A) 4.0 A
B) 8.0 A
C) \[\frac{20}{\sqrt{12}}A\]
D) 2.0 A
Correct Answer: A
Solution :
If\[\omega =50\times 2\pi \]then\[\omega L=20\,\Omega \] Similarly \[\omega '=100\times 2\pi \]then\[\omega '=L=40\,\Omega \] \[i=\frac{200}{Z}=\frac{200}{\sqrt{{{R}^{2}}+{{(\omega 'L)}^{2}}}}\] \[=\frac{200}{\sqrt{{{(30)}^{2}}+{{(40)}^{2}}}}=4A\]You need to login to perform this action.
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