An air-conditioner of 3.2 kW power rating is connected to a domestic electric circuit having a current rating of 10 A. The voltage of power supply is 220 V. What will happen when this air-conditioner is switched on?
Explain your answer.
At first, we will calculate the current drawn by this air-conditioner.
Given:
Power, P = 3.2 kW = 3.2 × 1000 W= 3200W (converted kilowatt into watt by multiplying with 1000)
Potential difference, V = 220 V
We know that-
$P=V\times I$
Substituting the given values we get-
$3200=220\times I$
$I=\frac{3200}{220}$
$I=14.54A$
Thus, the current drawn by the air-conditioner is 14.5 A which is very high and the fuse in this circuit is of capacity 10 A.
Therefore, when a very high current of 14.5 A flows through 10 A fuse, the fuse wire gets heated due to which it melts and break the circuit and thereby cutting off the power supply.
Hence, if the given 3.2 kW power rating air-conditioner is switched on, the fuse will blow cutting off the power supply.
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