^2Rt$ and the dissipated energy is $72\text{ J}$. > Common mistake: Forgetting to square the current in the heating formula or calculating current incorrectly. ## Related pages - [Electricity: NCERT solutions](https://www.swavid.com/science/class/10/chapter/electricity/ncert-solutions) - [All CBSE Class 10 Science papers](https://www.swavid.com/cbse/class-10/science/previous-year-papers) Solutions written by SwaVid, a personal AI tutor for Class 6 to 10 Maths and Science. Practise this chapter free: https://www.swavid.com/start/student