A) R + 2pfL
B) \[\sqrt{{{R}^{2}}+4{{\pi }^{2}}{{f}^{2}}{{L}^{2}}}\]
C) \[\sqrt{{{R}^{2}}+{{L}^{2}}}\]
D) \[\sqrt{{{R}^{2}}+2\pi fL}\]
Correct Answer: B
Solution :
\[Z=\sqrt{{{R}^{2}}+X_{L}^{2}},\ \ {{X}_{L}}=\omega L\]and \[\omega =2\pi f\] \[\therefore Z=\sqrt{{{R}^{2}}+4{{\pi }^{2}}{{f}^{2}}{{L}^{2}}}\]
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