A series LCR circuit is connected to an alternating source of emf E. The current amplitude at resonant frequency is $I_0$. If the value of resistance R becomes twice of its initial value then amplitude of current at resonance will be

A bulb and a capacitor are connected in series across an ac supply. A dielectric is then placed between the plates of the capacitor. The glow of the bulb :

A coil of negligible resistance is connected in series with $$90 \Omega$$ resistor across $$120 \mathrm{~V}, 60 \mathrm{~Hz}$$ supply. A voltmeter reads $$36 \mathrm{~V}$$ across resistance. Inductance of the coil is :

A LCR circuit is at resonance for a capacitor C, inductance L and resistance R. Now the value of resistance is halved keeping all other parameters same. The current amplitude at resonance will be now: