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Haryana PMT Haryana PMT Solved Paper-2006

  • question_answer
    An AC is represented by e = 220 sin (100 \[\pi \]) t V        and is applied over a resistance of 110 \[\Omega \]. The                 heat produced in 7 min is :

    A)                  \[11\times {{10}^{3}}cal\]                           

    B)  \[22\times {{10}^{3}}cal\]

    C)                  \[33\times {{10}^{3}}cal\]                           

    D)  \[25\times {{10}^{3}}cal\]

    Correct Answer: B

    Solution :

                    \[H=\frac{I_{v}^{2}RT}{J}cal\] \[=\frac{{{(\sqrt{2})}^{2}}\times 110\times 7\times 60}{4.2}\] \[\left( \because \,{{I}_{rms}}=\frac{{{I}_{o}}}{\sqrt{2}}=\frac{220/110}{\sqrt{2}}=\sqrt{2} \right)\]                 \[=22\times {{10}^{3}}cal\]


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