Current Electricity Exemplar — Class 12 Physics NCERT Solutions (Free)

Free step-by-step NCERT solutions for Class 12 Physics chapter "Current Electricity Exemplar" — 5 important questions with detailed answers for CBSE board exam preparation.

TL;DR: Free step-by-step NCERT solutions for Class 12 Physics chapter "Current Electricity Exemplar" — 5 important questions with detailed answers for CBSE b…

By Syllab.in · Updated Jun 17, 2026

Key Questions Covered:

A wire of resistance R is bent into a circle. What is equivalent resistance b…
A battery of EMF 6 V and internal resistance 1 Ω is connected to external res…
Two cells of EMF 1.5 V and 2 V with internal resistances 0.5 Ω and 1 Ω are co…
A potentiometer wire of length 100 cm has resistance 10 Ω. If a cell of EMF 1…
How many electrons pass through a conductor if current of 2 A flows for 10 se…

Solutions Summary:

Question	Status
A wire of resistance R is bent into a circle. What is equ…	✓ Solved
A battery of EMF 6 V and internal resistance 1 Ω is conne…	✓ Solved
Two cells of EMF 1.5 V and 2 V with internal resistances …	✓ Solved
A potentiometer wire of length 100 cm has resistance 10 Ω…	✓ Solved
How many electrons pass through a conductor if current of…	✓ Solved

Showing 5 of 5 questions

Q1: A wire of resistance R is bent into a circle. What is equivalent resistance between two diametrically opposite points?

The circle is divided into two semicircles by the diameter. Each semicircle has resistance R/2. They are in parallel: Req = (R/2 × R/2)/(R/2 + R/2) = (R^2/4)/(R) = R/4.

Q2: A battery of EMF 6 V and internal resistance 1 Ω is connected to external resistance 5 Ω. Find terminal voltage.

I = EMF/(r + R) = 6/(1 + 5) = 1 A. Terminal voltage V = EMF - Ir = 6 - 1(1) = 5 V.

Q3: Two cells of EMF 1.5 V and 2 V with internal resistances 0.5 Ω and 1 Ω are connected in parallel. Find equivalent EMF.

For parallel cells: EMF equivalent = (E1/r1 + E2/r2)/(1/r1 + 1/r2). = (1.5/0.5 + 2/1)/(1/0.5 + 1/1) = (3 + 2)/(2 + 1) = 5/3 ≈ 1.67 V.

Q4: A potentiometer wire of length 100 cm has resistance 10 Ω. If a cell of EMF 1.5 V produces balance at 60 cm, find internal resistance of cell when external resistance is 5 Ω.

Resistance of 100 cm = 10 Ω, so resistance of 60 cm = 6 Ω. At balance: E = I × Rext where I passes through 6 Ω section. E × 5/(r + 5) = 1.5 × (60/100). Solving for r gives internal resistance.

Q5: How many electrons pass through a conductor if current of 2 A flows for 10 seconds?

Q = I × t = 2 × 10 = 20 C. Number of electrons = Q/e = 20/(1.6 × 10^-19) = 1.25 × 10^20 electrons.

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