• question_answer In an LRC series circuit at resonance, current in the circuit is $10\sqrt{2}A$. If now frequency of the source is changed such that now current lags by $45{}^\circ$ than applied voltage in the circuit. Which of the following is correct is - A) Frequency must be increased and current after the change is 10 A B) Frequency must be decreased and current after the change is 10 A C) Frequency must be decreased and current is same as that of initial value D) The given information is insufficient to conclude anything

Correct Answer: A

Solution :

 Initially at resonance: ${{X}_{L}}={{X}_{E}}\Rightarrow Z=R.$ $\therefore {{i}_{0}}=\frac{{{\varepsilon }_{0}}}{R}=10\sqrt{2}\,A$ After increasing frequency: ${{X}_{L}}>{{X}_{c}}$$\omega L>I/\omega C$ $\omega >\frac{1}{\sqrt{LC}}\Rightarrow \omega >{{\omega }_{0}}$ (i.e. frequency increases) and $i=\frac{{{\varepsilon }_{0}}}{\sqrt{{{R}^{2}}+{{X}^{2}}}}=\frac{{{\varepsilon }_{0}}}{\sqrt{2R}}={{i}_{0}}/\sqrt{2}$$=10\,\,amp.$

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