JEE Mains · Physics · STD 12 - 3. current electricity
In a building there are \(15\) bulbs of \(45\; \mathrm{W}, 15\) bulbs of \(100\; \mathrm{W}, 15\) small fans of \(10 \;\mathrm{W}\) and \(2\) heaters of \(1 \;\mathrm{kW}\). The voltage of electric main is \(220\; \mathrm{V}\). The minimum fuse capacity (rated value) of the building will be: .......... \(A\)
- A \(10\)
- B \(25\)
- C \(15\)
- D \(20\)
Answer & Solution
Correct Answer
(D) \(20\)
Step-by-step Solution
Detailed explanation
\(220 \mathrm{I}=\mathrm{P}=15 \times 45+15 \times 100+15 \times 10+2 \times 10^{3}\) \(I=\frac{4325}{220}=19.66\) \(\mathrm{I} \simeq 20 \mathrm{A}\)
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