Question 1:
easyA transformer having efficiency of 90% is working on 200 V and 3 kW power supply. If the current in the secondary coil is 6 A, the voltage across the secondary coil and the current in the primary coil respectively are:
(2014)
Input power is $P_{in} = 3000 W$, so primary current $I_p = \frac{P_{in}}{V_p} = \frac{3000}{200} = 15 A$.
Transformer efficiency is $\eta = \frac{P_{out}}{P_{in}} = \frac{V_s I_s}{P_{in}}$.
Substitute values: $0.90 = \frac{V_s \times 6}{3000}$, which gives secondary voltage $V_s = \frac{2700}{6} = 450 V$.