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Electricity

NCERT Class 10 · Science Based on NCERT Class 10 Science textbook · Free CBSE study kit

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Chapter Notes

Overview
This chapter introduces the fundamental concepts of electricity, including electric current, potential difference, and circuits. It explains Ohm's law, the factors affecting resistance, and the heating effect of electric current. The big idea is that electric current is the flow of charge, and energy is dissipated as heat when current flows through a resistor. Students must understand how to analyze series and parallel circuits and apply formulas for power and energy.
Key Concepts (15)
Electric current
The amount of charge flowing through a particular area in unit time; the rate of flow of electric charges.
e.g. When you switch on a torch, the cells provide a flow of charges (electric current) through the bulb to make it glow.
Electric circuit
A continuous and closed path of an electric current.
e.g. A torch circuit with a cell, bulb, and switch forms a closed path when the switch is on.
Ammeter
An instrument used to measure electric current in a circuit, always connected in series.
e.g. In a school lab, an ammeter is connected in series with a resistor to measure the current flowing through it.
Potential difference
The work done to move a unit charge from one point to another in an electric circuit.
e.g. A 6 V battery provides a potential difference of 6 V across its terminals, meaning 6 J of work is done per coulomb of charge.
Voltmeter
An instrument used to measure potential difference, always connected in parallel across the points between which the potential difference is to be measured.
e.g. To measure the voltage across a bulb, a voltmeter is connected in parallel with the bulb.
Ohm's law
The potential difference across the ends of a given metallic wire is directly proportional to the current flowing through it, provided its temperature remains the same.
e.g. For a nichrome wire at constant temperature, doubling the voltage doubles the current.
Resistance
The property of a conductor to resist the flow of charges through it.
e.g. A thin wire has higher resistance than a thick wire of the same material, so it allows less current.
Rheostat
A device used to change the resistance in a circuit to regulate current without changing the voltage source.
e.g. In a laboratory, a rheostat is used to vary the current in a circuit while studying Ohm's law.
Resistivity
A characteristic property of a material; the resistance of a conductor of unit length and unit cross-sectional area.
e.g. Copper has low resistivity (1.62 × 10⁻⁸ Ω m), making it a good conductor for electrical wires.
Joule's law of heating
The heat produced in a resistor is directly proportional to the square of the current, the resistance, and the time for which the current flows.
e.g. An electric iron becomes hot because current flows through its high-resistance coil, producing heat according to H = I²Rt.
Electric power
The rate at which electric energy is dissipated or consumed in an electric circuit.
e.g. A 100 W bulb consumes 100 J of electrical energy per second.
Kilowatt hour
The commercial unit of electric energy, equal to 3.6 × 10⁶ J.
e.g. An electric meter at home measures energy consumption in kilowatt hours (units).
Fuse
A safety device consisting of a piece of wire made of a metal or alloy of appropriate melting point, placed in series with the device to protect circuits from unduly high electric current.
e.g. A 5 A fuse is used in an electric iron circuit to prevent damage if current exceeds 5 A.
Series combination
Resistors joined end to end such that the same current flows through each resistor.
e.g. Three resistors of 2 Ω, 3 Ω, and 5 Ω connected in series have a total resistance of 10 Ω.
Parallel combination
Resistors connected together between two points such that the potential difference across each resistor is the same.
e.g. Household appliances are connected in parallel so each gets the same 220 V supply.
Formulas & Equations (12)
I = Q/t
Reads as: Current equals charge divided by time
Variables: I = electric current (ampere, A), Q = net charge flowing (coulomb, C), t = time (second, s)
Use when: Calculating current from charge and time, or finding charge from current and time.
⚠ Forgetting to convert time to seconds (e.g., using minutes instead of seconds).
V = W/Q
Reads as: Potential difference equals work done divided by charge
Variables: V = potential difference (volt, V), W = work done (joule, J), Q = charge (coulomb, C)
Use when: Calculating potential difference from work and charge, or finding work done from potential difference and charge.
⚠ Confusing work done with energy; they are the same in this context.
V = IR
Reads as: Potential difference equals current times resistance
Variables: V = potential difference (volt, V), I = current (ampere, A), R = resistance (ohm, Ω)
Use when: Applying Ohm's law to find voltage, current, or resistance in a circuit.
