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Electricity — Question 14

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Question 14

Compare the power used in the 2 Ω resistor in each of the following circuits:

(i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and

(ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors.

Answer

(i) The potential difference is 6 V and the resistors 1 Ω and 2 Ω are connected in series:

Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors. NCERT Class 10 Science CBSE Solutions.

Rs = 1 Ω + 2 Ω = 3 Ω.

Applying Ohm's law:

V=IR∴I=VR⇒I=63⇒I=2 A\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{6}{3} \\[1em] \Rightarrow \text{I} = 2 \text{ A}

Power across 2 Ω resistor = I2R = 2 x 2 x 2 = 8 W

Hence, the power consumed by the 2 Ω resistor is 8 W.

(ii) In parallel circuit, voltage across the resistors remains the same. So, voltage across 2 Ω resistor is 4 V:

Compare the power used in the 2 Ω resistor in each of the following circuits: (i) a 6 V battery in series with 1 Ω and 2 Ω resistors, and (ii) a 4 V battery in parallel with 12 Ω and 2 Ω resistors. NCERT Class 10 Science CBSE Solutions.

Power across 2 Ω resistor = V2R=4×42\dfrac{\text{V}^2}{{\text{R}}} = \dfrac{4 \times 4}{2} = 8 W

The power consumed by the 2 Ω resistor is 8 W.

Hence, same power is consumed by 2 Ω resistor in both the circuits.

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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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