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
     An electric bulb is designed to draw power P0 at voltage V0. If the voltage is V it draws a power P. Then [KCET 2001]

    A)            \[P={{\left( \frac{{{V}_{0}}}{V} \right)}^{2}}{{P}_{0}}\]       

    B)            \[P={{\left( \frac{V}{{{V}_{0}}} \right)}^{2}}{{P}_{0}}\]

    C)            \[P=\left( \frac{V}{{{V}_{0}}} \right)\,{{P}_{0}}\]     

    D)            \[P=\left( \frac{{{V}_{0}}}{V} \right)\,{{P}_{0}}\]

    Correct Answer: B

    Solution :

               \[P\propto {{V}^{2}}\]Þ \[\frac{P}{{{P}_{0}}}={{\left( \frac{V}{{{V}_{0}}} \right)}^{2}}\Rightarrow P={{\left( \frac{V}{{{V}_{0}}} \right)}^{2}}{{P}_{0}}\]


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