⚠ Forgetting that Ohm's law applies only at constant temperature.
R = ρ l/A
Reads as: Resistance equals resistivity times length divided by area of cross-section
Variables: R = resistance (ohm, Ω), ρ = resistivity (ohm metre, Ω m), l = length (metre, m), A = area of cross-section (square metre, m²)
Use when: Calculating resistance of a conductor given its material, length, and cross-sectional area.
⚠ Using diameter instead of radius when calculating area, or forgetting to convert units (e.g., mm to m).
Rs = R1 + R2 + R3 + ...
Reads as: Total series resistance equals sum of individual resistances
Variables: Rs = equivalent resistance in series (ohm, Ω), R1, R2, R3 = individual resistances (ohm, Ω)
Use when: Finding total resistance of resistors connected in series.
⚠ Adding resistances in parallel using this formula instead of the reciprocal formula.
1/Rp = 1/R1 + 1/R2 + 1/R3 + ...
Reads as: Reciprocal of parallel resistance equals sum of reciprocals of individual resistances
Variables: Rp = equivalent resistance in parallel (ohm, Ω), R1, R2, R3 = individual resistances (ohm, Ω)
Use when: Finding total resistance of resistors connected in parallel.
⚠ Forgetting to take the reciprocal of the sum to get Rp, i.e., writing Rp = 1/R1 + 1/R2.
H = VIt
Reads as: Heat produced equals voltage times current times time
Variables: H = heat produced (joule, J), V = potential difference (volt, V), I = current (ampere, A), t = time (second, s)
Use when: Calculating heat produced when current flows through a resistor with known voltage.
⚠ Using hours instead of seconds for time.
H = I²Rt
Reads as: Heat produced equals current squared times resistance times time
Variables: H = heat produced (joule, J), I = current (ampere, A), R = resistance (ohm, Ω), t = time (second, s)
Use when: Calculating heat produced using current and resistance (Joule's law).
⚠ Forgetting to square the current.
P = VI
Reads as: Power equals voltage times current
Variables: P = electric power (watt, W), V = potential difference (volt, V), I = current (ampere, A)
Use when: Calculating power consumed by a device given voltage and current.
⚠ Using resistance instead of current in the formula.
P = I²R
Reads as: Power equals current squared times resistance
Variables: P = electric power (watt, W), I = current (ampere, A), R = resistance (ohm, Ω)
Use when: Calculating power when current and resistance are known.
⚠ Forgetting to square the current.
P = V²/R
Reads as: Power equals voltage squared divided by resistance
Variables: P = electric power (watt, W), V = potential difference (volt, V), R = resistance (ohm, Ω)
Use when: Calculating power when voltage and resistance are known.
⚠ Forgetting to square the voltage.
1 kWh = 3.6 × 10⁶ J
Reads as: One kilowatt hour equals 3.6 times 10 to the power 6 joules
Variables: kWh = kilowatt hour (commercial unit of energy), J = joule (SI unit of energy)
Use when: Converting between kilowatt hours and joules.
⚠ Using 1000 × 3600 = 3.6 × 10⁶ correctly but forgetting the unit conversion.
How It Works
Flow of electric current in a circuit
  1. Step 1: A cell or battery creates a potential difference across its terminals.
  2. Step 2: When a closed circuit is formed (e.g., by closing a switch), the potential difference sets charges in motion.
  3. Step 3: Electrons flow from the negative terminal to the positive terminal of the cell.
  4. Step 4: Conventionally, current is taken as the flow of positive charges from positive to negative terminal.
  5. Step 5: The current is measured by an ammeter connected in series.
💡 Conventional current direction is opposite to electron flow; electrons are negative charges.
Ohm's law verification (Activity 11.1)
  1. Step 1: Set up a circuit with a nichrome wire, ammeter, voltmeter, and cells.
  2. Step 2: Use one cell, note ammeter reading (I) and voltmeter reading (V).
  3. Step 3: Repeat with two, three, and four cells.
  4. Step 4: Calculate V/I for each pair; observe it is constant.
  5. Step 5: Plot a graph of V vs I; it is a straight line through the origin.
💡 The V-I graph is a straight line only if temperature remains constant.
Factors affecting resistance (Activity 11.3)
  1. Step 1: Connect a nichrome wire of length l in a circuit with an ammeter.
  2. Step 2: Replace with a wire of same material and thickness but double the length (2l); note current decreases.
  3. Step 3: Replace with a thicker wire of same length (l); current increases.
  4. Step 4: Replace with a copper wire of same length and thickness; current changes due to different material.
💡 Resistance is directly proportional to length and inversely proportional to area of cross-section; it also depends on material.
Resistors in series (Activity 11.4 and 11.5)
  1. Step 1: Connect three resistors (e.g., 1 Ω, 2 Ω, 3 Ω) end to end in series with a battery and ammeter.
  2. Step 2: Note ammeter reading; it is the same at any point in the circuit.
  3. Step 3: Measure potential difference across each resistor (V1, V2, V3) and across the combination (V).
