Question
An ideal efficient transformer has a primary power input of $$10\,kW.$$ The secondary current when the transformer is on load is $$25\,A.$$ If the primary : secondary turns ratio is $$8: 1,$$ then the potential difference applied to the primary coil is
A.
$$\frac{{{{10}^4} \times {8^2}}}{{25}}V$$
B.
$$\frac{{{{10}^4} \times 8}}{{25}}V$$
C.
$$\frac{{{{10}^4}}}{{25 \times 8}}V$$
D.
$$\frac{{{{10}^4}}}{{25 \times {8^2}}}V$$
Answer :
$$\frac{{{{10}^4} \times 8}}{{25}}V$$
Solution :
$$\eqalign{
& P = {V_1}{i_1} = {V_2}{i_2} \cr
& 10 \times {10^3} = 25 \times {V_2} \cr
& {V_2} = \frac{{{{10}^4}}}{{25}} \cr
& {\text{Now}}\,\,{V_1} = \frac{{{n_1}}}{{{n_2}}} \times {V_2} = \frac{8}{1} \times \frac{{{{10}^4}}}{{25}}V \cr} $$