Current Electricity — Class 12 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 12 Physics chapter "Current Electricity" — 8 important questions with detailed answers for CBSE board exam preparation.

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TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Current Electricity" — 8 important questions with detailed answers for CBSE board exam…

Written & reviewed by the Syllab.in Academic Team (CBSE/NCERT subject experts) · Updated Jul 23, 2026

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Key Questions Covered:

Define electric current. State its SI unit and symbol.
State Ohm's law. A wire has resistance 5 Ω. If potential difference across it…
Define resistance and resistivity. Write formula relating resistance to resis…
A copper wire has length 100 m and cross-sectional area 1 mm². If resistivity…
Define emf and internal resistance of a cell. Write expression relating V, em…
Derive expression for combined resistance when resistances are in series.
+ 2 more questions in the full chapter

Solutions Summary:

Question	Status
Define electric current. State its SI unit and symbol.	✓ Solved
State Ohm's law. A wire has resistance 5 Ω. If potential …	✓ Solved
Define resistance and resistivity. Write formula relating…	✓ Solved
A copper wire has length 100 m and cross-sectional area 1…	✓ Solved
Define emf and internal resistance of a cell. Write expre…	✓ Solved
Derive expression for combined resistance when resistance…	✓ Solved

Showing 6 of 8 questions

Q1: Define electric current. State its SI unit and symbol.

Electric Current (I): Rate of flow of electric charge through a conductor. I = Q/t (charge per unit time) SI Unit: Ampere (A) = Coulomb per second (C/s) Symbol: I Direction: Conventional current flows from positive to negative terminal (opposite to electron flow). One Ampere: When charge of 1 coulomb flows through conductor in 1 second 1 A = 1 C/s Microscopic definition: I = nAve where n = number density of charge carriers A = cross-sectional area ve = drift velocity e = magnitude of charg...

Q2: State Ohm's law. A wire has resistance 5 Ω. If potential difference across it is 10 V, find the current.

Ohm's Law: At constant temperature, current through conductor is directly proportional to applied potential difference. I ∝ V (at constant temperature) V = IR or I = V/R where V = potential difference (V) I = current (A) R = resistance (Ω) Given: R = 5 Ω, V = 10 V Step 1: I = V/R = 10/5 Step 2: I = 2 A Final Answer: Current = 2 A Note: Ohm's law is valid only for ohmic conductors (metallic conductors at constant temperature). Semiconductors and gases do not follow Ohm's law.

Q3: Define resistance and resistivity. Write formula relating resistance to resistivity.

Resistance (R): Opposition offered by conductor to flow of electric current. R = V/I (potential difference per unit current) SI Unit: Ohm (Ω) = V/A Resistivity (ρ): Inherent property of material that opposes current flow. SI Unit: Ohm-meter (Ω⋅m) Symbol: ρ (rho) Relation between R and ρ: R = ρL/A where ρ = resistivity of material L = length of conductor A = cross-sectional area Typical resistivity values at 20°C: Copper: ρ ≈ 1.7 × 10⁻⁸ Ω⋅m Aluminum: ρ ≈ 2.6 × 10⁻⁸ Ω⋅m Nichrome: ρ ≈ 1 × 10...

Q4: A copper wire has length 100 m and cross-sectional area 1 mm². If resistivity of copper is 1.7 × 10⁻⁸ Ω⋅m, find its resistance.

Given: L = 100 m A = 1 mm² = 1 × 10⁻⁶ m² ρ = 1.7 × 10⁻⁸ Ω⋅m Formula: R = ρL/A Step 1: R = (1.7 × 10⁻⁸) × 100 / (1 × 10⁻⁶) Step 2: R = 1.7 × 10⁻⁶ / (1 × 10⁻⁶) Step 3: R = 1.7 Ω Final Answer: Resistance = 1.7 Ω

Q5: Define emf and internal resistance of a cell. Write expression relating V, emf, and r.

EMF (ε - epsilon): Total energy per unit charge supplied by the cell. Emf: E = W/Q (work done per unit charge) SI Unit: Volt (V) Internal Resistance (r): Resistance of the cell due to electrolyte and electrodes. SI Unit: Ohm (Ω) Relation when current I flows through cell: V = ε - Ir where V = terminal voltage across cell ε = emf of cell I = current flowing through circuit r = internal resistance When cell is in open circuit (I = 0): V = ε (terminal voltage equals emf) When current flows ...

Q6: Derive expression for combined resistance when resistances are in series.

Series Connection: Resistances connected end-to-end so same current flows through each. Consider three resistances R₁, R₂, R₃ connected in series across potential difference V. Same current I flows through all: I = I₁ = I₂ = I₃ Voltage distribution: V₁ = IR₁ V₂ = IR₂ V₃ = IR₃ Total voltage: V = V₁ + V₂ + V₃ = IR₁ + IR₂ + IR₃ = I(R₁ + R₂ + R₃) Let equivalent resistance be R_s: V = IR_s Comparing: R_s = R₁ + R₂ + R₃ For n resistances in series: R_s = R₁ + R₂ + R₃ + ... + Rₙ Properties of s...

Showing 6 of 8 questions. Visit the full page for complete solutions.