  4. Step 4: Observe that V = V1 + V2 + V3.
  5. Step 5: The equivalent resistance Rs = R1 + R2 + R3.
💡 In series, current is same through all resistors, but voltage divides.
Resistors in parallel (Activity 11.6)
  1. Step 1: Connect three resistors (e.g., 5 Ω, 10 Ω, 30 Ω) in parallel across a battery with an ammeter and voltmeter.
  2. Step 2: Measure total current I and potential difference V across the combination.
  3. Step 3: Measure current through each resistor (I1, I2, I3) separately.
  4. Step 4: Observe that I = I1 + I2 + I3.
  5. Step 5: The equivalent resistance is given by 1/Rp = 1/R1 + 1/R2 + 1/R3.
💡 In parallel, voltage is same across each resistor, but current divides.
Heating effect of electric current
  1. Step 1: A current I flows through a resistor of resistance R for time t.
  2. Step 2: Work done by the source is VIt, which is dissipated as heat in the resistor.
  3. Step 3: Using Ohm's law (V = IR), heat H = I²Rt (Joule's law).
  4. Step 4: This heat is used in devices like electric heaters, irons, and bulbs.
  5. Step 5: Fuses use this effect to break the circuit if current exceeds a safe value.
💡 Heat produced is proportional to I², not I; doubling current quadruples heat.
Exam Traps — Don't Make These Mistakes
⚠ Direction of current vs electron flow
✗ Current flows from negative to positive terminal because electrons flow that way.
→ Conventionally, current is taken as the flow of positive charges, opposite to electron flow.
✓ Conventional current flows from the positive terminal to the negative terminal of the cell, opposite to the direction of electron flow.
⚠ Series vs parallel resistance calculation
✗ For resistors in parallel, total resistance is the sum of individual resistances.
→ That formula is for series; for parallel, the reciprocal of total resistance equals sum of reciprocals.
✓ For resistors in parallel, 1/Rp = 1/R1 + 1/R2 + 1/R3, and Rp is always less than the smallest individual resistance.
⚠ Unit conversion in heat formula
✗ H = I²Rt, using time in minutes directly.
→ Time must be in seconds because the joule is defined per second.
✓ Convert time to seconds before using H = I²Rt. For example, 10 minutes = 600 seconds.
⚠ Power formula confusion
✗ Power = V²/R is always correct for any circuit.
→ P = V²/R applies only when the device obeys Ohm's law and resistance is constant; it is derived from P = VI and V = IR.
✓ Use P = VI as the basic formula; P = I²R and P = V²/R are derived forms valid for ohmic conductors.
⚠ Ammeter and voltmeter connections
✗ Ammeter is connected in parallel to measure current.
→ Ammeter has very low resistance; connecting it in parallel would cause a short circuit and damage it.
✓ An ammeter is always connected in series in the circuit. A voltmeter is always connected in parallel across the component.
Compare & Contrast
Current flow
Series combination
Parallel combination
In series, the same current flows through each resistor. In parallel, the current divides among the branches, and the total current is the sum of the branch currents.
Dependence on dimensions
Resistance
Resistivity
Resistance depends on the length, area of cross-section, and material of the conductor. Resistivity is a material property and does not depend on the dimensions of the conductor.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA-I (2 marks), 1 SA-II (3 marks), 1 LA (5 marks) — total 12 marks typically
Most asked: Ohm's law and its applications · Series and parallel combinations of resistors · Heating effect of electric current (Joule's law) · Electric power and energy calculations · Factors affecting resistance
Don't skip: Numerical problems on series and parallel resistor combinations; Numerical problems on power and energy (including cost calculation); Derivation of equivalent resistance for series and parallel combinations; Explanation of Joule's law with examples
One-Line Revision (read aloud before the exam)
  1. Electric current is the rate of flow of charge, I = Q/t, measured in amperes.
  2. One ampere is one coulomb per second.
  3. Potential difference is work done per unit charge, V = W/Q, measured in volts.
  4. Ohm's law: V = IR, provided temperature remains constant.
  5. Resistance of a conductor depends on length, area of cross-section, and material: R = ρl/A.
  6. Resistivity ρ is a material property; SI unit is ohm metre (Ω m).
  7. In series, Rs = R1 + R2 + R3; current is same, voltage divides.
  8. In parallel, 1/Rp = 1/R1 + 1/R2 + 1/R3; voltage is same, current divides.
  9. Joule's law of heating: H = I²Rt, heat produced in a resistor.
  10. Electric power P = VI = I²R = V²/R, measured in watts.
  11. Commercial unit of energy is kilowatt hour (kWh); 1 kWh = 3.6 × 10⁶ J.
  12. Ammeter is connected in series; voltmeter is connected in parallel.
  13. Conventional current flows from positive to negative terminal, opposite to electron flow.
  14. Alloys like nichrome have higher resistivity than pure metals and are used in heating devices.
  15. A fuse protects circuits by melting when current exceeds a safe value.

