A) 0.51 W
B) 0.67 W
C) 0.76 W
D) 0.89 W
Correct Answer: A
Solution :
[a] Given: L=20mH; \[C=50\mu F\]; \[R=40\Omega \] \[V=10\,\sin \,340\,t\] \[\therefore \,\,\,{{V}_{runs}}=\frac{10}{\sqrt{2}}\] \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{340\times 50\times {{10}^{-6}}}=58.8\Omega \] \[{{X}_{L}}=\omega L=340\times 20\times {{10}^{-3}}=6.8\Omega \] Impedance, \[Z=\sqrt{{{R}^{2}}+{{({{X}_{C}}-{{X}_{L}})}^{2}}}\] \[=\sqrt{{{40}^{2}}+{{(58.8-6.8)}^{2}}}=\sqrt{4304}\,\Omega \] Power loss in A.C. circuit, \[P=i_{rms}^{2}R={{\left( \frac{{{V}_{rms}}}{Z} \right)}^{2}}\] \[R={{\left( \frac{10/\sqrt{2}}{\sqrt{4304}} \right)}^{2}}\times 40\] \[=\frac{50\times 40}{4304}\simeq 0.51\,W\]
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