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