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Electricity — Question 2

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Question 2

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it?

Answer

Given,

Potential difference across circuit = 220 V

Resistance of electric lamp = 100 Ω

Resistance of toaster = 50 Ω

Resistance of water filter = 500 Ω

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it? NCERT Class 10 Science CBSE Solutions.

Equivalent resistance across the three appliances connected in parallel

1Rp=1100+150+1500=5+10+1500=16500\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{100}} + \dfrac{1}{\text{50}}+ \dfrac{1}{\text{500}} = \dfrac{5+10+1}{\text{500}} = \dfrac{16}{\text{500}}

Hence, Rp = 50016=1254=31.25 Ω\dfrac{500}{\text{16}} = \dfrac{125}{\text{4}} = 31.25 \space \Omega

The resistance of the electric iron connected to the same source will be 31.25 Ω as same current flows through it as that across the three appliances (connected in parallel).

An electric lamp of 100 Ω, a toaster of resistance 50 Ω, and a water filter of resistance 500 Ω are connected in parallel to a 220 V source. What is the resistance of an electric iron connected to the same source that takes as much current as all three appliances, and what is the current through it? NCERT Class 10 Science CBSE Solutions.

Current across electric iron = ?

Applying Ohm's law

V = IR

Substituting we get,

220 = I x 31.25

I = 22031.25\dfrac{220}{31.25} = 7.04 A

Therefore, current through the electric iron will be 7.04 A.

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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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