MCQs — 10 Questions with Answers

Q1. A student is setting up a circuit to light a bulb using a battery, ammeter, and switch. She connects the ammeter in parallel with the bulb instead of in series. What will happen?

  • A. The ammeter will burn out due to very high current flowing through it ✓
  • B. The circuit will work normally and the ammeter will measure current correctly
  • C. The bulb will not glow because ammeter blocks current flow
  • D. The voltmeter will not work in the circuit

Answer: A — An ammeter has very low resistance and must be in series; connecting it in parallel creates a short circuit with extremely high current that damages the ammeter, whereas incorrect option B ignores this fundamental requirement.

Q2. In a laboratory, a charge of 4 coulombs flows through a wire in 8 seconds. A student calculates the current and then predicts that if the same charge flows for 4 seconds instead, the current will be half. Is this prediction correct?

  • A. Yes, because current is inversely proportional to time
  • B. No, because current would be double since I = Q/t and time is halved ✓
  • C. Yes, because less time means proportionally less current
  • D. No, because charge and time are independent of current

Answer: B — Current I = Q/t; if Q remains 4 C but t becomes 4 s instead of 8 s, then I = 4/4 = 1 A (double the original 0.5 A), not half, making option A a common misconception about inverse relationships.

Q3. A battery is connected to a circuit with a bulb. The voltmeter reads 12 V across the bulb and 2 joules of work is done to move charge through the bulb. How much charge flowed through the bulb?

  • A. 0.167 coulombs ✓
  • B. 6 coulombs
  • C. 24 coulombs
  • D. 14 coulombs

Answer: A — Using V = W/Q, we get Q = W/V = 2 J / 12 V = 0.167 C, whereas option B incorrectly multiplies instead of dividing.

Q4. A household circuit operates at 220 V. A technician observes that when an electric heater is switched on, the current drawn is 5 A. Based on this observation, what can be concluded about the work done by the heater in 2 hours?

  • A. 7920 J because W = V × I × t = 220 × 5 × 2
  • B. 7920 kJ because W = V × I × t = 220 × 5 × 7200 seconds ✓
  • C. 2200 J because work depends only on voltage
  • D. 220 J because work depends only on current

Answer: B — Work W = V × I × t where t must be in seconds; 2 hours = 7200 seconds, so W = 220 × 5 × 7200 = 7,920,000 J = 7920 kJ, whereas option A incorrectly converts time.

Q5. In a physics class demonstration, the teacher shows that when the switch is open, no current flows through the bulb even though the cell is connected. She then asks why closing the switch completes the circuit and allows current to flow. Which explanation best describes this phenomenon?

