A household supply is 220 , V with a fuse of | vedclass

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(D) The total power available in the circuit is given by $P_{total} = V 	imes I$.Given $V = 220 \, V$ and $I = 9 \, A$,the total power is $P_{total}

Vedclass · Accepted Answer

$33$

Vedclass · Answer

(D) The total power available in the circuit is given by $P_{total} = V 	imes I$.Given $V = 220 \, V$ and $I = 9 \, A$,the total power is $P_{total} = 220 	imes 9 = 1980 \, W$.Let $n$ be the maximum number of $60 \, W$ bulbs that can be connected in parallel.The total power consumed by $n$ bulbs is $P_{consumed} = n 	imes 60 \, W$.For the fuse not to blow,the total power consumed must be less than or equal to the total power available: $n 	imes 60 \leq 1980$.Solving for $n$: $n \leq \frac{1980}{60} = 33$.Therefore,the maximum number of bulbs is $33$.

$A$ household supply is $220 \, V$ with a fuse of $9 \, A$. What is the maximum number of $60 \, W$ bulbs that can be connected in parallel and operated simultaneously?

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