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Electricity - ELECTRIC POWER11.8 ELECTRIC POWER11.8 ELECTRIC POWER11.8 ELECTRIC POWER11.8 ELECTRIC POWER

Grade 10CBSE

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

🔑Concepts

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Electric Power (PP) is defined as the rate at which electrical energy is consumed or dissipated in an electric circuit. Mathematically, it is the product of potential difference (VV) and current (II): P=VIP = VI.

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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}. Therefore, 1 W=1 Volt×1 Ampere1\text{ W} = 1\text{ Volt} \times 1\text{ Ampere}.

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Power can also be expressed in terms of resistance (RR) using Ohm's Law (V=IRV = IR): P=I2RP = I^2R or P=V2RP = \frac{V^2}{R}.

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The commercial unit of electrical energy is the kilowatt-hour (kWhkWh), commonly referred to as a 'unit'.

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The relationship between the commercial unit and the SI unit of energy (Joule) is: 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}.

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Total electrical energy consumed (EE) depends on the power of the appliance and the duration for which it is used: E=P×tE = P \times t.

📐Formulae

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

P=V×IP = V \times I

P=I2RP = I^2 R

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 connected to a 220 V220\text{ V} generator. The current is 0.50 A0.50\text{ A}. What is the power of the bulb?

Solution:

Given: V=220 VV = 220\text{ V}, I=0.50 AI = 0.50\text{ A}. Using the formula P=VIP = VI: P=220 V×0.50 A=110 WP = 220\text{ V} \times 0.50\text{ A} = 110\text{ W}

Explanation:

The power is calculated by multiplying the potential difference provided by the generator with the current flowing through the bulb.

Problem 2:

An electric oven rated 2000 W2000\text{ W} is operated for 2 hours2\text{ hours} daily. Calculate the energy consumed in 30 days30\text{ days} and the total cost if the rate is Rs. 5Rs.\ 5 per unit.

Solution:

Power P=2000 W=2 kWP = 2000\text{ W} = 2\text{ kW}. Time per day t=2 ht = 2\text{ h}. Total time for 30 days=2×30=60 h30\text{ days} = 2 \times 30 = 60\text{ h}. Energy E=P×t=2 kW×60 h=120 kWhE = P \times t = 2\text{ kW} \times 60\text{ h} = 120\text{ kWh}. Cost calculation: 120×5600\begin{array}{r} 120 \\ \times 5 \\ \hline 600 \end{array} Total cost = Rs. 600Rs.\ 600.

Explanation:

First, convert power to kilowatts and find the total hours of operation. Multiply power and time to find the energy in kWhkWh (units), then multiply by the rate per unit to find the total cost.