CBSE Class 10 Science

Electricity — 15 Most Important Questions
for Board Exam 2025-26

Carefully selected questions covering all topics from the Electricity chapter — original, exam-pattern aligned, with full explanations.

📘 Chapter: Electricity 🎯 Board Pattern: CBSE ❓ 15 Questions 💡 Explanation on Every Question
Topics Covered: Electric Current Potential Difference Ohm's Law Resistance & Resistivity Series & Parallel Circuits Heating Effect Electric Power P, V, I, R Interrelation
📌 How to Use This Page Read each question carefully as you would in a board exam. Try to write your answer first, then click "Show Explanation" to check a detailed, step-by-step answer with formulas and key points.
1 Mark

Section A — Very Short Answer Questions (Q1–Q5)

1
Define electric current. What is its SI unit?

✅ Answer & Explanation

Electric current is defined as the rate of flow of electric charge through a conductor.

I = Q / t

Where I = electric current, Q = charge (in coulombs), t = time (in seconds).

SI unit of electric current is Ampere (A). 1 Ampere = 1 Coulomb per second (1 C/s).
2
What does the slope of a V-I graph represent for a metallic conductor?

✅ Answer & Explanation

For a metallic conductor, when voltage (V) is plotted on the Y-axis and current (I) on the X-axis, the graph is a straight line passing through the origin (Ohm's Law).

Slope = V / I = Resistance (R)
The slope of the V-I graph represents the Resistance (R) of the conductor. A steeper slope means higher resistance.
3
Two resistors of 4 Ω and 6 Ω are connected in parallel. What is their equivalent resistance?

✅ Answer & Explanation

For resistors in parallel:

1/R = 1/R₁ + 1/R₂ = 1/4 + 1/6 = 3/12 + 2/12 = 5/12
Equivalent Resistance R = 12/5 = 2.4 Ω

Note: In parallel combination, equivalent resistance is always less than the smallest individual resistance.

4
State one factor that does NOT affect the resistance of a conductor.

✅ Answer & Explanation

Resistance depends on: length of conductor, cross-sectional area, nature of material (resistivity), and temperature.

The colour (or shape/colour/mass) of the conductor does NOT affect resistance. Resistance is purely a function of geometry and material properties.
5
An electric bulb is rated 60 W, 220 V. What is the resistance of its filament?

✅ Answer & Explanation

Using the power formula:

P = V² / R → R = V² / P = (220)² / 60 = 48400 / 60
Resistance R = 806.67 Ω ≈ 807 Ω
2 Marks

Section B — Short Answer Questions (Q6–Q9)

6
State Ohm's Law. Write its mathematical expression and draw a neat labelled circuit diagram to verify it.

✅ Answer & Explanation

Ohm's Law: At constant temperature, the electric current flowing through a conductor is directly proportional to the potential difference across its ends.

V ∝ I → V = IR → R = V/I

Circuit Diagram components: Battery (EMF source), Rheostat (variable resistance), Key (switch), Ammeter (in series to measure current), Voltmeter (in parallel across the resistor R).

When voltage is increased using the rheostat, the ammeter reading increases proportionally, confirming V/I = constant = R.
7
Explain why parallel combination of resistors is preferred over series combination in household electric circuits.

✅ Answer & Explanation

Parallel combination is preferred because:

  • Each appliance gets the same voltage (equal to the supply voltage), ensuring proper functioning.
  • Each appliance can be switched on/off independently without affecting others.
  • If one appliance fails, the others continue to work (no circuit break).
  • The overall resistance is reduced, allowing more current to flow as needed.
In series, all appliances share the voltage and one failure breaks the whole circuit — making it unsuitable for homes.
8
Define resistivity. Write its SI unit. On what factors does resistivity depend?

✅ Answer & Explanation

Resistivity (ρ) is the resistance offered by a conductor of unit length and unit cross-sectional area.

R = ρ × (L / A) → ρ = R × A / L

SI Unit: Ohm-metre (Ω·m)

Resistivity depends only on the nature of the material and temperature. It does NOT depend on length or area of the conductor.
9
A current of 2 A flows through a resistor for 5 minutes. If the resistance is 10 Ω, calculate the heat produced.

✅ Answer & Explanation

Given: I = 2 A, t = 5 min = 300 s, R = 10 Ω

Using Joule's Law of Heating:

H = I² × R × t = (2)² × 10 × 300 = 4 × 10 × 300
Heat produced H = 12,000 J = 12 kJ
4 Marks

Section C — Long Answer Type I (Q10–Q13)

10
Derive the expression for equivalent resistance when three resistors R₁, R₂, and R₃ are connected in series. Also state one practical application.

✅ Answer & Explanation

Series Combination: Resistors are connected end-to-end so the same current (I) flows through each.

Let the potential difference across each be V₁, V₂, and V₃.