  • A. The switch creates potential difference only when closed
  • B. The switch provides a complete conducting path between the cell terminals through the bulb, allowing the existing potential difference to move charges ✓
  • C. The switch increases the voltage of the cell when closed
  • D. The switch generates electrons inside the bulb

Answer: B — The potential difference is created by the cell whether the switch is open or closed; the switch simply completes the conducting path for charges to flow, whereas option A incorrectly suggests the cell generates voltage only when the switch closes.

Q6. Assertion (A): An ammeter should always be connected in series with a circuit component to measure the current flowing through it. Reason (R): An ammeter has very high resistance so it must be placed in series to minimize its effect on the circuit.

  • A. Both A and R are true and R is the correct explanation of A
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false ✓
  • D. A is false but R is true

Answer: C — Assertion A is correct—ammeters must be in series—but Reason R is false because ammeters have very LOW (not high) resistance, making option B a common misconception about ammeter properties.

Q7. Assertion (A): When 2 coulombs of charge flows through a conductor in 4 seconds, the current is 0.5 amperes. Reason (R): Current is defined as the rate of flow of charge, calculated using I = Q/t.

  • A. Both A and R are true and R is the correct explanation of A ✓
  • B. Both A and R are true but R is not the correct explanation of A
  • C. A is true but R is false
  • D. A is false but R is true

Answer: A — Both A and R are true; I = Q/t = 2 C / 4 s = 0.5 A, and R correctly explains why A is true by defining current.

Q8. Assertion (A): Electric potential difference is conventionally defined as the work done per unit positive charge moved from lower to higher potential. Reason (R): The direction of conventional current is defined as the flow of positive charges, opposite to electron flow.

  • A. Both A and R are true and R is the correct explanation of A
  • B. Both A and R are true but R is not the correct explanation of A ✓
  • C. A is true but R is false
  • D. A is false but R is true

Answer: B — Both A and R are scientifically true, but R does not directly explain why potential difference is defined as work per unit charge; they address different conventions in electricity.

Q9. A student connects a voltmeter across a battery's terminals and reads 9 V. When the student connects a light bulb to the battery, the voltmeter reading drops to 8 V. What does this observation indicate about the circuit?

  • A. The battery lost charge and now provides less voltage
  • B. The internal resistance of the battery causes a voltage drop when current flows through the circuit ✓
  • C. The voltmeter is connected incorrectly in the circuit
  • D. The light bulb creates additional voltage in the circuit

Answer: B — When current flows, some voltage is 'lost' due to the battery's internal resistance, so the terminal voltage drops from 9 V to 8 V, whereas option A incorrectly suggests charge is lost rather than voltage being divided across internal resistance.

Q10. An electrical device operates at 230 V and draws a current of 2 A for 5 hours. How much electrical energy is consumed by the device?

  • A. 2300 J
  • B. 8,280,000 J (or 8.28 MJ) ✓
  • C. 1150 J
  • D. 23,000 J

Answer: B — Energy W = V × I × t = 230 × 2 × (5 × 3600) = 230 × 2 × 18,000 = 8,280,000 J, whereas option A incorrectly forgets to convert hours to seconds.

Flashcards

What is electric current?

Electric current is the rate of flow of electric charge through a conductor, expressed as I = Q/t in amperes.

Define one coulomb of charge.

One coulomb is the total charge contained in approximately 6 × 10¹⁸ electrons.

What is the SI unit of current and who is it named after?

The SI unit is ampere (A), named after French scientist André-Marie Ampère.

How should an ammeter be connected in a circuit?

An ammeter must always be connected in series with the component whose current is being measured.

Define electric potential difference.

Electric potential difference is the work done per unit charge to move a charge from one point to another in a circuit, V = W/Q.

What does 1 volt mean?

One volt is the potential difference when 1 joule of work is done to move a charge of 1 coulomb between two points.

How should a voltmeter be connected in a circuit?

A voltmeter must always be connected in parallel across the two points between which potential difference is to be measured.

What is an electric circuit?

An electric circuit is a continuous and closed path through which electric charge flows from the positive terminal of a cell through external components back to the negative terminal.

Why does conventional current flow opposite to electron flow?

Conventional current was defined as the flow of positive charges before electrons were discovered, so current direction is opposite to actual electron movement.

What is the charge of one electron?

One electron carries a negative charge of 1.6 × 10⁻¹⁹ coulombs.

