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 beYou need to login to perform this action.
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