question_answer

A series LCR circuit is tuned to resonance. The impedance of the circuit at resonance is :

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.