Laminated core provide less area of cross-section for the current to flow. Because of this, resistance of the core increases and current decreases thereby decreasing the eddy current losses.

$$\eqalign{ & {i_1} = \frac{{dq}}{{dt}} = \frac{d}{{dt}}\left( {2\cos \,4t} \right) = - 8\sin \,4t, \cr & {i_2} = \frac{{dq}}{{dt}} = \frac{d}{{dt}}\left( {4\cos \,t} \right) = - 4\sin \,t, \cr & {i_3} = \frac{{dq}}{{dt}} = \frac{d}{{dt}}\left( {3\cos \,4t} \right) = - 12\sin \,4t, \cr & {i_4} = \frac{{dq}}{{dt}} = \frac{{d\left( {4\cos \,2t} \right)}}{{dt}} = - 8\sin \,2t. \cr} $$
Clearly, amplitude of $${i_3}$$ is greatest.

Here, $${X_L} = \omega L = 2\pi fL = 2\pi \times 50 \times \frac{{0.4}}{\pi } = 40\Omega $$
$$\eqalign{ & R = 30\Omega \cr & \therefore Z = \sqrt {{R^2} + X_L^2} = \sqrt {{{30}^2} + {{40}^2}} = 50\Omega \cr & {I_{rms}} = \frac{{{V_{rms}}}}{Z} = \frac{{200}}{{50}} = 4A \cr} $$

As $${X_C} = \frac{1}{{\omega C}},$$   so with the increase in frequency, $${X_C}$$ decreases and so $${i_C}$$ increases.

The efficiency of transformer $$ = \frac{{{\text{Energy obtained from the secondary coil}}}}{{{\text{Energy given to the primary coil}}}}$$
$${\text{or}}\,\,\eta = \frac{{{\text{Output power}}}}{{{\text{Input power}}}}$$
$$\eqalign{ & {\text{or}}\,\,\eta = \frac{{{V_s}{I_s}}}{{{V_p}{I_p}}}\,\,\left[ {{\text{as}}\,{\text{power}} = VI} \right] \cr & {\text{Given,}}\,\,{V_s}{I_s} = 100\,W,\,{V_p} = 220\,V,\,{I_p} = 0.5\,A \cr & {\text{Hence,}}\,\eta = \frac{{100}}{{220 \times 0.5}} = 0.90 = 90\% \cr} $$

$$\eqalign{ & {N_P} = 400,{N_S} = 2000\,\,{\text{and}}\,\,{V_S} = 1000\,V. \cr & \frac{{{V_P}}}{{{V_S}}} = \frac{{{N_P}}}{{{N_S}}}\,\,{\text{of,}}\,\,{V_P} = \frac{{{V_S} \times {N_P}}}{{{N_S}}} = \frac{{1000 \times 400}}{{2000}} = 200\,V. \cr} $$

$$\eqalign{ & {B_0} = \sqrt {B_0^2 + B_1^2} = \sqrt {{{30}^2} + {2^2}} \times {10^{ - 6}} \cr & \approx 30 \times {10^{ - 6}}T \cr & \therefore {E_0} = cB = 3 \times {10^8} \times 30 \times {10^{ - 6}} \cr & = 9 \times {10^3}V/m \cr & {E_{rms}} = \frac{{{E_0}}}{{\sqrt 2 }} = \frac{9}{{\sqrt 2 }} \times {10^3}V/m \cr} $$
Force on the charge,
$$F = {E_{rms}}Q = \frac{9}{{\sqrt 2 }} \times {10^3} \times {10^{ - 4}} \simeq 0.64N$$

$$\eqalign{ & {i_3} = {i_1} + {i_2} = 3\sin \omega t + 4\cos \omega t \cr & = 5\sin \left( {\omega t + \phi } \right), \cr & {\text{where}}\,\tan \phi = \frac{3}{4}\,\,or\,\,\phi = {53^ \circ }. \cr} $$

$$RMS$$  value over one cycle $$ = RMS$$   value over $$AB.$$
$$\eqalign{ & 0 \leqslant t \leqslant 15 \cr & i\left( t \right) = \sqrt 5 \,{t^2}c \cr & {i_{rms}} = \sqrt { < {i^2} > } = \sqrt {\frac{{5\int\limits_0^1 {{t^4}dt} }}{{\int\limits_0^1 {dt} }}} = 1\,mA \cr} $$

Current in inductor $$ = \frac{E}{R}$$
∴ Its energy $$ = \frac{1}{2}\frac{{L{E^2}}}{{{R^2}}}$$
Same energy is later stored in capacitor
$$\eqalign{ & \frac{{{Q^2}}}{{2C}} = \frac{1}{2}\frac{{L{E^2}}}{{{R^2}}} \cr & \Rightarrow Q = \sqrt {LC} \frac{E}{R} \cr} $$