A) \[40\,\Omega \]
B) \[80\,\Omega \]
C) \[40\sqrt{2}\,\Omega \]
D) \[2\sqrt{40}\,\Omega \]
Correct Answer: C
Solution :
Pure resistive, \[R=\frac{{{E}_{0}}}{{{l}_{0}}}=\frac{200}{5}=40\,\Omega \] As current lags behind the applied voltage by \[{{90}^{o}},\]therefore element Y must be pure inductor. \[{{X}_{L}}=\frac{{{E}_{0}}}{{{l}_{0}}}\] \[=\frac{200}{5}=40\,\Omega \] \[\therefore \] Total impedance, \[Z=\sqrt{{{R}^{2}}+X_{L}^{2}}\] \[=\sqrt{{{40}^{2}}+{{40}^{2}}}=40\sqrt{2}\,\Omega \]You need to login to perform this action.
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