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Electricity — Question 4

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Question 4

Two conducting wires of the same material and of equal lengths and equal diameters are first connected in series and then parallel in a circuit across the same potential difference. The ratio of heat produced in series and parallel combinations would be

  1. 1:2
  2. 2:1
  3. 1:4
  4. 4:1
Answer

1:4

Reason — Let Rs and Rp be the equivalent resistance of the wires when connected in series and parallel respectively.

Rs = R + R = 2R

1Rp=1R+1R=2R\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{R}} + \dfrac{1}{\text{R}} = \dfrac{2}{\text{R}}

Rp = R2\dfrac{\text{R}}{2}

Let Ps and Pp be the power consumed in series and parallel circuits, respectively.

Power (P) = V2R\dfrac{\text{V}^2}{\text{R}}

Ps = V22R\dfrac{\text{V}^2}{\text{2R}}

and

Pp = V2R2=2V2R\dfrac{\text{V}^2}{\dfrac{\text{R}}{2}} = \dfrac{2\text{V}^2}{\text{R}}

For the same potential difference V, the ratio of the heat produced in the circuit is given by ratio of Ps and Pp hence

Hs:Hp=V22R2V2R=V2×R2R×2V2=14\text{H}_s : \text{H}_p = \dfrac{\dfrac{\text{V}^2}{\text{2R}}}{\dfrac{2\text{V}^2}{\text{R}}} \\[1em] = \dfrac{\text{V}^2 \times \text{R}}{\text{2R} \times 2\text{V}^2} \\[1em] = \dfrac{1}{4}

Hence, the ratio of the heat produced is 1:4.

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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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