The output of a step-down transformer is measured to be 24 V when connected to a 12 W Light bulb. The value of the peak current is

Given,
Power associated with secondary P_s = 12 W
Secondary voltage, V_s=24 V 
Current in the secondary, ${\mathrm{I}}_{\mathrm{s}}=\frac{{\mathrm{P}}_{\mathrm{s}}}{{\mathrm{V}}_{\mathrm{s}}}=\frac{12}{24}=0.5 \mathrm{A}$
Peak value of the current in the secondary 
${\mathrm{I}}_0={\mathrm{I}}_{\mathrm{s}}\sqrt{2}=(0.5)(1.414)=0.707 \mathrm{or} \frac{1}{\sqrt{2}}\mathrm{A}$