A) 0, 31 H
B) 0.35 H
C) 1 H
D) 2 H
Correct Answer: A
Solution :
For minimum impedence \[{{X}_{L}}={{X}_{C}}\] \[=\frac{1}{2\pi \,fC}=\frac{1}{2\pi \times f\times 25\times {{10}^{-6}}}2\pi \,fL=140\] \[\therefore \] \[L=0.42\] \[0.1+x=0.42\] \[x=0.31H\]
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