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JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors / वर्तमान बिजली, चार्ज और कैपेसिटर का निर Question Bank Heating Effect of Current

  • question_answer
    A heater coil connected to a supply of a 220 V is dissipating some power \[{{P}_{1}}.\] The coil is cut into half and the two halves are connected in parallel. The heater now dissipates a power \[{{P}_{2}}.\] The ratio of power \[{{P}_{1}}\,\,:\,\,{{P}_{2}}\] is         [AFMC 2004]

    A)            2 : 1                                          

    B)            1 : 2

    C)            1 : 4                                          

    D)            4 : 1

    Correct Answer: C

    Solution :

                       \[P=\frac{{{V}^{2}}}{R}\]. If resistance of heater coil is R, then resistance of parallel combination of two halves will be \[\frac{R}{4}\] So  \[\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{{{R}_{2}}}{{{R}_{1}}}=\frac{R/4}{R}=\frac{1}{4}\]


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