22
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.
(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.
Show SolutionHide Solution↓
(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)
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)