RAJASTHAN ­ PET Rajasthan PET Solved Paper-2006

  • question_answer
    A heater coil is labelled 100 W, 220 V. The coil is cut into two equal halves and the two pieces are joined in parallel to the same source. The energy now liberated per second is

    A)  50 J             

    B)  40 J

    C)  4000 J           

    D)  400 J

    Correct Answer: D

    Solution :

     The coil is cut into two equal halves, then resistance of each part\[=\frac{R}{2}\] If the both part are joined in parallel, then resistance \[R'=\frac{R}{4}\] The energy per second \[P=\frac{{{V}^{2}}}{R}\] Or \[P\propto \frac{1}{R}\] \[\frac{P'}{P}=\frac{R}{R'}=4\] \[P'=4P=400J/s\]


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