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.

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.

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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…

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NCERT Solutions for Class 12 Physics — Current Electricity Exemplar. Step-by-step answers to all 5 textbook questions from this chapter, written for CBSE board preparation and free to use.

Current Electricity Exemplar — All 5 Questions Solved

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.

How to Use These Solutions

Attempt each question yourself first and only then compare with the worked answer. Marks in the CBSE board exam are awarded for the METHOD as much as the final result, so reproduce the steps rather than memorising the last line. Where a solution states a law, a formula or a definition, learn that wording — examiners look for it.

Related practice for this chapter: chapter MCQs, previous-year questions, revision notes and sample papers.

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