A) 5.5 mH
B) 0.55 mH
C) 55 mH
D) 0.55 H
Correct Answer: D
Solution :
\[V=100V\] \[i=1amp\] \[V=iR\] \[R=\frac{100}{1}\] \[=100\Omega \] \[Va.c.=100V\] \[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}\] \[i=0.5amp.\] \[V=\sqrt{({{R}^{2}}+X{{L}^{2}})}\] \[\frac{1000}{5}=\sqrt{{{R}^{2}}+X_{L}^{2}}\] \[{{(200)}^{2}}={{100}^{2}}+X{{L}^{2}}\] \[40,000-10,000=X_{L}^{2}\] \[\sqrt{30,000}={{X}_{L}}\] \[{{X}_{L}}=173.2\Omega \] \[{{X}_{L}}=2\pi f\,L\] \[173.2=2\pi \times 50L\] \[L=0.551H\]You need to login to perform this action.
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