205 Marks · March 2026 · Standardopen ↗
(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.
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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.
215 Marks · March 2026 · Standardopen ↗
(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).
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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)
225 Marks · March 2026 · Standardopen ↗
Attempt either subpart (a) or (b) :
(a) (i) In a house, $\text{5}$ fans, rated $\text{100 W}$; $\text{220 V}$ each are used for $\text{4}$ hours daily and an electric heater rated $\text{1000 W}$; $\text{220 V}$ is used for $\text{5}$ hours daily. Calculate the cost of electrical energy consumed by the fans and the electric heater in a month of $\text{30}$ days if the cost of $\text{1}$ unit of electrical energy is $\text{\text{Rs}5}$.
(ii) You have two fuse wires X and Y with electric current rating $\text{5 A}$ and $\text{10 A}$ respectively. Which of the two would you use with a heater rated $\text{2000 W}$; $\text{220 V}$ ? Justify your answer.
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(a) (i) E=Pxt (½ Mark)
E (fan) = $5 \times 100 \times 4$ (½ Mark)
= $2000$ Wh (½ Mark)
= $2$ kWh (½ Mark)
E(heater) = $1000 \times 5$ (½ Mark)
= $5000$ Wh (½ Mark)
= $5$ kWh (½ Mark)
Total energy, E = E(fan) + E(heater) (½ Mark)
= $2 + 5 = 7$ kWh (½ Mark)
Cost = $7 \times 30 \times 5$= Rs $1050$ (1 Mark)
(ii) P=VI (½ Mark)
Current required for heater
$I = \frac{P}{V}$ (½ Mark)
$I = \frac{2000}{220}$ (½ Mark)
$I = 9.09$ A (½ Mark)
Current rating of Y wire is slightly greater than required value. Hence fuse wire Y will be used. (1 Mark)