Important Board Questions

Define electric current and give its SI unit. How is an ammeter connected in a circuit to measure current? [2 marks]

State that electric current = rate of flow of charge (Q/t). SI unit is ampere. Ammeter must be connected in series with the component being measured.

A charge of 4 coulombs flows through a bulb in 2 seconds. Calculate the current through the bulb. If the potential difference across the bulb is 6 V, find the work done in moving this charge. [3 marks]

Use I = Q/t to find current (answer: 2 A). Then use W = VQ to find work done (answer: 24 J). Show both formula substitutions clearly.

Explain why an electric circuit must be a closed continuous path for current to flow. Also explain the difference between conventional current direction and the direction of electron flow in a metallic conductor with an example. [5 marks]

Explain that current only flows when circuit is closed (broken circuit = no complete path = no current, no light). For electron flow: electrons move from negative to positive terminal (negative charges attracted to positive), but conventional current is defined as positive charge flow from positive to negative (opposite direction) because this convention was established before electrons were discovered.

Assertion-Reason Questions for Electricity

Q1. Assertion (A):

An ammeter should always be connected in series with a circuit component.

Reason (R):

An ammeter has very low resistance and connecting it in series ensures accurate current measurement without significantly altering the circuit.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both the assertion and reason are true. An ammeter must be connected in series to measure the current flowing through a component, and the reason correctly explains why—its low resistance prevents it from affecting the circuit. This is the proper procedure for measuring current.

Q2. Assertion (A):

A voltmeter is connected in parallel across a conductor to measure potential difference.

Reason (R):

If a voltmeter is connected in series, it would form a closed loop and prevent proper measurement of potential difference.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — The assertion is true—a voltmeter is indeed connected in parallel. However, the reason is not the correct explanation. A voltmeter has very high resistance and, if connected in series, it would draw negligible current and provide inaccurate readings. The parallel connection allows it to measure potential difference between two points without disturbing the main circuit current significantly.

Q3. Assertion (A):

A potential difference of 5 V means that 5 joules of work is done on each coulomb of charge.

Reason (R):

Potential difference is defined as work done per unit charge, and 1 volt equals 1 joule per coulomb.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both assertion and reason are true, and the reason correctly explains the assertion. This is the precise definition of potential difference: V = W/Q. A 5 V potential difference supplies 5 joules of energy to each coulomb of charge passing through it.

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Case-Based Questions (CBQ) for Electricity

Case Study:

A student performed an experiment to study the relationship between current and potential difference using a nichrome wire. She set up a circuit with a variable power supply (0-12 V), an ammeter, a voltmeter, and a nichrome wire. She recorded the following readings: At 1 V, current was 0.25 A; at 2 V, current was 0.50 A; at 3 V, current was 0.75 A; at 4 V, current was 1.0 A. The ammeter was connected in series with the wire, while the voltmeter was connected in parallel across the wire. The student plotted these readings and observed a straight-line graph passing through the origin.

Q1. Why was the ammeter connected in series in this experiment?

  • A. To measure the potential difference across the wire
  • B. To measure the current flowing through the wire ✓
  • C. To protect the circuit from overload
  • D. To increase the current in the circuit

Answer: B

Q2. Why was the voltmeter connected in parallel across the nichrome wire?

  • A. To measure the current flowing through the circuit
  • B. To measure the potential difference across the wire ✓
  • C. To reduce the resistance of the circuit
  • D. To increase the current flowing through the wire

Answer: B

Q3. What is the relationship between voltage and current shown by the student's straight-line graph?

  • A. Current is inversely proportional to voltage
  • B. Current is proportional to the square of voltage
  • C. Current is independent of voltage
  • D. Current is directly proportional to voltage ✓

Answer: D

Q4. If the student applies a potential difference of 6 V across the same nichrome wire, what current would be expected?

  • A. 1.5 A ✓
  • B. 1.25 A
  • C. 2.0 A
  • D. 2.5 A

Answer: A

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Previous Year Questions (PYQ) for Electricity

Real CBSE board exam questions from past papers — free to attempt.

CBSE 2026 · Set 31/4/1 · 2 marks

(A) A fuse in electric circuit is rated 4 A. Can it be used with an electric heater of rating 2 kW, 200 V? Explain your answer.

