JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर JEE PYQ-Current Electricity Charging and Discharging of Capacitors

  • question_answer
    In a large building, there are 15 bulbs of 40 W, 5 bulbs of 100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage of the electric mains is 220 V. The minimum capacity of the main fuse of the building will be:                          [JEE MAIN 2014]

    A) 12 A

    B) 14 A

    C) 8 A

    D) 10 A

    Correct Answer: A

    Solution :

    [a] \[\frac{1}{{{\operatorname{R}}_{eq}}}=\frac{1}{{{\operatorname{R}}_{1}}}+\frac{1}{{{\operatorname{R}}_{2}}}\]
    \[P=\frac{{{V}^{2}}}{{{\operatorname{R}}_{eq}}}=\frac{{{V}^{2}}}{{{\operatorname{R}}_{1}}}+\frac{{{V}^{2}}}{{{\operatorname{R}}_{2}}}+.......\]
    \[=15\times 40+5\times 100+80\times 5+1\times 1000=2500w\]
    \[P=VI\]
    \[P=2500=220I\]
    \[I=\frac{250}{22}=\frac{125}{11}\]


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