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 turnsratio is 8 : 1, then the potential difference applied to the primary coil is

Options

A. `(10^(4) xx 8^(2))/(25)V`
B. `(10^(4) xx 8)/(25)V`
C. `(10^(4))/(25 xx 8) V`
D. `(10^(4))/(25 xx 8^(2))` V

Answer

B

Explanation

`P = V_(1)i_(1) = V_(2)i_(2) 10 xx 10^(3) = 25 xx V_(2) V_(2) = (10^(4))/(25)`
Now `V_(1) = (n_(1))/(n_(2)) xx V_(2) = (8)/(1) xx (10^(4))/(25)V`