A) 4.0 A
B) 8.0 A
C) \[\frac{20}{\sqrt{13}}\,A\]
D) 2.0 A
Correct Answer: A
Solution :
[a] If \[\omega =50\times 2\pi \] then \[\omega L=20\Omega \] If \[\omega '=100\times 2\pi \] then \[\omega 'L=40\Omega \] Current flowing in the coil is \[I=\frac{200}{Z}=\frac{200}{\sqrt{{{R}^{2}}+{{(\omega 'L)}^{2}}}}=\frac{200}{\sqrt{{{(30)}^{2}}+{{(40)}^{2}}}}\]
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