If a capacitor of capacitance $$C$$ is connected with an $$AC$$  signal, then reactance of that circuit is purely capacitive.
The capacitive reactance is
$$\eqalign{ & X = \frac{1}{{\omega C}} = \frac{1}{{2\pi fC}}\,\,\left( {\omega = 2\pi f} \right) \cr & {\text{or}}\,\,X \propto \frac{1}{{fC}} \cr} $$
Considering two different situations of frequency and capacitance, we have
$$\eqalign{ & \frac{{X'}}{X} = \frac{{fC}}{{f'C'}} = \frac{{f \times C}}{{2f \times 2C}}\,\,\left[ {\because f' = 2f\,{\text{and}}\,\,C' = 2C} \right] \cr & {\text{or}}\,\,\frac{{X'}}{X} = \frac{1}{4} \cr & {\text{or}}\,\,X' = \frac{X}{4} \cr} $$

Just after switch is closed, inductor acts like an open switch (open path) and capacitor acts like a closed switch (closed path) because in $$D.C.$$  circuit inductive resistance becomes zero.
Just after switch is closed, given circuit is equivalent to the circuit shown below.

So, equivalent resistor $$ = \frac{R}{2} = \frac{9}{2}ohms$$
Battery emf, $$V = 18\,volts$$
∴ Current in circuit $$ = \frac{V}{R} = \frac{{18 \times 2}}{9} = 4\,A$$

$$\eqalign{ & i = \sqrt 5 A \cr & \frac{{q_m^2}}{{2C}} = \frac{{{q^2}}}{{2C}} + \frac{1}{2}L{i^2} \Rightarrow {q_{\max }} = 6C \cr} $$

$$\eqalign{ & q = C{V_1}\cos \omega t \Rightarrow i = \frac{{dq}}{{dt}} = - \omega C{v_1}\sin \omega t \cr & {\text{Also,}}\,{\omega ^2} = \frac{1}{{LC}}\,{\text{and}}\,V = {V_1}\cos \omega t \cr & {\text{At}}\,t = {t_1},V = {V_2}\,{\text{and}}\,i = - \omega C{V_1}\sin \omega {t_1} \cr & \therefore \cos \omega {t_1} = \frac{{{V_2}}}{{{V_1}}}\left( { - ve{\text{ sign gives direction}}} \right) \cr & {\text{Hence,}}\,i = {V_1}\sqrt {\frac{C}{L}} {\left( {1 - \frac{{V_2^2}}{{V_1^2}}} \right)^{\frac{1}{2}}} = {\left( {C\frac{{\left( {V_1^2 - V_2^2} \right)}}{L}} \right)^{\frac{1}{2}}} \cr} $$

$$\eqalign{ & {x_C} - {x_L} = 31.85 - 2\left( {6.28} \right) = 19.29 \cr & {I_m} = \frac{{10}}{{19.29}} = 0.52 \cr & {\text{Hence}}\,\,I = 0.52\sin \left( {314\,t + \frac{\pi }{2}} \right) = 0.52\,\cos 314\,t \cr} $$

$${X_C} = \frac{1}{{\omega C}} \Rightarrow {X_C}\alpha \frac{1}{\omega }\,{\text{for}}\,{\text{given}}\,C.$$

$$\eqalign{ & Q = \frac{{EC}}{2}\left( {1 - {e^{\frac{{ - t}}{\tau }}}} \right) \cr & {I_2} = \frac{{\varepsilon C}}{{2r}}{e^{\frac{{ - t}}{\tau }}} = \frac{\varepsilon }{R}{e^{\frac{{ - t}}{\tau }}} \cr & {I_1} = \frac{{Vc}}{R} = \frac{{\frac{\varepsilon }{2}}}{R}\left( {1 - {e^{\frac{{ - t}}{\tau }}}} \right) \cr & {I_1} = \frac{\varepsilon }{R} - {e^{\frac{{ - t}}{\tau }}} + \frac{\varepsilon }{{2R}} - \frac{\varepsilon }{{2R}}{\varepsilon ^{\frac{{ - t}}{\tau }}} \cr & \frac{\varepsilon }{{2R}}\left( {1 + {e^{\frac{{ - t}}{\tau }}}} \right) = \frac{\varepsilon }{{2R}}\left( {1 + {e^{\frac{{ - 2t}}{{CR}}}}} \right) \cr} $$

Power factor of $$AC$$ circuit is given by
$$\cos \phi = \frac{R}{Z}\,\,......\left( {\text{i}} \right)$$
where, $$R$$ is resistance and $$Z$$ is the impedance of the circuit and is given by
$$Z = \sqrt {{R^2} + \left( {{X_L} - X_{_C}^2} \right)} \,......\left( {{\text{ii}}} \right)$$
$${{X_L}} =$$  inductive reactance
$${{X_C}} =$$  capacitive reactance
Eqs. (i) and (ii) meet to give,
$$\cos \phi = \frac{R}{{\sqrt {{R^2} + {{\left( {{X_L} - {X_C}} \right)}^2}} }}\,......\left( {{\text{iii}}} \right)$$
Given, $$R = 8\,\Omega ,{X_L} = 31\,\Omega ,{X_C} = 25\,\Omega $$
$$\eqalign{ & \therefore \cos \phi = \frac{8}{{\sqrt {{{\left( 8 \right)}^2} + {{\left( {31 - 25} \right)}^2}} }} = \frac{8}{{\sqrt {64 + 36} }} \cr & {\text{Hence,}}\,\,\cos \phi = 0.80 \cr} $$

$$\eqalign{ & {\text{If}}\,\omega = 50 \times 2\pi \,{\text{then}}\,\omega L = 20\Omega \cr & {\text{If}}\,\omega ' = 100 \times 2\pi \,{\text{then}}\,\omega 'L = 40\Omega \cr} $$
Current flowing in the coil is
$$\eqalign{ & I = \frac{{200}}{Z} = \frac{{200}}{{\sqrt {{R^2} + {{\left( {\omega 'L} \right)}^2}} }} = \frac{{200}}{{\sqrt {{{\left( {30} \right)}^2} + {{\left( {40} \right)}^2}} }} \cr & I = 4A. \cr} $$

$$\eqalign{ & 2\pi nt = 400\pi t \cr & \therefore n = 200 \cr & {I_0} = 50\sqrt 2 \,amp. \cr & r.m.s.{\text{ current}} = \frac{{{I_0}}}{{\sqrt 2 }} = 50\,amp. \cr} $$