A) \[{{\left[ {{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}} \right]}^{{1}/{2}\;}}\]
B) \[{{\left[ {{R}^{2}}+{{(\omega L)}^{2}}+{{\left( \frac{1}{\omega C} \right)}^{2}} \right]}^{{1}/{2}\;}}\]
C) \[{{\left[ {{R}^{2}}+{{\left( \frac{1}{\omega C} \right)}^{2}} \right]}^{{1}/{2}\;}}\]
D) R
Correct Answer: D
Solution :
In a series L-C-R circuit, the impedance Z is given by \[Z=\sqrt{{{R}^{2}}+{{\left( \omega L-\frac{1}{\omega C} \right)}^{2}}}\] At resonance Inductive reactance = capacitance reactance \[\therefore \] \[{{X}_{L}}={{X}_{C}}\] \[\Rightarrow \] \[\omega L=\frac{1}{\omega C}\] and \[\tan \phi =0,\]i.e.,\[\phi =0,\]the emf and current are in phase. \[\therefore \] \[Z=\sqrt{{{R}^{2}}+0}=R\]which is the minimum value Z can have.You need to login to perform this action.
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