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Electricity - Electric power

Grade 10CBSE

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Electric power (PP) is defined as the rate at which electrical energy is consumed or dissipated in an electric circuit: P=WtP = \frac{W}{t}.

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The SI unit of electric power is the Watt (WW). 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}, i.e., 1 W=1 V×1 A1\text{ W} = 1\text{ V} \times 1\text{ A}.

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Power can be expressed using Ohm's Law (V=IRV = IR) in three main forms: P=VIP = VI, P=I2RP = I^2R, and P=V2RP = \frac{V^2}{R}.

•

The power rating of an appliance (e.g., 100 W,220 V100\text{ W}, 220\text{ V}) indicates the power it consumes when connected to the specified voltage. The resistance of such a device is constant and can be calculated using R=V2PR = \frac{V^2}{P}.

•

The commercial unit of electrical energy is the kilowatt-hour (kWhkWh), commonly known as a 'unit'. It is related to Joules by: 1 kWh=3.6×106 J1\text{ kWh} = 3.6 \times 10^6\text{ J}.

•

Electrical energy consumed (EE) is the product of power (PP) and time (tt): E=P×tE = P \times t.

📐Formulae

P=WtP = \frac{W}{t}

P=V×IP = V \times I

P=I2RP = I^2R

P=V2RP = \frac{V^2}{R}

E=P×tE = P \times t

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

💡Examples

Problem 1:

An electric bulb is rated 220 V220\text{ V} and 100 W100\text{ W}. When it is operated on 110 V110\text{ V}, what will be the power consumed?

Solution:

  1. Find the resistance of the bulb: R=V2P=2202100=48400100=484 ΩR = \frac{V^2}{P} = \frac{220^2}{100} = \frac{48400}{100} = 484\,\Omega.
  2. Calculate new power at 110 V110\text{ V}: Pnew=Vnew2R=1102484=12100484=25 WP_{new} = \frac{V_{new}^2}{R} = \frac{110^2}{484} = \frac{12100}{484} = 25\text{ W}.

Explanation:

Resistance remains constant for the filament. Since power is proportional to the square of the voltage (P∝V2P \propto V^2), reducing the voltage by half reduces the power to one-fourth.

Problem 2:

Calculate the energy consumed by a 250 W250\text{ W} TV set in 1 hour1\text{ hour} and a 1200 W1200\text{ W} electric toaster in 10 minutes10\text{ minutes}. Which uses more energy?

Solution:

For TV: ETV=P×t=250 W×1 h=250 WhE_{TV} = P \times t = 250\text{ W} \times 1\text{ h} = 250\text{ Wh}. For Toaster: Etoaster=1200 W×1060 h=200 WhE_{toaster} = 1200\text{ W} \times \frac{10}{60}\text{ h} = 200\text{ Wh}. Comparing the two: 250 Wh>200 Wh250\text{ Wh} > 200\text{ Wh}.

Explanation:

Energy is the product of power and time. Even though the toaster has higher power, the TV consumes more energy because it is operated for a significantly longer duration.

Problem 3:

An electric motor takes 5 A5\text{ A} from a 220 V220\text{ V} line. Determine the power of the motor and the energy consumed in 2 h2\text{ h}.

Solution:

  1. Power P=VI=220 V×5 A=1100 WP = VI = 220\text{ V} \times 5\text{ A} = 1100\text{ W}.
  2. Energy E=P×t=1100 W×2 h=2200 WhE = P \times t = 1100\text{ W} \times 2\text{ h} = 2200\text{ Wh} or 2.2 kWh2.2\text{ kWh}.

Explanation:

First, use the current and voltage to find the power in Watts. Then, multiply by time in hours to get the energy in Watt-hours, or convert to kilowatt-hours for standard units.

Electric power - Revision Notes & Key Formulas | CBSE Class 10 Science