Q12
3 marksShort AnswerSection Section B

(a) List the factors on which the resistance of a uniform cylindrical conductor of a given material depends.

(b) The resistance of a wire of 0.01 cm0.01\text{ cm} radius is 10Ω10\Omega. If the resistivity of the wire is 50×108Ω m50 \times 10^{-8}\Omega\text{ m}, find the length of this wire.

Electricity
Electric Power and Energy
Official Answer

OR


(a) Electric Power: Electric power is the rate at which electrical energy is consumed or dissipated in an electric circuit.

Power (P)=Work Done (W)Time (t)=VI\text{Power } (P) = \frac{\text{Work Done } (W)}{\text{Time } (t)} = VI


SI Unit: The SI unit of electric power is watt (W). One watt is the power consumed by a device that carries 1 A1\text{ A} of current when operated at a potential difference of 1 V1\text{ V}.


(b) Calculation of Energy Consumed:

Given:

  • Power of the kettle, P=2 kWP = 2\text{ kW}
  • Time of use, t=2 ht = 2\text{ h}

(i) In kilowatt-hour (kWh):

Energy (E)=Power (P)×Time (t)\text{Energy } (E) = \text{Power } (P) \times \text{Time } (t)

E=2 kW×2 h=4 kWhE = 2\text{ kW} \times 2\text{ h} = 4\text{ kWh}


(ii) In Joules (J):

We know that 1 kWh=3.6×106 J1\text{ kWh} = 3.6 \times 10^6\text{ J}.

E=4×3.6×106 JE = 4 \times 3.6 \times 10^6\text{ J}

E=14.4×106 J=1.44×107 JE = 14.4 \times 10^6\text{ J} = 1.44 \times 10^7\text{ J}


Therefore, the energy consumed is 4 kWh4\text{ kWh} or 1.44×107 J1.44 \times 10^7\text{ J}.

rate of consumption of electrical energywattkilowatt-hour4 kWh1.44 x 10^7 J

Marking Scheme

  • 1Define electric power and state its SI unit (Watt): 1 Mark
  • 2Calculate energy in kWh (4 kWh4\text{ kWh}): 1 Mark
  • 3Convert energy to Joules (1.44×107 J1.44 \times 10^7\text{ J}): 1 Mark

Hint

Energy is the product of power and time. For (i), keep power in kW and time in hours. For (ii), multiply the result of (i) by 3.6×1063.6 \times 10^6.

Quick Oral Answer

Q: What is the relation between the commercial unit of energy and the SI unit of energy? A: The commercial unit of energy is the kilowatt-hour (kWh) and the SI unit is the Joule (J). The relation is: 1 kWh=3.6×106 J1\text{ kWh} = 3.6 \times 10^6\text{ J} (or 3.6 MJ3.6\text{ MJ}).

Analysis & Explanation

Electric power represents how fast work is being done by an electric current. The commercial unit of electrical energy is the kilowatt-hour (kWh), commonly known as a 'unit'.


Since a Joule is a very small unit of energy (1 J=1 W1 s1\text{ J} = 1\text{ W} \cdot 1\text{ s}), it is inconvenient for expressing household electricity consumption. Hence, we use kilowatt-hours (1 kWh=1000 W×3600 s=3.6×106 J1\text{ kWh} = 1000\text{ W} \times 3600\text{ s} = 3.6 \times 10^6\text{ J}). In this problem, the 2 kW2\text{ kW} kettle running for 2 hours2\text{ hours} consumes 44 commercial units of electricity, which is equivalent to a massive 1.44×107 Joules1.44 \times 10^7\text{ Joules} of energy.

Common Mistakes

  1. 1Confusing power (PP) with energy (EE).
  2. 2Using incorrect conversion factors for converting hours to seconds (e.g., multiplying by 60 instead of 3600).
  3. 3Writing the unit of energy in Joules as W/s\text{W/s} instead of Ws\text{W} \cdot \text{s}.

Interesting Facts

The unit 'kilowatt-hour' is what electricity boards use to calculate your monthly electricity bill. One 'unit' on your meter corresponds exactly to 1 kWh1\text{ kWh} of energy consumption.

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Frequently Asked Questions

How many marks does this question carry in CBSE Class 10 Science 2022?

This question carries 3 marks in the CBSE Class 10 Science 2022 examination.

Which chapter does this question come from in Science?

This question is from the chapter "Electricity" in the CBSE Class 10 Science syllabus.

What topic does this question cover in Science?

This question covers the topic "Electric Power and Energy" from CBSE Class 10 Science.

What type of question is this in the CBSE Class 10 Science 2022 paper?

This is a Short Answer question from Section Section B in the CBSE Class 10 Science 2022 paper.