RAJASTHAN ­ PET Rajasthan PET Solved Paper-2011

  • question_answer
    A 750 Hz, 20 V source is connected to a resistance of\[100\,\Omega ,\]an inductance of 0.1803 H and a capacitance of 10\[\mu \]F all in series. What is the time in which resistance (thermal capacity\[=2\text{ }J/{}^\circ C\]) will get heated by\[10{}^\circ C\]?

    A)  380 s         

    B)  384 s

    C)  470 s         

    D)  479 s

    Correct Answer: B

    Solution :

     \[{{X}_{L}}=\omega L=2\times 3.14\times 750\times 0.1803=850\,\Omega \] \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2\times 3.14\times 750\times {{10}^{-4}}}=21.2\,\Omega \] Impedance \[Z=\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}\] \[=\sqrt{{{(100)}^{2}}+{{(580-21.2)}^{2}}}\] \[=835\,\Omega \] Power dissipated\[P={{E}_{rms}}\times {{I}_{rms}}\cos \phi \] \[={{E}_{rms}}\times \frac{{{E}_{rms}}}{Z}\times \frac{R}{Z}=\frac{E_{rms}^{2}\times R}{{{Z}^{2}}}\] \[=\frac{{{(20)}^{2}}\times 100}{{{(835)}^{2}}}=0.0574\,W\] Produced resistance \[H=2\times 10=20\text{ }J\] Let time taken to produce this heat is t, then\[Pt=20\] \[t=\frac{20}{0.0574}=384\,s\]


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