A) 3.14m A
B) 6.28 mA
C) 1.51mA
D) 7.36 mA
Correct Answer: A
Solution :
(a) 3.14 mA \[{{X}_{C}}=\frac{1}{2\pi vC}\] \[=\frac{1}{2\times 3.14\times 50\times 0.1\times {{10}^{-6}}}\] \[=3.2\times {{10}^{4}}\Omega \] \[Z=\sqrt{{{R}^{2}}+X_{C}^{2}}=\sqrt{100+10.28\times {{10}^{8}}}\] \[=3.2\times {{10}^{4}}\Omega \] \[{{l}_{rms}}=\frac{{{v}_{rms}}}{Z}=\frac{100}{3.2\times {{10}^{4}}}\] \[=3.14\times {{10}^{-3}}A=3.14\,mA\]
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