As change in flux of primary and secondary coil is proportional to the no. of turns in primary and secondary coil respectively.
$$\eqalign{ & {\text{So,}}\,\,\frac{{{\phi _p}}}{{{N_p}}} = \frac{{{\phi _s}}}{{{N_s}}} \cr & {\text{or}}\,\,\frac{1}{{{N_p}}} \cdot \frac{{d{\phi _p}}}{{dt}} = \frac{1}{{{N_s}}}\frac{{d{\phi _s}}}{{dt}} \cr & \therefore \frac{{{V_s}}}{{{V_p}}} = \frac{{{N_s}}}{{{N_p}}}\,\,\left( {{\text{as}}\,\,V \propto \frac{{d\phi }}{{dt}}} \right) \cr} $$
For no loss of power,
$$\eqalign{ & Vi = {\text{constant}} \cr & \therefore i = \frac{1}{V} \times {\text{constant}} \cr & {\text{or}}\,\,\frac{{{i_p}}}{{{i_s}}} = \frac{{{V_s}}}{{{V_p}}}\,\,{\text{or}}\,\,\frac{{{i_p}}}{{{i_s}}} = \frac{{{N_s}}}{{{N_p}}} \cr} $$
$${{i_p}}$$ and $${{i_s}}$$ are currents in primary and secondary coils
$$\eqalign{ & {\text{Here,}}\,\,\frac{{{N_p}}}{{{N_s}}} = \frac{1}{{25}} \cr & {i_s} = 2\,A \cr & \therefore \frac{{{i_p}}}{2} = \frac{{25}}{1} \cr & {\text{or}}\,\,{i_p} = 25 \times 2 = 50\,A \cr} $$

Given, root mean square value of electric field,
$${E_{rms}} = 6\;V/m$$
We know that, peak value of electric field,
$$\eqalign{ & {E_0} = \sqrt 2 {E_{rms}} \cr & \Rightarrow {E_0} = \sqrt 2 \times 6\;V/m \cr} $$
Also, we know that, $$c = \frac{{{E_0}}}{{{B_0}}}$$
where, $$c =$$  speed of light in vacuum
$${B_0} =$$  peak value of magnetic field
$$\eqalign{ & \Rightarrow {B_0} = \frac{{{E_0}}}{c} \Rightarrow {B_0} = \frac{{\sqrt 2 \times 6}}{{3 \times {{10}^8}}} \cr & \Rightarrow {B_0} = \frac{{8.48}}{3} \times {10^{ - 8}} \Rightarrow {B_0} = 2.83 \times {10^{ - 8}}\;T \cr} $$

$$\eqalign{ & {\text{Efficiency}}\,\eta = \frac{{{V_s}{I_s}}}{{{V_p}{I_p}}} \Rightarrow 0.9 = \frac{{{V_s}\left( 6 \right)}}{{3 \times {{10}^3}}} \Rightarrow {V_s} \cr & = 450\,V \cr & {\text{As}}\,\,{V_p}{I_p} = 3000 \cr & {\text{so}}\,\,{I_p} = \frac{{3000}}{{{V_p}}} = \frac{{3000}}{{200}}A = 15\,A \cr} $$

$$\eqalign{ & V = I\sqrt {{{\left( {20 + 10} \right)}^2} + {{\left( {{X_L} + {X_L} - {X_C}} \right)}^2}} \cr & 100 = 2\sqrt {{{\left( {30} \right)}^2} + {{\left( {30 + {X_L} - 40} \right)}^2}} \cr & {(50)^2} = {(30)^2} + {\left( {{X_L} - 10} \right)^2} \cr & {\left( {{X_L} - 10} \right)^2} = 80 \times 20:{X_L} - 10 = 40;{X_L} = 50\,\Omega \cr} $$

NOTE : Since current leads emf (as seen from the graph), therefore, this is an $$R - C$$  circuit.
$$\eqalign{ & \tan \phi = \frac{{{X_C} - {X_L}}}{R} \cr & {\text{Here}}\,\phi = {45^ \circ } \cr & \therefore {X_C} = R\,\,\,\,\,\,\left[ {{X_L} = 0{\text{ as there is no inductor}}} \right] \cr & \frac{1}{{\omega C}} = R \Rightarrow RC\omega = 1 \cr & \therefore RC = \frac{1}{{100}}{s^{ - 1}} \cr} $$

$$\eqalign{ & {\text{Using}}\,\,I = {I_0}\left( {1 - {e^{\frac{{ - t}}{\tau }}}} \right) \cr & {\text{But}}\,\,{I_0} = \frac{V}{R}\,\,{\text{and}}\,\,\tau = \frac{L}{R} \cr & \therefore I = \frac{V}{R}\left( {1 - {e^{\frac{{ - Rt}}{L}}}} \right) \cr & = \frac{{12}}{6}\left[ {1 - {e^{\frac{{ - 6t}}{{8.4 \times {{10}^{ - 3}}}}}}} \right] \cr & = 1\,\,\left( {{\text{given}}} \right) \cr & \therefore t = 0.97 \times {10^{ - 3}}\,s \cr & \approx 1\,ms \cr} $$

$${R_{{\text{eq}}}} = \frac{{5R}}{6} \Rightarrow I = \frac{{6E}}{{5R}} = 1A$$

Phase difference between current and voltage in $$LCR$$  series circuit is given by
$$\tan \phi = \frac{{\omega L - \frac{1}{{\omega C}}}}{R}\,\,\left[ {_{\frac{1}{{\omega C}}\, = \,\,{\text{capacitive reactance}}}^{\omega L \,= \,\,{\text{inductive reactance}}}} \right]$$
$$\phi $$ being the angle by which the current leads the voltage.
Given, $$\phi = {45^ \circ }$$
$$\eqalign{ & \therefore \tan {45^ \circ } = \frac{{\omega L - \frac{1}{{\omega C}}}}{R} \cr & \Rightarrow 1 = \frac{{\omega L - \frac{1}{{\omega C}}}}{R} \cr & \Rightarrow R = \omega L - \frac{1}{{\omega C}} \cr} $$
$$\eqalign{ & \Rightarrow \omega C = \frac{1}{{\left( {\omega L - R} \right)}} \cr & \Rightarrow C = \frac{1}{{\omega \left( {\omega L - R} \right)}} \cr & = \frac{1}{{2\pi f\left( {2\pi fL - R} \right)}} \cr} $$

$$\eqalign{ & {\text{Given,}}\,\,V = 120\sin \left( {100\pi t} \right)\cos \left( {100\pi t} \right) \cr & = 60\left[ {2\sin \left( {100\pi t} \right) \cdot \cos \left( {100\pi t} \right)} \right] = 60\sin 200\,\pi t. \cr & {\text{Thus,}}\,{V_0} = 60\,V,\omega = 200\,\pi \,\,{\text{or}}\,\,f = \frac{{200\pi }}{{2\pi }} = 100\,Hz. \cr} $$

$$\eqalign{ & V = \frac{{{V_0}}}{{\frac{T}{4}}}t \Rightarrow V = \frac{{4{V_0}}}{T}t \cr & \Rightarrow {V_{rms}} = \sqrt { < {V^2} > } = \frac{{4{V_0}}}{T}\sqrt { < t > } \cr & = \frac{{4{V_0}}}{T}\left\{ {\frac{{\int\limits_0^{\frac{T}{4}} {{t^2}dt} }}{{\int\limits_0^{\frac{T}{4}} {dt} }}} \right\} = \frac{{{V_0}}}{{\sqrt 3 }} \cr} $$