The equivalent primary load is
$${R_1} = {\left( {\frac{{{N_1}}}{{{N_2}}}} \right)^2}{R_2} = {\left( {\frac{{20}}{1}} \right)^2}\left( {6.0} \right) = 2400\,\Omega $$
Current in the primary coil $$ = \frac{{240}}{{{R_1}}} = \frac{{240}}{{2400}} = 0.1\,A$$

The three branches of the circuits carry currents
$$i = {i_0},i = {i_1}{e^{\frac{t}{{RC}}}}\,{\text{and}}\,i = {i_2}2{e^{\frac{{ - t}}{{2\,RC}}}}\,{\text{respectively}}{\text{.}}$$
The current through the cell, $$i\left( t \right)$$  can be found by using Kirchhoff's current law (or mode law).

Efficiency is defined as the ratio of output power and input power
i.e. $$\eta \% = \frac{{{P_{{\text{out}}}}}}{{{P_{{\text{input}}}}}} \times 100 = \frac{{{V_s}{i_s}}}{{{V_p}{i_p}}} \times 100$$
$$80 = \frac{{2 \times 440}}{{220 \times {i_p}}} \times 100 \Rightarrow {i_p} = 5A$$

$$\eqalign{ & V = - L\frac{{di}}{{dt}} \cr & {\text{Here}}\,\,\frac{{di}}{{dt}}\,{\text{is}}\, + ve\,{\text{for}}\,\frac{T}{2}\,{\text{time}}\,{\text{and}}\,\frac{{di}}{{dt}}\,{\text{is}}\, - ve\,{\text{for}}\,{\text{next}}\,\frac{T}{2}{\text{time}} \cr} $$

As we know that phase difference for $$L,C,R$$  series circuit is given by
$$\tan \phi = \frac{{{X_L} - {X_C}}}{R}$$
When $$L$$ is removed $$\tan \frac{\pi }{3} = \frac{{{X_C}}}{R}$$
$$\eqalign{ & \sqrt 3 = \frac{{{X_C}}}{R} \cr & \Rightarrow {X_C} = R\sqrt 3 \cr} $$
When $$C$$ is removed, $$\tan \frac{\pi }{3} = \sqrt 3 = \frac{{{X_L}}}{R}$$
$$ \Rightarrow {X_L} = R\sqrt 3 $$
Hence, in resonant circuit
$$\tan \phi = \frac{{\sqrt 3 R - \sqrt 3 R}}{R} = 0$$
So, $$\phi = 0$$
Power factor, $$\cos \phi = 1$$
It is the condition of resonance, therefore, phase difference between voltage and current is zero and power factor, $$\cos \phi = 1.$$

Power of source $$ = EI = 240 \times 0.7 = 166$$
$$ \Rightarrow {\text{Efficiency}} = \frac{{140}}{{166}} \Rightarrow \eta = 83.3\% $$

$$\eqalign{ & \frac{{{n_p}}}{{{n_s}}} = \frac{{{E_p}}}{{{E_s}}} = \frac{1}{{25}} \cr & \therefore {E_s} = 25\,{E_p} \cr & {\text{But}}\,\,{E_s}{I_s} = {E_p}{I_p} \cr & \Rightarrow {I_p} = \frac{{{E_S} \times {I_S}}}{{{E_p}}} \cr & \Rightarrow {I_p} = 50\,A \cr} $$

The quantity $$\tau = CR$$   is called time constant or capacitive time constant of $$CR$$  circuit. This is because dimensions of $$RC$$ are those of time and for a given $$CR$$  circuit, its value is constant.
Alternative
As $$R = \frac{V}{i}\,\,{\text{and}}\,\,C = \frac{q}{V}$$
$$\eqalign{ & \therefore RC = \frac{V}{i}\frac{q}{V} = \frac{q}{i} \cr & = \frac{{i \times t}}{i} = t \cr & \therefore \left[ {RC} \right] = \left[ t \right] = \left[ T \right] \cr} $$

Initial power $$= 3000\,W$$
As efficiency is $$90\% ,$$  then final power $$ = 3000 \times \frac{{90}}{{100}} = 2700\;W$$
\[ \Rightarrow \left. {\begin{array}{*{20}{c}} {{V_1}{I_1} = 3000\;W}\\ {{V_2}{I_2} = 2700\;W} \end{array}} \right]\,\,......\left( {\rm{i}} \right)\]
$$\eqalign{ & {V_2} = \frac{{2700}}{6} = \frac{{900}}{2}\,\,\left[ {\because {I_1} = 6A} \right] \cr & \Rightarrow {V_2} = 450\,V \cr & {\text{and}}\,\,{I_1} = \frac{{3000}}{{200}}\,\,\left[ {\because {V_1} = 200\;V} \right] \cr & \Rightarrow {I_1} = 15\,A \cr} $$

Quality factor $$Q = \frac{{{\omega _0}}}{{2\Delta \omega }} = \frac{{{\omega _0}L}}{R}$$