A) \[10\sqrt{2}A\]
B) \[10\text{ }A\]
C) \[30\sqrt{11}A\]
D) \[30\sqrt{11}\text{ }A\]
E) 5 A
Correct Answer: B
Solution :
\[e=300\sqrt{2}\sin \omega t\] ... (i) \[{{I}_{0}}=\frac{{{e}_{0}}}{Z}=\frac{300\sqrt{2}}{\sqrt{{{(30)}^{2}}+{{(10-10)}^{2}}}}\] \[\{\because Z=\sqrt{{{R}^{2}}+{{X}_{L}}-{{X}_{C}}{{)}^{2}}}\}\] \[=\frac{300\sqrt{2}}{30}=10\sqrt{2}A\] \[\therefore \] Current\[I=\frac{{{I}_{0}}}{\sqrt{2}}=10\,A\]
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