V = V₁ + V₂ + V₃ V = IR₁ + IR₂ + IR₃ V = I(R₁ + R₂ + R₃) ∴ Rs = R₁ + R₂ + R₃

Key Properties:

  • Same current flows through all resistors.
  • Total voltage is distributed across resistors.
  • Equivalent resistance is always greater than any individual resistance.
Practical Application: Christmas/decorative series light bulbs. If one bulb fuses, the whole chain goes off (a disadvantage that highlights why parallel is preferred for home wiring).
11
Derive the expression for equivalent resistance of three resistors connected in parallel. How is the current distributed among them?

✅ Answer & Explanation

Parallel Combination: All resistors are connected across the same two points, so the same potential difference (V) applies to each.

I = I₁ + I₂ + I₃ V/Rp = V/R₁ + V/R₂ + V/R₃ 1/Rp = 1/R₁ + 1/R₂ + 1/R₃

Current Distribution: Each branch carries a current inversely proportional to its resistance.

I₁ = V/R₁ , I₂ = V/R₂ , I₃ = V/R₃
The branch with the lowest resistance carries the highest current. Equivalent resistance (Rp) is always less than the smallest individual resistance.
12
What is the heating effect of electric current? Derive Joule's Law of Heating and mention three appliances that work on this effect.

✅ Answer & Explanation

When electric current flows through a resistor, the free electrons collide with the lattice ions, converting electrical energy into heat energy. This is the heating effect of electric current.

Derivation of Joule's Law:

Work done in moving charge Q through potential difference V:

W = V × Q = V × I × t Since V = IR: H = I²Rt (Also: H = V²t/R = VIt)

Joule's Law: Heat produced is directly proportional to I², R, and t.

Appliances using heating effect: 1. Electric iron (tungsten/nichrome element) 2. Electric heater / room heater 3. Electric geyser / water heater (Also: toaster, electric bulb filament, electric kettle)
13
Three resistors of 3 Ω, 6 Ω, and 9 Ω are connected in parallel across a 18 V battery. Find (i) equivalent resistance (ii) total current drawn from the battery (iii) current through each resistor.

✅ Answer & Explanation

Given: R₁ = 3 Ω, R₂ = 6 Ω, R₃ = 9 Ω, V = 18 V

(i) 1/Rp = 1/3 + 1/6 + 1/9 = 6/18 + 3/18 + 2/18 = 11/18 Rp = 18/11 ≈ 1.64 Ω (ii) I = V/Rp = 18 ÷ (18/11) = 11 A (iii) I₁ = 18/3 = 6 A I₂ = 18/6 = 3 A I₃ = 18/9 = 2 A
Verification: I₁ + I₂ + I₃ = 6 + 3 + 2 = 11 A ✓ (matches total current)
5 Marks

Section D — Long Answer Type II (Q14–Q15)

14
Define electric power. Derive the relationship between P, V, I, and R. A household uses the following appliances: one 1000 W iron for 2 hours, two 60 W bulbs for 8 hours, and one 500 W TV for 4 hours per day. Calculate (i) total energy consumed in a day and (ii) monthly electricity bill at ₹5 per unit.

✅ Answer & Explanation

Electric Power: The rate at which electrical energy is consumed or dissipated by a device.

P = Work / Time = W/t W = VQ = V × I × t ∴ P = VI Since V = IR: P = (IR) × I = I²R Since I = V/R: P = V × (V/R) = V²/R

Summary of Power Formulas: P = VI = I²R = V²/R

Energy Calculation (per day):

Iron: 1000 W × 2 h = 2000 Wh = 2 units Bulbs: 2 × 60 W × 8 h = 960 Wh = 0.96 units TV: 500 W × 4 h = 2000 Wh = 2 units ───────────────────────────────────── Total per day = 4.96 units ≈ 4.96 kWh
(i) Daily consumption ≈ 4.96 kWh (ii) Monthly consumption = 4.96 × 30 = 148.8 units Monthly Bill = 148.8 × ₹5 = ₹744
15
Write a detailed note on the factors affecting resistance of a conductor. A wire of resistance 8 Ω is stretched uniformly to double its original length. What is its new resistance? Justify your answer.

✅ Answer & Explanation

Factors Affecting Resistance:

  • Length (L): R ∝ L — Longer wire → higher resistance (more collisions)
  • Cross-sectional area (A): R ∝ 1/A — Thicker wire → lower resistance (more paths for electrons)
  • Nature of material (ρ): R = ρL/A — Each material has a unique resistivity
  • Temperature: For metals, resistance increases with temperature
R = ρ × L / A

When wire is stretched to double length:

Volume remains constant: L × A = L' × A' → 2L × A' = L × A → A' = A/2

R' = ρ × 2L / (A/2) = ρ × 2L × 2/A = 4 × (ρL/A) = 4R R' = 4 × 8 = 32 Ω
New Resistance = 32 Ω. Doubling the length and halving the area (constant volume) increases resistance by 4 times.
⚠️ Disclaimer These questions have been created for educational purposes based on the CBSE Class 10 Science syllabus. While every effort has been made for accuracy, there may be unintentional typographical or calculation errors. Students are advised to cross-verify all answers with their NCERT textbook, official CBSE resources, or a qualified teacher before the examination. EduBrightPages is not responsible for any errors or omissions.