NEET NEET SOLVED PAPER 2018

  • question_answer
    An inductor 20 mH, a capacitor \[100\text{ }\mu \text{F}\] and a resistor \[50\text{ }\Omega \]are connected in series across a source of emf, V=10 sin 314 t. The power loss in the circuit is [NEET - 2018]

    A)  2.74 W                       

    B)  0.43 W

    C)  0.79 W                       

    D)  1.13 W

    Correct Answer: C

    Solution :

    \[{{\text{P}}_{\text{av}}}\text{=}{{\left( \frac{{{\text{V}}_{\text{RMS}}}}{\text{Z}} \right)}^{\text{2}}}\text{R}\] \[\text{Z=}\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}=56\Omega \] \[\therefore {{P}_{av}}={{\left( \frac{10}{\left( \sqrt{2} \right)56} \right)}^{2}}\times 50=0.79\,\,\text{W}\]


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