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Electricity — Question 11

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Question 11

Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω.

Answer

If we connect resistors in series, R = 6 Ω + 6 Ω + 6 Ω = 18 Ω.

If we connect all resistors in parallel,

1R=16+16+16=36=12\dfrac{1}{\text{R}} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{3}{6} = \dfrac{1}{2}

Hence, R = 2Ω

We can obtain the desired value by connecting two of the resistors in either series or parallel.

Case (i)

If two resistors are connected in parallel, then their equivalent resistance is

1Rp=16+16=26=13\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3}

Hence, Rp = 3Ω

The third resistor in series, then we get equivalent resistance

Rp + 6 = 3 + 6 = 9Ω

∴ To get a resistance of 9 Ω, two 6 Ω resistors should be connected in parallel and the third 6 Ω resistor should be connected in series with the combination as shown below:

Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω. NCERT Class 10 Science CBSE Solutions.

Case (ii)

When two resistors are connected in series, their equivalent resistance is

Rs = 6 Ω + 6 Ω = 12 Ω

The third resistor is connected in parallel with 12 Ω. Hence the equivalent resistance is:

1R=112+16=1+212=312=14\dfrac{1}{\text{R}} = \dfrac{1}{12} + \dfrac{1}{6} = \dfrac{1+2}{12} = \dfrac{3}{12} = \dfrac{1}{4}

⇒ R = 4 Ω

∴ To get a resistance of 4 Ω, two 6 Ω resistors should be connected in series and the third 6 Ω resistor should be connected in parallel with the combination as shown below:

Show how you would connect three resistors, each of resistance 6 Ω, so that the combination has a resistance of (i) 9 Ω, (ii) 4 Ω. NCERT Class 10 Science CBSE Solutions.
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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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