A) \[40\Omega \]
B) \[80\Omega \]
C) \[40\sqrt{2}\Omega \]
D) \[2\sqrt{40}\Omega \]
Correct Answer: C
Solution :
Pure resistive,\[R=\frac{{{E}_{0}}}{{{I}_{0}}}=\frac{200}{5}=40\Omega \] As current lags behind the applied voltage by\[{{90}^{o}}\], therefore element \[Y\] must be pure inductor. \[{{X}_{L}}=\frac{{{E}_{0}}}{{{I}_{0}}}=\frac{200}{5}=40\Omega \] \[\therefore \] 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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