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 \[I=\frac{{{V}_{rms}}}{{{Z}_{C}}}={{V}_{rms}}\times \omega C\] \[=\frac{{{V}_{0}}}{\sqrt{2}}\times \omega C\] \[=\frac{200\sqrt{2}}{\sqrt{2}}\times 100\times 1\times {{10}^{-6}}\] \[=20\times {{10}^{-3}}A\] \[=20\,mA\]
You need to login to perform this action.
You will be redirected in
3 sec
