Q. A capacitor and a coil in series are connected to a $6 \,V$ 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 $6 \,V$ internal resistance of $2 \,\Omega$, the current through it will be

Solution:

At resonance, $I_{\max }=600 \,m A=\frac{V}{Z}$
$z=\frac{V}{I}=\frac{6}{0.6}=10 \, \Omega=R$
In $D C$ circuit the current through the coil is
$I=\frac{E}{R+r}=\frac{6}{10+2}=0.5 \,A$