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(i) Define the SI unit of power. Express the electric power of an electric appliance in terms of potential difference and current flowing through it.
(ii) In a house, 3 bulbs of $100\text{ W}$ each, are lit for $5\text{ hours}$ daily and an electric heater of $1.0\text{ kW}$ is used for half an hour daily. Calculate the total energy consumed in a month of 30 days and its cost at the rate of ₹ $3.60$ per $\text{kWh}$.
(iii) Convert the commercial unit of electric energy into Joule (J).
(ii) In a house, 3 bulbs of $100\text{ W}$ each, are lit for $5\text{ hours}$ daily and an electric heater of $1.0\text{ kW}$ is used for half an hour daily. Calculate the total energy consumed in a month of 30 days and its cost at the rate of ₹ $3.60$ per $\text{kWh}$.
(iii) Convert the commercial unit of electric energy into Joule (J).
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(i) • Power consumed by a device that carries 1 A of current when operated at a potential difference of 1 V / If one joule energy is consumed in one second then power of instrument is said to be 1 watt / 1 W = 1 volt $\times$ 1 ampere (1 Mark)
• $P = VI$ (1 Mark)
(ii) $E = P \times t$ (½ Mark)
Energy (3 bulbs) $= 3 \times 100 \times 5 = 1500\text{ Wh} = 1.5\text{ kWh}$
Energy (electric heater) $= 1.0 \times 0.5 = 0.5\text{ kWh}$
Total energy consumed (1 day) $= 1.5 + 0.5 = 2\text{ kWh}$ (½ Mark)
Total energy consumed (30 days) $= 30 \times 2 = 60\text{ kWh} = 60\text{ units}$
Total cost = Units $\times$ Rate $= 60 \times 3.60 = \text{\text{Rs} } 216$ (1 Mark)
(ii) $1\text{ kW h} = 1000\text{ watt} \times 3600\text{ second} = 3.6 \times 10^6\text{ watt second} = 3.6 \times 10^6\text{ joule (J)}$ (1 Mark)
• $P = VI$ (1 Mark)
(ii) $E = P \times t$ (½ Mark)
Energy (3 bulbs) $= 3 \times 100 \times 5 = 1500\text{ Wh} = 1.5\text{ kWh}$
Energy (electric heater) $= 1.0 \times 0.5 = 0.5\text{ kWh}$
Total energy consumed (1 day) $= 1.5 + 0.5 = 2\text{ kWh}$ (½ Mark)
Total energy consumed (30 days) $= 30 \times 2 = 60\text{ kWh} = 60\text{ units}$
Total cost = Units $\times$ Rate $= 60 \times 3.60 = \text{\text{Rs} } 216$ (1 Mark)
(ii) $1\text{ kW h} = 1000\text{ watt} \times 3600\text{ second} = 3.6 \times 10^6\text{ watt second} = 3.6 \times 10^6\text{ joule (J)}$ (1 Mark)