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BHU PMT BHU PMT Solved Paper-2004

  • question_answer
    If a 2 kW boiler is used everyday for 1 h, then electrical energy consumed by boiler in thirty days is:                                                                                                                                                                                         [BHU PMT-2004]

    A)  15 unit                

    B)  60 unit

    C)  120 unit                              

    D)  240 unit

    Correct Answer: B

    Solution :

                     Key Idea: The rate at which electrical energy \[\left( W \right)\] is dissipated into other from of energy is called electric power \[\left( P \right)\] Power=rate of electrical energy \[\Rightarrow \]                               \[P\frac{W}{t}\] \[\Rightarrow \]               \[W=pt=2\,kW\times 1\,h=2\,kWh\] Also 1 kWh or 1 unit is the quantity of electric energy which is dissipated in 1 h in a circuit when the electric power in the circuit is 1 kWh.                 \[\therefore \]Energy consumed in 30 days                 \[=2\times 30=60\,kWh=60\,units\].


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