A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is $$I_{\circ}$$. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be

A series LCR circuit is connected to an AC source with emf $$E$$, and at resonance the current amplitude is $$I_0$$. Since the inductive and capacitive reactances cancel ($$X_L = X_C$$), the impedance equals the resistance alone, so $$Z = R$$ and the current amplitude at resonance is $$I_0 = \frac{E}{R}$$.

The resonance condition depends on $$L$$ and $$C$$ but not on $$R$$, so at the same resonant frequency with $$R' = 2R$$ the current becomes $$I' = \frac{E}{R'} = \frac{E}{2R} = \frac{I_0}{2}$$. From this, the new amplitude of current at resonance is $$\frac{I_0}{2}$$, which corresponds to Option 3.