A) 10 mA
B) 20 mA
C) 40 mA
D) 80 mA
Correct Answer: B
Solution :
Given\[E=200\sqrt{2}\sin (100t)\] ... (i) Comparing Eq. (i) with \[E={{E}_{0}}\sin \omega t\] Peak voltage\[{{E}_{0}}=200\sqrt{2},\,\,\omega =100\] \[C=1\mu F=1\times {{10}^{-6}}F\] Reading of ammeter,\[i=\frac{{{E}_{rms}}}{{{X}_{C}}}\] \[=\frac{{{E}_{0}}{{\omega }_{C}}}{\sqrt{2}}=\frac{{{E}_{rms}}}{1/\omega C}=\frac{{{E}_{0}}\omega C}{\sqrt{2}}\] \[=\frac{200\sqrt{2}\times 100\times 1\times {{10}^{-6}}}{\sqrt{2}}\] \[=20\times {{10}^{-3}}A\] \[=20\,\,mA\]
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