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Concept Topic GuideCBSE & ICSEClass 10PhysicsChapter 11Verified Faculty Notes & Solutions

Electricity Class 10 – Ohm's Law, Resistors & Joule's Heating NCERT Solutions

Master Class 10 Physics Electricity: Ohm's law, series and parallel circuits, resistivity, electrical power, Joule's law of heating, and solved numericals.

6 students readUpdated 30 August 2026

Electricity – Complete Chapter Revision & Solutions

Electricity is a controllable and convenient form of energy. Electric current (I) is defined as the rate of flow of electric charges across any cross-section of a conductor.

Ohm's Law & Resistance

Ohm's Law Statement: The potential difference (V) across the ends of a metallic wire is directly proportional to the current (I) flowing through it, provided its temperature remains constant.
V = I × R (where R is the electrical resistance, in Ohms, Ω)

Factors Affecting Resistance & Resistivity

R = ρ × (l / A)
Where: l = length of conductor (m), A = cross-sectional area (m²), ρ = specific electrical resistivity of material (Ω·m).

Combination of Resistors

FeatureSeries CombinationParallel Combination
Equivalent FormulaR_eq = R₁ + R₂ + R₃1/R_eq = 1/R₁ + 1/R₂ + 1/R₃
Current (I)Same through every resistorDivides among branches: I = I₁ + I₂ + I₃
Voltage (V)Divides across resistors: V = V₁ + V₂ + V₃Same voltage across all parallel branches
Overall ResistanceGreater than the largest resistorSmaller than the smallest individual resistor

Joule's Law of Heating & Electric Power

Heat Generated: H = I²Rt = VIt = (V² / R) × t (Joules)
Electric Power: P = V × I = I²R = V² / R (Watts)
Commercial Unit: 1 kilowatt-hour (1 kWh) = 3.6 × 10⁶ Joules = 1 Unit of electricity.
QUESTION • CBSE Class 10 Board Exam Problem (3 Marks)

An electric lamp of resistance 20 Ω and a conductor of 4 Ω resistance are connected in series to a 6 V battery. Calculate:
(a) The total resistance of the circuit,
(b) The current flowing through the circuit,
(c) The potential difference across the lamp and the conductor.

SOLUTION • Step-by-Step Calculation

(a) Total Resistance in Series:
R_total = R_lamp + R_conductor = 20 Ω + 4 Ω = 24 Ω

(b) Current in Circuit (Ohm's Law):
I = V / R_total = 6 V / 24 Ω = 0.25 A

(c) Potential Difference Across Each:
• Across Lamp: V₁ = I × R_lamp = 0.25 A × 20 Ω = 5.0 V
• Across Conductor: V₂ = I × R_conductor = 0.25 A × 4 Ω = 1.0 V
Check: V₁ + V₂ = 5.0 V + 1.0 V = 6.0 V (Matches battery voltage!)

Indexed Topics & Examination Keywords

#CBSE Class 10 Physics#ICSE Class 10 Physics#Electricity Class 10#Ohms Law#Series and Parallel#Resistivity#Joules Heating#Electric Power#NCERT Solutions#Board Exam 2026#SL Arora Physics
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