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Electricity — Question 6

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Question 6

A copper wire has diameter 0.5 mm and resistivity of 1.6 × 10–8 Ω m. What will be the length of this wire to make its resistance 10 Ω? How much does the resistance change if the diameter is doubled?

Answer

Given,

diameter = 0.5 mm

area = πd24=3.14×(5×10−4)24\dfrac{πd^2}{4} = \dfrac{3.14 \times (5 \times 10^{-4})^2}{4} = 19.625 x 10–8 m

resistivity = 1.6 × 10–8 Ω m

R = 10 Ω

l = ?

We know, R = ρ lA\dfrac{l}{A}

Substituting we get,

10 = 1.6 × 10–8 x l19.625×10−8\dfrac{l}{19.625 \times 10^{-8}}

l = 10×19.6251.6\dfrac{10 \times 19.625}{1.6 } = 122.7 m

The length of the wire is 122.72 m.

When the diameter is doubled

According to the formula,

R = ρ lA\dfrac{l}{A} where A = πd24\dfrac{πd^2}{4}

Hence, R ∝ 1d2\dfrac{1}{d^2}

When diameter is double,

Resistance will change by 1(2d)2\dfrac{1}{(2d)^2} = 14d2\dfrac{1}{4d^2}

∴ Resistance will become 14th\dfrac{1}{4}^{\text{th}} of its original value.

Hence, new resistance = 14×10\dfrac{1}{4} \times 10 = 2.5 Ω

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Science | Chapter 11: ElectricityWeb Content

Chapter 11: Electricity — Quick Revision Guide

Introduction

Electricity powers modern life. This chapter covers electric current, potential difference, Ohm's law, resistance, series and parallel circuits, and electrical power and energy calculations.

Key Points at a Glance

  1. Current I = Q/t (ampere); potential difference V = W/Q (volt); Ohm's law V = IR
  2. Resistance R = V/I (ohm); factors: R ∝ l, R ∝ 1/A, material (ρ), temperature; R = ρl/A
  3. Conductors: low ρ (~10−8 Ω·m); alloys: higher ρ, used in heating elements (nichrome)
  4. Series: Req = R1 + R2 + ...; same current; voltage divides; one failure breaks all
  5. Parallel: 1/Req = 1/R1 + 1/R2 + ...; same voltage; current divides; independent operation; used in homes
  6. Power: P = VI = I2R = V2/R (watt); Energy: E = Pt (joule); 1 kWh = 3.6 × 106 J
  7. Heating effect: H = I2Rt; applications: heater, iron, fuse (low m.p. alloy, series with live wire)
  8. Electric bill: units (kWh) = power(kW) × time(h); cost = units × rate

Real-World Connections

Household wiring in parallel allows independent appliance use; fuses and MCBs prevent fire; LED bulbs save energy (lower power for same brightness); electricity bill management.

Quick Self-Test (5 Questions)

  1. What is the most important concept you learned from this chapter?
  2. Can you write three key equations/formulae from this chapter from memory?
  3. Draw a labelled diagram relevant to this chapter without looking at your notes.
  4. Explain one real-world application of a concept from this chapter.
  5. What is one common mistake students make in this chapter, and how can you avoid it?

Further Study

  • NCERT Textbook Chapter 11
  • NCERT Exemplar Problems
  • Bright Tutorials Detailed Notes: ch11-electricity.html
  • Bright Tutorials Practice Questions: ch11-electricity.html
  • Previous Year CBSE Board Papers

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