JEE Main & Advanced Physics Current Electricity, Charging & Discharging Of Capacitors Question Bank Heating Effect of Current

  • question_answer Two electric bulbs rated \[{{P}_{1}}\] watt V volts and \[{{P}_{2}}\] watt V volts are connected in parallel and V volts are applied to it. The total power will be [MP PMT 2001; MP PET 2002]

    A)            \[{{P}_{1}}+{{P}_{2}}watt\]   

    B)            \[\sqrt{{{P}_{1}}{{P}_{2}}}\]watt

    C)            \[\frac{{{P}_{1}}{{P}_{2}}}{{{P}_{1}}+{{P}_{2}}}watt\]         

    D)            \[\frac{{{P}_{1}}+{{P}_{2}}}{{{P}_{1}}{{P}_{2}}}watt\]

    Correct Answer: A

    Solution :

                       If resistances of  bulbs are \[{{R}_{1}}\] and \[{{R}_{2}}\] respectively then in parallel \[\frac{1}{{{R}_{P}}}=\frac{1}{{{R}_{1}}}+\frac{1}{{{R}_{2}}}\]Þ \[\frac{1}{\left( \frac{{{V}^{2}}}{{{P}_{p}}} \right)}=\frac{1}{\left( \frac{{{V}^{2}}}{{{P}_{1}}} \right)}+\frac{1}{\left( \frac{{{V}^{2}}}{{{P}_{2}}} \right)}\] Þ \[{{P}_{P}}={{P}_{1}}+{{P}_{2}}\]

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