A) 0.35 mH
B) 35 mH
C) 3.5 mH
D) Zero
Correct Answer: A
Solution :
Capacitance of wire \[C=0.014\times {{10}^{-6}}\times 200=2.8\times {{10}^{-6}}F=2.8\mu F\] For impedance of the circuit to be minimum \[{{X}_{L}}={{X}_{C}}\] Þ \[2\pi \nu L=\frac{1}{2\pi \nu C}\] Þ\[L=\frac{1}{4{{\pi }^{2}}{{\nu }^{2}}C}=\frac{1}{4{{(3.14)}^{2}}\times {{(5\times {{10}^{3}})}^{2}}\times 2.8\times {{10}^{-6}}}\] \[=0.35\times {{10}^{-3}}H=0.35\,mH\]
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