A 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 

Given, effi ciency is 90%,
$\mathrm{\eta }=0.9$
We know,
$\mathrm{\eta }=\frac{{\mathrm{P}}_{\text{output }}}{{\mathrm{P}}_{\text{input }}}=0.9$
Where Poutput = V_S i_S and Pinput = V_P i_P Here, V_P and V_S are the voltages across primary and secondary coils and i_P and i_S are the respective currents.
$\Rightarrow \frac{{\mathrm{V}}_{\mathrm{s}}{\mathrm{i}}_{\mathrm{S}}}{{\mathrm{V}}_{\mathrm{p}}{\mathrm{i}}_{\mathrm{p}}}=0.9\dots \left(\mathrm{l}\right)$
Given, source power, P_input = V_Pi_P= 3 x 103 W
and source voltage,V_P = 200 V
$\begin{array}{l}\Rightarrow \text{ Source current, }{\mathrm{i}}_{\mathrm{p}}=\frac{{\mathrm{P}}_{\text{input }}}{{\mathrm{V}}_{\mathrm{p}}}\\ =\frac{3\times {10}^3}{200}=15\mathrm{A}\end{array}$
From (1),
$\begin{array}{l}{\mathrm{V}}_{\mathrm{s}}{\mathrm{i}}_{\mathrm{s}}=0.9\times {\mathrm{V}}_{\mathrm{p}}{\mathrm{i}}_{\mathrm{p}}\\ =0.9\times 3\times {10}^3\mathrm{W}\\ \Rightarrow {\mathrm{V}}_{\mathrm{s}}=\frac{0.9\times 3\times {10}^3}{{\mathrm{i}}_{\mathrm{s}}}\left(\text{ Here, }{\mathrm{i}}_{\mathrm{s}}=6\mathrm{A}\right)\\ =\frac{0.9\times 3\times {10}^3}{6}=450\mathrm{V}.\end{array}$