Figure shows a circuit that contains four identical resistors with resistance R = $2.0\Omega ,$  two identical inductors with inductance L = 2.0mH and an ideal battery with emf E = 9V. The current 'i' just after switch 'S' is closed will be:

At resonance V_L = V_C

_{$\Rightarrow cos?=\frac{V_R}{\sqrt[]{V_R^2+{\left(V_L-V_C\right)}^2}}=1$}

$X_C=\frac{1}{\omega C}$

On increasing $\omega$ $$ XC decreases so Z decreases, current in circuit increases.

$\frac{H_{D.C.}}{H_{A.C.}}=\frac{I^{2}R}{I_{rms}^{2}R}=1$

$E=\sqrt[]{V_R^2+{\left(V_L-V_C\right)}^2}$

$E=\sqrt[]{{\left(60\right)}^2+{\left(100-20\right)}^2}$

E = 100V

Since current is in phase with voltage, it means circuit is in resonance, so we can write

f = $\frac{1}{2\pi \sqrt[]{LC}}=\frac{1}{2\pi \sqrt[]{\left(0.5\times 10^{-3}\right)\times \left(200\times 10^{-6}\right)}}$

$\Rightarrow f=\frac{10^4}{2\pi \sqrt[]{10}}\approx 5\times 10^{2}Hz$ $\left[Taking\pi =\sqrt[]{10}\right]$