A current of 5 A is flowing at 220 V in the primary coil of a transformer. If the voltage produced in the secondary coil is 2200 V and 50% of power is lost, then the current in the secondary coil will be

0.25 A

$\begin{aligned} & \text { Efficiency }\eta =\frac{P_{\text {out }}}{P_{\text {in }}}=\frac{V_0 I_0}{V_{\text {in }} I_{\text {in }}} \\ \newline & \frac{50}{100}=\frac{(2200)\left(I_0\right)}{(220)(5)} \\ \newline & \left(\frac{1}{2}\right)\left(\frac{5}{10}\right)=I_0 \\ \newline & I_0=\frac{1}{4} A=0.25 \mathrm{~A}\end{aligned}$