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(i) How much electric current will an electric iron draw from $220\,\text{V}$ source if the resistance of its heating element when hot, is $55\,\Omega$ ? Calculate the power consumed by the electric iron when it is operated at $220\,\text{V}$.
(ii) In a house, 3 bulbs of $100\,\text{watt}$ 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 $30\,\text{days}$ and its cost at the rate of ₹ $3.60$ per $\text{kWh}$.
(iii) With reason explain, why are alloys commonly used to make elements of electrical heating devices.
(ii) In a house, 3 bulbs of $100\,\text{watt}$ 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 $30\,\text{days}$ and its cost at the rate of ₹ $3.60$ per $\text{kWh}$.
(iii) With reason explain, why are alloys commonly used to make elements of electrical heating devices.
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$I = \frac{V}{R}$
$I = \frac{220}{55}$
$I = 4 A$
Power of electric iron $P = VI$
$P = 220 \times 4$
$P = 880 W$
$E=Pxt$
Energy (3 bulbs) $= 3 \times 100 \times 5$
$= 1500$
$= 1.5 \text{ kWh}$
Energy (electric heater) $= 1.0 \times 0.5$
$= 0.5 \text{ kWh}$
Total energy consumed (1 day) $= 1.5 + 0.5 = 2 \text{ kWh}$
Total energy consumed (30 days) $= 30 \times 2$
$= 60 \text{ kWh}$
$= 60 \text{ units}$
Total cost = Units $\times$ Rate
$= 60 \times 3.60$
$= 216$
The resistivity of an alloy is generally higher than that of its constituent metals. / Alloys do not oxidise (burn) readily at high temperatures.
$I = \frac{220}{55}$
$I = 4 A$
Power of electric iron $P = VI$
$P = 220 \times 4$
$P = 880 W$
$E=Pxt$
Energy (3 bulbs) $= 3 \times 100 \times 5$
$= 1500$
$= 1.5 \text{ kWh}$
Energy (electric heater) $= 1.0 \times 0.5$
$= 0.5 \text{ kWh}$
Total energy consumed (1 day) $= 1.5 + 0.5 = 2 \text{ kWh}$
Total energy consumed (30 days) $= 30 \times 2$
$= 60 \text{ kWh}$
$= 60 \text{ units}$
Total cost = Units $\times$ Rate
$= 60 \times 3.60$
$= 216$
The resistivity of an alloy is generally higher than that of its constituent metals. / Alloys do not oxidise (burn) readily at high temperatures.