JEE Main & Advanced JEE Main Paper Phase-I (Held on 07-1-2020 Evening)

  • question_answer
    In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W, 15 small fans of 10 W and 2 heaters of 1 kW. The voltage of electric main is 220 V. The minimum fuse capacity (rated value) of the building will be [JEE MAIN Held on 07-01-2020 Evening]

    A) 15 A    

    B) 10 A

    C) 20 A

    D) 25 A

    Correct Answer: C

    Solution :

    [c] P = VI \[\Rightarrow \,\,\,\,\,{{I}_{main}}=15\times \frac{45}{220}+15\times \frac{100}{220}+15\times \frac{10}{220}+2\times \frac{{{10}^{3}}}{220}\] \[\Rightarrow \,\,\,\,\,{{I}_{main}}=\frac{15\times 155+2000}{220}=19.66\,A\] Answer is 20 A

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