NEET Physics Alternating Current / प्रत्यावर्ती धारा NEET PYQ-Alternating Current

An inductor 20 mH, a capacitor \[50\,\mu F\] and a resistor \[40\,\,\Omega \] are connected in series across a source of emf \[V=10\text{ }\sin \text{ }340t\]. The power loss in A.C. circuit is :-                                    [NEET - 2016]

A) 0.51 W

B)      0.67 W

C) 0.76 W

D)      0.89 W

Correct Answer: A

Solution :

[a] \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{340\times 50\times {{10}^{-6}}}=58.8\,\Omega \]

\[{{X}_{L}}=\omega L=340\times 20\times {{10}^{-3}}=6.8\,\Omega \]

\[Z=\sqrt{{{R}^{2}}+{{({{X}_{C}}-{{X}_{L}})}^{2}}}\]

\[=\sqrt{{{40}^{2}}+{{(58.8-6.8)}^{2}}}=\sqrt{4304\,\Omega }\]

\[P=i_{rms}^{2}R={{\left( \frac{{{V}_{rms}}}{Z} \right)}^{2}}R\]

\[={{\left( \frac{10/\sqrt{2}}{\sqrt{4304}} \right)}^{2}}\times 40=\frac{50\times 40}{4304}=0.47\,W\]

So best answer (nearest answer) will be [a]