A) 10.7
B) 11.7
C) 14.7
D) 12.7
Correct Answer: D
Solution :
Impedance \[Z=\sqrt{{{R}^{2}}+4{{\pi }^{2}}{{\nu }^{2}}{{L}^{2}}}\] \[=\sqrt{{{(12)}^{2}}+4\times {{(3.14)}^{2}}\times {{(50)}^{2}}\times (0.04)}\]= 17.37 A Now current \[i=\frac{V}{Z}\]\[=\frac{220}{17.37}=12.7\Omega \]
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