First slide

Thermal effect of current

Question

In a building there are 15 Bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10W and 2 heater of 1 kW. The voltage of electric main is 220V. the minimum fuse capacity (rated value) of the building will be:

Moderate

A

10 A
B

20 A
C

25 A
D

15 A

Solution

All the components are connected in parallel with the voltage source. Voltage across each each component is 220V
Total power consumed = $(15 \times 45) + (15 \times 100) + (15 \times 10) + (2 \times 1000)$ $= 4325 W$
So current is $i = \frac{P}{V} = \frac{4325}{220} = 19.66 A$
$∴ A f u s e o f c a p a c i t y 20 A s h o u l d b e u s e d .$

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