CBSE 2026 · Set 31/4/1 · 2 marks

(a) Define volt, the unit of potential difference.

CBSE 2026 · Set 31/4/1 · 5 marks

(A) (i) The given electric circuit is a part of an electrical device. Use the information given in the electric circuit diagram to calculate: (I) Potential difference across the ends of resistor R2. (II) Value of resistor R 2. (III) Value of resistor R 1. (ii) Write the factors on which resistance of a conductor depends and derive the formula for resistance of a given conductor.

CBSE 2026 · Set 31/4/2 · 2 marks

What do the symbols given below represent in an electric circuit? Write one function of each. (i) [Diagram-based question — skipped] (ii) [Diagram-based question — skipped]

CBSE 2026 · Set 31/4/2 · 5 marks

(A) V-I graph for a nichrome wire is shown in the diagram. (i) Prove that the graph follows Ohm's law and calculate the resistance of nichrome wire from the given V-I graph. (ii) [Diagram-based question — skipped] (I) Calculate the equivalent resistance of the combination of resistors connected in the given circuit. (II) Find the value of the total current drawn from the battery.

CBSE 2026 · Set 31/4/3 · 2 marks

A fuse in electric circuit is rated 4 A. Can it be used with an electric heater of rating 2 kW, 200 V? Explain your answer.

CBSE 2026 · Set 31/4/3 · 2 marks

(a) State the law which gives the relation between the electric current flowing through a given conductor and the potential difference applied across it. (b) Draw V-I graph for a given conductor as per the law stated in above question (a).

CBSE 2026 · Set 31/4/3 · 5 marks

(i) Derive an expression for the equivalent resistance of three resistors connected in parallel with a battery as shown in figure. (ii) Find the values of the electric currents flowing through the resistors R1, R2 and R3 in the given electric circuit.

CBSE 2026 · Set 31/5/1 · 3 marks

(a) Why does an electric bulb become dim when an electric heater in parallel circuit is switched ON? (b) How to connect three resistors each of resistance 8 Ω, so that the equivalent resistance of the combination is 12 Ω? Draw diagram of the combination and justify your answer.

CBSE 2026 · Set 31/5/1 · 4 marks

Three students Shweta, Ayesha and Samridhi were performing an experiment to understand the factors on which the resistance of a conductor depends. Each one of them completed electric circuit with the help of a cell, an ammeter, a plug key and wire. Shweta put nichrome wire of length 'l' in the circuit and after plugging the key, noted current in the ammeter. Ayesha put nichrome wire of same thickness but twice the length i.e. '2l' in the circuit and after plugging the key, noted current in the ammeter. Samridhi took copper wire of length 'l' and same thickness in the circuit and after plugging the key, noted current in the ammeter.

(a) If the ammeter reading is X ampere with nichrome wire of length 'l', then what will be the ammeter reading if the length of nichrome wire is doubled with same area of cross-section?

CBSE 2026 · Set 31/5/2 · 3 marks

(a) Why does an electric bulb become dim when an electric heater in parallel circuit is switched ON ? (b) How to connect three resistors each of resistance 8 Ω, so that the equivalent resistance of the combination is 12 Ω ? Draw diagram of the combination and justify your answer.

CBSE 2026 · Set 31/5/2 · 4 marks

Three students Shweta, Ayesha and Samridhi were performing an experiment to understand the factors on which the resistance of a conductor depends. Each one of them completed electric circuit with the help of a cell, an ammeter, a plug key and wire. Shweta put nichrome wire of length 'l' in the circuit and after plugging the key, noted current in the ammeter. Ayesha put nichrome wire of same thickness but twice the length i.e. '2l' in the circuit and after plugging the key, noted current in the ammeter. Samridhi took copper wire of length 'l' and same thickness in the circuit and after plugging the key, noted current in the ammeter. (a) If the ammeter reading is X ampere with nichrome wire of length 'l', then what will be the ammeter reading if the length of nichrome wire is doubled with same area of cross-section ? (b) What happens to the ammeter reading if the area of cross-section of nichrome wire is doubled, keeping the length of wire 'l' the same ? (c) Define 'resistivity'. Write its SI unit. Compare the resistivity of an alloy with its constituents metals.

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