2. Solved Papers
  3. NEET

NEET NEET SOLVED PAPER 2016 Phase-I

  • question_answer
    An inductor 20 mH, a capacitor \[50\mu F\] and a resistor \[\text{40}\,\Omega \] are connected in series across a source of emf V = 10 sin 340 t. The power loss in A.C. circuit is :-                                                                                

    A)  0.51 W                

    B)  0.67 W

    C)  0.76 W                

    D)  0.89 W

    Correct Answer: A

    Solution :

                     \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{340\times 50\times {{10}^{-6}}}=58.8\,\Omega \]                 \[{{X}_{L}}=\omega L=340\times 20\times {{10}^{-3}}=6.8\,\Omega \]                 \[Z=\sqrt{{{R}^{2}}+{{({{X}_{C}}-{{X}_{L}})}^{2}}}\]                 \[=\sqrt{{{40}^{2}}+{{(58.8-6.8)}^{2}}}=\sqrt{4304\Omega }\] \[P=i_{rms}^{2}R={{\left( \frac{{{V}_{rms}}}{Z} \right)}^{2}}R\] \[={{\left( \frac{10/\sqrt{2}}{\sqrt{4304}} \right)}^{2}}\times 40=\frac{50\times 40}{4304}=0.47\,W\] So best answer (nearest answer) will be


You need to login to perform this action.
You will be redirected in 3 sec spinner