A) \[10\,\,mA\]
B) \[20\,\,mA\]
C) \[40\,\,mA\]
D) \[80\,\,mA\]
Correct Answer: B
Solution :
Given,\[V=200\sqrt{2}\sin 100t\] Comparing this equation with \[V={{V}_{0}}\sin \omega t\], we have\[{{V}_{0}}=200\sqrt{2}V\] and \[\omega =100\,\,rad/s\] The current in the capacitor is \[I=\frac{{{V}_{rms}}}{{{Z}_{C}}}={{V}_{rms}}\times \omega C=\frac{{{V}_{0}}}{\sqrt{2}}\times \omega C\] \[=\frac{200\sqrt{2}}{\sqrt{2}}\times 100\times 1\times {{10}^{-6}}\] \[=20\times {{10}^{-3}}A=20\,\,mA\]
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