A) 0 V, 0.47 A
B) 1.68 V, 0.47 A
C) 0 V, 1.4 A
D) 5.6 V, 1.4 A
Correct Answer: D
Solution :
\[Z=\sqrt{{{(R)}^{2}}+{{({{X}_{L}}-{{X}_{C}})}^{2}},}\] \[R=100\,\Omega ,\,\,{{X}_{L}}=\omega L=2000\times 5\times {{10}^{-3}}=10\,\Omega \] \[{{X}_{C}}=\frac{1}{\omega C}=\frac{1}{2000\times 50\times {{10}^{-6}}}=10\,\,\Omega \] i.e. \[Z=10\,\,\Omega \] Maximum current \[{{i}_{0}}=\frac{{{V}_{0}}}{Z}=\frac{20}{10}=2\,A\] Hence \[{{i}_{rms}}=\frac{2}{\sqrt{2}}=1.4\,A\] and \[{{V}_{rms}}=4\times 1.41\times 5.64\,V\]
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