A step-up transformer operates on a $$230\,V$$  line and supplies current of $$2\,A$$  to a load. The ratio of the primary and secondary windings is $$1:25.$$  The current in the primary coil is

Solution :
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
$$\eqalign{ & {\text{Here,}}\,\,\frac{{{N_p}}}{{{N_s}}} = \frac{1}{{25}} \cr & {i_s} = 2\,A \cr & \therefore \frac{{{i_p}}}{2} = \frac{{25}}{1} \cr & {\text{or}}\,\,{i_p} = 25 \times 2 = 50\,A \cr} $$