A capacitor and a coil having a resistance R are in series are connected to a 6 Volt AC source. By varying the frequency of the source, maximum current of 600 mA is observed. If the same coil is now connected to a cell of emf 6 volt and internal resistance of 2 ohm, the current through it will be:

0.5 A

The correct answer is Option (1) → 0.5 A

The current at reasonance is,

$I_{max}=\frac{V_{source}}{R}$

$⇒R=\frac{V_{source}}{I_{max}}=\frac{6}{0.6}=10Ω$

Total Resistance, $R_{total}=R+r$

$=(10+2)=12Ω$

Now,

$I_{DC}=\frac{V_{DC}}{R_{total}}=\frac{6}{12}=0.5A$