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:

1. \(\dfrac{1}{\sqrt{2}}~\text{A}\) 2. \(\sqrt{2}~\text{A}\)
3. \(2~\text{A}\) 4. \(2\sqrt{2}~\text{A}\)
(a) Hint: The peak value of the current can be calculated by the RMS value of the current.
Secondary voltage, Vs=24V
Step 1: The power associated with secondary; Ps=12 W
                                               IS=PSVS=1224
=12A=0.5 A
Step 2: The peak value of the current in the secondary;
I0=IS2
   =(0.5)(1.1414)=0.707=12A