A) 30 ohm
B) 40 ohm
C) 50 ohm
D) 60 ohm
Correct Answer: C
Solution :
\[Z=\sqrt{{{R}^{2}}+{{(2\pi \nu L)}^{2}}}\] \[=\sqrt{{{(40)}^{2}}+4{{\pi }^{2}}\times {{(50)}^{2}}\times {{(95.5\times {{10}^{-3}})}^{2}}}=50\,\,ohm\]
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